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A p VI' IT) = (1 )" pIT) VI' ' In fact,
§3.7 IrTeducible 8a.re8 in Group Algebra
97
denoted by Pj for cOlwcnicncc, arc et A)
=Po = (E + A )/2,
e(8)
= PI
= (E 
A )/2.
(3.137)
4>tU.
Let 1>t~} be proportiolilil to P" RP,," Take R = E to calculate Then, taking R to be, say D , which does not appear in 4>~U, one obtains
4>~:} = 4Pv Ep,. = 26,,1' {E + ( I)I'A} ,
pt~} = 4P"DPI' = D + (1)"+1' F + (_ 1)1' B
(3.138)
+ ( I )"C.
The mutrix of C"l in rpt~} is
c2f,t.:l = C"lrpl~~
= 2(1)1'
{E+
D+ F} + 2 {A + B + C}
 ( 1)' [4>(1) 1'1' + 2,p(2)] 1'1"
C2ft~ When I" =
1.1,
= 0,
/I
f
11.
the mlltrix OfC2 in the basis functions
4>1~~!
The eigenvector of C2 with the eigenvalue 3( _1 )1' is that wit.h t.he eigenval ue 0 is [
b~)
=
~{
) b(/J ll
""'
I { J6
(I) Pll
j
=
is (1)1'
( ~ ~).
[ 4>J~:l + 2 4>,~~} J and

(2)} 1 24>11 = J6{E
t· D t F + A + B +C}, j
, D + F  A  B C},
{(I)
(E) 1 (l) } 1 boo = 2V3
b\7·)= CI1 When
1/
=I
11. ,
{
 1,g)}
=
Cll
{2E D F2A + B+C}.
bOI = (E l
(2l COl
= CoI{ DF  B+C) ,
blO = CIOP IO = CIO{D  F + B  C}.
(3.139)
(3.140)
T he represent.ation DE depends on the choice of the coefficients t hrough &js. (3. 125) and (3.133). Conversely, if the repr('Sentation (2 .12 ) is takclI,
(3.141)
98
Chop. :1 Th l
Thus, the coefficients are calculated , COl = C IO = 1/ 2 and Gil = ti M. T he calculations of the standa rd irreducible basis vcclon; bf,,, for t he groups T a nd 0 can be found in P robs. 22 a nd 23 of Chap. 3 of [ ~ I a and Gil (200,1)]. The property of I is introduced in §2.5A. The chnracter table is given in Table 3.14 and calcula ted in Prob. 18 of Chap 3. The representation matrices of the genera tors To and SI arc given ill Prob. 21 of Chap. 3 and calculate<1 in Prob. 12 of Chap. 4. The CSCOS of I arc (5 and the left and rightmuliiplications with 1\ 5fo1d rotation To.
,
(5 =
L
(Tk
+ r:) ,
k=O
u A = 12 ,
aTI
= 4p l ,
C5&{", = bf.,,,Cs = oT. =  4p,
uilJf""
oe =  3,
(3.1<12) ( I ll
= 0,
where p = 1/ + /,1, p I = _'1 2 _ 11 2, and 1/ = Cxp{ i27rj5). First calculate the eigenfunctions ~ ~:! of the idcHlpotcnt.s e(I') of ti ,e cyclic gTOUp C:; gencnlted by To. e( " ) is dellotoo by I~, for cOllvcniencc,
I'"
I'
p
",
(3 .143)
II, v mod 5.
Tttke n (1 ) = E to calculate 1>W. Then , taking R (2) in 1, MY 5 11 , which docs not appear ill 1>}}J, one calculates 1:> }?J. III this way, olle calculates
"'"
~~
=
=
<")~!
=
I'}
(E + I, " To + 1, 2WrJ + 1,21'Tf} + "I'T;}) / /5,
+ /, "51 4 +1J 2"SI2 + 11 21 '8 15 + 1,"5 13 ) 1V5, {(85 + ,,I' H; + 11  21'TjI + ,,21'T4 + '11'11.1) + r/I' "} (84 + ,, I' R~ + 1}  21'T~ + 1,21'T3 + J/I'R3) + 1/2{1' ,,) (5.1 + 1, 1' R~ + I) 21'T:i + 1}21'T2 + ., /1' R2) + 1] 2(,. ,,) (82 + '1  1' R~ + J}  2I'T; + .,}21'TI + 1)1' RJ) + 71  (1'  ") (51 + IJ  I' R~ + '1  2"T24 + 1}2"Tr; + 1}"Rr;)} 15, { (810 + 'I I'T? + 1}  21' R~ + "/}21' fig + I}I'T;) + 1,(1'  ") (S!;I + ,, I'TS1 + 1/_ 2" R?o + r?" Rg + li'r.n + 1/ 2{1' "1 (S8 + ,, I'T43 + ,}  21' R~ + .,}21' R7 + ry'Tj) t 71  2 (1'  ") (S7 I 1,  I'Tll,, 2"R~ j ,P"R6 + ,"'Tl) + ,, (1' ") (St; + I, "T? + 1, 21' m+ 1/2" R IO + 71"T~)} 15, (5 11
(3.1 '1'1)
§3.7 1rrt:<J.tu:ible
B(j~e~
in Group Algebra
99
8(:'(:ond, calculate t he matrix of the class operator C ~ in (l>~~ and its cigCIl\'(:'(:lors. Due to Eq. (3. 1'1'1), only the coefficients of the te rms E , 5 11 , 8~, and SIO in C!.
n = T~Jl.J. ,
T3 = T6 RI , T4 = TC1 R4 , T~= TJS1' = TJSl . TJ = ToJij , TJ = TJR2, Tt = ToRI.
Til = TJR4 ,
T;
S Il = S l2T6 = TOS l2TJ = TJSl2TJ = TJSl2 = TJSl2TO,
RJTo' = ToR'1 = TJR~TJ = TJ R2TJ = TOISI TO' SIO  R6TJ = ToRloTo = TJSsTJ = TJ R~ = 'l!: R1oT£:· Thcll , say, 8 11 :"": TJS12TJ = (TJ Ri)( R 2 S 12 n) = T 1(R iOn) = TI R~, 85 =
E = TjTj' = T}Tj,
TOS13 = TdS14 = T1R~ = Tt R~ = T2T'f = ~TJ T3Tr = TtT.~ = T4RS = T.tR6 = T5SS = rtS6 ,
8 11 S5
0 :5 j :5 5,
= ToR.I = nR~ = TITa = Tt T? = T 2S4 = TtSiO = T 3 S 9 = TiSI = T4Tg = Tt TJ = TsRs = ~Rl.
510 = ToTl = TilT? = TIRs = Ti'R~o = T2 S S = T~Sl 3 T3T? = TtTr = T4S L!. = T 44S l = TsR IO = TtRg. Thus, t he matrix of the class operator C~ in CP~,:! is obtained,
C5 (1)~~l = [(I}" + ,/ ") D + (17 2" + 17 2,,) S5 + . . .J /.j5
IS (1/2" + '/ 2,,)
= (')"
C5 (l>;~ = =
+ ,/ ")(I) ;,~l +
Cs'ilW = [05"" ('12" + ,, 2,,) E + (1/" + ,/ 1' + 1/" + ,/ " + '/" "11/" ")85 I (I = 6,,,,../5 (1/2"
+ '/
")(1 + I()Sw + .. ·1/5
+ 1, 2,,) (l» ,~l + ('/" + 1/ " + ,!" + 1/ "
+1/' V+ r( ")(l>},:~ + (J + '/" )( I + ln
C5(JI)~) = [05, ,(_ v) (1, " I 1/ 2,,) 8 11 =
+ ( I + 1/" ) (I + 1] ") S[, + (')" + ,/ " + I}" + '} " + ,/,d ·" + '/ I' ") 510 + .. .1/5 05, ,,)../5 ( ,, /' + 1, 2,,) 'I);~ + (1 + 'I") (1 + 1/ ") <[I~3J I(
+ (,/" + I) /' + 1( + '/ "+ 1/,'1" + 17 " ")
HlO
Chap. S Theory of Representations
It s cigcnvcctor~ an~ the irn,,"{iucible bases 11.,,, of I as listed in Table 2.1 5.
For example, when II = J) = 0, the matrix of ( 5 in ~) and its eigenvectors with the eigenvalues 12 , 4pl,  4p, a nd 0, res pectively, a rc
~ ~
(o
2J5 0
't' 2~) 6
"
2,;5 4
6
'
bob, b'{;o, blJ, a nd b{k, res pect ively.
Normalizing' t he eigenvectors, one obtains \\llen It = 0 a ud
2 (
+ II'"
b'{;~) b;r~, and
""
_ 1/ Z" _
btL,
(I>&~ a ud
0 olle has the ll ul.t ric of ( 5 ill
1 + 1}v
_ l /Z"  '1]21.'
I
/1
) .
Normalizing the eigeuvectors, olle obtains
1,2"
res pectively.
T a ble 3.15
Ir r educible b ases
'1."
in the gr oup s p ace of I
1/ = exp(  i2ro /5), «=1
bA 00  ( '1>(1) 00
+ <}> DO (2) + v'51>(3) '51,(4)) a 110 + V'' 00 / V'f2 L: ·
, ,, ", ,, , I'
e
0
,0,
0 0 0
0 0 0
," _<,_I
'I~
TJ 2 p
,1 _,,_ ' p_l
, ,
'I
'1 2
0 0' 0 0
0 '
_,,2
0 0 0
,
0',
N
,
~
0
, , , , , 0
0 0
~
,
.,
_'/pI
2 4 2
_,,_2
'I
,
,
_ p_l
4 2 4 2
DC,
"2 0
2 2 0
2
e
2 2 2 0 0 0
2 0
 2
 2
 2
2
,
0,
"0
0,
,"
0 0 0
0
'1_ 2
_ '1 1
"
'Il,  I
0 0 0
 '7'
"
'11 pl
0 0
0' ,
, , ,
0
,
,,'
, ,,_2
'1
1
1'
N 4 2 4
2 , , _,,_I , , 'I
2
,/2p
'I _ p_l
2
§S ,7 Irreducible Bases in Group Algebra
0, ",, ,," , 0
0, 0 0 0
,, ,,, , , ,T ,, ,' T 2
0 0 0
2
, ,, ,
T T T T
T
," ,"
T 2
",
,
0
T
",,
" ,,, , ,, 2
0,
v'
3
0
,',
0
'I'P
_'1 2 pl
_ '1 2pl
 'I ' P
,,'
,' 0 0 0
"0
0
_'1  '
0
 ,;5'I ~
_'1 2 p 2
 ,;',
0 0 0
,
T T
0
T
T
T T
2
",, ," ,T ,," 0
,"
",'
0 0 0 0 0
,
,"
0 0 0
V5
V5 0 0 0 0
"
V5
 ,;5,f
0 0 0
,;5'1_' 0 0
"0
'r
,
'II?
V5
3
_'I~pl
,
 'IP
,;5"
0
T
,'
'1'1''
0 0 0
,
,"
0 0 0
0 0 0 0 0
0 0 0
,
,
3 3
, ,T, ,, 0 0 0 0 0
'IP
_ '1 2
 OP
V5
0
,
'1 ' p
0
,
_,, ~p l
3 3 3 3 3 3 3
'1 _'
0
0 0 0
_ '1 2 p 1
 'IIp
_ '1 2 _ ,,21' 1 '1 2 p'
0
0
N
 OP
0 0
0 0
,
,
0,
_ ,,1 p l
0
2
",
'"
,' ,,2
l'
"
,,'
" ,
,'2
'1 '1 2 1' _ ,fp 2
.,
"
 'I'p 1
3 3 3 3 3
N
,11'
,'
",
 '11' '
3
'1 _ ~p _ 2
" ", ", , ",,
,.,' p
,'
 ,;'
'I'P> 'IP' " ,,~
, ,,'
'1 _ ' _ '1'1''  ",' 'I  ~
p_2
,', ,
3
3
3
3
", " ", 3
'1 ' p2 _'1  2 1'
,fp2 '1  'p  '
" ,,_2
11  2 1'
3
"
"
'11' '
3
Chap. S Theory of Representations
102
3.8
E xe rcises
1. Let C be a nonAbelian group, D(C) he a faithfnl repre;entatiOIl of the group C , and D(R) be the representation matrix of the element R. Tf the clement R in C corresponds to the following matrix in the set, please decide whether the following set forms Ii representation of the gl"Oup C. For example, in (a), if R <> D(R)t , plelL<;e decide whether the set D(G)t composed of D(R)t forms a representation of C. la) DIR)t; Ib) D I R )T; Ie) D IR  ' ); Id) DIR) ' ; Ie) DIR ' )t; If) det DIR); Ig) T< DIR). 2. The homogeneous function space of degTee 2 spanned by the basis functions l/>J(X,y) = x'l, ·I .h(x,y) = xy, and W3(X,y) = y2 is invariant in the following rotations R in the twodimensional coordinate space. Calculate the matrix form D( Il) of the cor n~po llding transformatiOlI operator Pn for scalar functions in the threedimensional function space:
la) Ib)
R ~!. ( 2
 1
J3
V3)  1
'
R ~ (10). 01 .
(e) R ~ (c~n sinn). Sllln cosn
By rcplacing the basis function "l{Jz(x, y) = xy with ..j2xy, how will the matrix form of the operator PR change? 3. P rove that the module of any representatio n matrix ill a onedimensional representation of a finite group is equal to 1. 4. Prove that any irreducible representation o f an infinite Abelian group is onedimensional . 5. P rove that the similarity transformation matrix between two equivalent irr<:.duciblc unitary representations of a finite group, if restricting its determinant to be 1, has to be unitary. 6 . Show that for a fi nit.e group G, the sum of the eharact.ers of all elements in any ilTeducible representation of G, excep t for the identical repres~ntation , is equal to zero. 7. Calculate t he similarity trans fo rmation matrix X between two equivalent regular representations D (R) and D(fl) , given in Eqs. (3.6) and (3.12), in the group space of the D3 group. How to generalize the res ult. to any other finile group?
'00
8. C" is a class in a finite group C .' and en is a vector in the group space composoo of the SlIlJI of all dements in Co.. Prove that the re presentation matrix of Co in an irreducible representat.ion is a constant mat.rix, and calculate this constant. 9. Prove that the number of selfreciprocal classes in a finite group G' is equal to the numbe r of inc<}uivaient. and irn.,
,,) ( y'
~
R
'
(x) y
'
where R is equal to the repr~entl;\tion matrix in the twodimellsiollal representation DE(03) . For the generators D a nd A in 0 3 , one has
D
~
DEID)
~ ~2
('
J3
v'3)  1
'
A
~
DEIA)
~
(' 0 ) . 0 1
The fourd imensioual functional space spanned by the following basis fUllctiolls is illvari(Hlt. in the grou p 03:
Calculate the representation matrices of the generators D and A of O:j in this re presentation . Then, rednce this representation into the
direct sum of the irreducible representations of 0 3 , a nd construct new basis funet.ions which arc the linear combinations of the original basis fundions and bdong to the irr(,.>{lucible representations. 12 , Calculate the characters and the representat.ion matrices of the proper symmetric gTOUp 0 of a cube with thc method of the quoticnt gTOUp and the method of coordinate transformations . 13. The lIlultipli{:at.ioll table for Il grou p G of o rder 12 is IJS follows.
10·1
Chap. S Theory of Representations
E ,\
E
A
E
A
B B
A
H
E
B
A
B C
C
E K
D F I
D F I
N D M
J
J
K L M N
f(
L
L C
M
J
N
F
L
C
D
F
C
D L
F K M
N D
J E
B
F
E
C
N
K
M
J
L
.4 N
L I E D
M
H
C K I D
M A F
F
N
K
J
B
J C A
J J N C AI
D L
J E
M
F L
N J<
A D 8 K
E F A C
B
J<
L
AI
N
J(
L
N
M
J
F N
D F
At F K J
A L C E .1
B D I
C
A
H
I
K
C
A
J(
D C
J B
A I E N M
B N F L
F D M I. E
(a) Find out the inverse of eaeh element in C; (b) Point out which elemellts can commute wi t h any element in the group; (c) List the period and the order of each element;
(d) Find out the elements in each class of C; (e) Fiud out all iuvariant subgroups in C. For each invariant subgroup, list its cosels and point out onto which gTOUp its quo ticnt gTOUp is isomorphic;
(f) Establish the character table of C; (g) Decide whether C is isomorphic onto the symmetric group T of the tetrahedron, or the symmetric group 0 6 of the regular sixsided polygon. 14 . Calculate t.he character table of the group G given ill P i"Ob. 17 of Chap.
2. 15 . Calculate all inequivalent in·educible representations of the group
D2n+1 with t.he method of iuduced represent.ation. 16 . Calculate all inequivalent irreducible representations of the group 02" with the method of induced represe ntation.
17 . Calculate all inequivalcnt irreducible representations of the symmetric gi"OUp 0 of a cube with the method of induced representation. 18. The regula!" icosahedron is showll in Fig. 2.3. Calculate the character table of the proper symmetric group I of nil icoonhedrOIl with the method of induced representation. 19 . Calculate thc reduction of the subduced representation from each irredu(;ible representation of the gronp I with respect to the s ubgrollps C s , Os , and T .
005
20. Calculate the reduction of the subduccd I'cpn..scntatioll from the regular representatio n of the improper symmetric grOllp h of an icooahedro n with respect to t.he subgroup C s ;, D"d, and T h. 21. Calculate the representation matrice; of Rij find SIZ o f J in its irreducible representat.ions by R6 = SITJS1 To 1 and 5 12 = ToS t TJR 6 (sec Prob . 4 ill C ha p. 4), where t he representa tion matrices of the generators To and 5 1 are (see Prob. 12 ill C hap. 4):
DTl{ Tol DG(To)
= diag =
h, 1, 1]I} ,
diag {r? , f/, 1,1, 1,2} ,
DT'{ Tol = diag hZ , I , II 'l }
DH (Tol
,
= dias' {,, 2 ,1} , I , 11 1" 12 } ,
, (1'/2' /2  "/2 )
DT,( S.) ~ ~ v5
I "" p  VL
P
_,
 p _ p l
DG(Sd=
~5
(
1  I' _ p l
1
I  I _pl 2p  l
( 21', '
,,'
 I  I'
',
 p
 I'  I
)
,;6  ,;6
_ p 2
,;6
1" ,;6
21'
 2"
DH (Sd=~ ~
 ,;6
 I
21'
_ p2
,;6
p'
_ 2p l
p'
 2p
J5
_ 2pl
p2
5
)
22. P lease use the projection operators to reduce the regular representation of ST OUp T. 23 . P lease use the proj('''CLioll operato rs to reOuce the reg ula!" representation of ST OUp O. 24. The lIlult ipli{;at.ioll table of the group G is givell
I!.S
folio"''!>.
HJ6
Chap. S Theory of R epres entati ons
~
E
A
B
C
F
K
M
N
F A B C F K AI N
"£
B M
C
F N
K
K
AI F
M B C
hi
N F A B C AI E K
left.
E A B C F K AI N
N
E
K
N /,
M C
F B
F A C
~ B
F
[(
E B C A
C P A B E N M
F A N {(
E
(a) Write the character table for the group C. (b) If we know that tlu;l represClltatiOIl matrices of the geIH~f1itors A and B in a twodimensional irreducible representation Dare D (A ) =
(~ ~l) '
D(B) = (~~) ,
and two set.s of htlSi.<; functiOlls lPl' and .pI' transform according to this twodimellSiollal representation of G, respectively:
PRJP!! =
L ",
t/!I"D1,,!,(R),
comhine the product fUllction tP1, t/J" such that the hllSi:;; function belongs to the irreducible representation of C. 25. Calculate tIle unitary similarity transformation matrix X for reducing the selfdirect product of the threedimensional irreducible unitary representation DT of the group T:
x ' (DT(R)
x DT ( R ) } X ~
L
" j D'(R ).
j
26. Calculate the Clebsch  Gordan series and the Clebsch Gordan coefficients for the direct product representation of each pail' of two in"edllcible repre;;cntat.ions of the group O .
27 . Calculate the Clebsch Gordan series and Clebscb  Gordall coefficients in the reduction of direct product re presentation in terms of the character table of the group I given ill P rob. 18: ( 1) D T ] x D T.; (2 ) DT! x D T,; (3) D T, x D T,; (4) D T, x DC;
(5) DTo x DC;
(6) DT, x DlI ; (7) DTo x Dllj
(9) DC x Dll ;
(10) DH x DlI.
(8) DC x DC;
C hapte r 4
THREEDIMENSIONAL ROTATION GROUP
A system with the spherical symmetry has a symmetric center, chosen as the origin, and the system is isotropic with respect to the origin. T he system is invariant. under any rotation around any axis ac roos the origin thro ugh any angle. T he symmetric gronp of the system is threedimensional rotation group 80 (3). The s phe rical sYlllmetry is the llloot important symmet ry studied in physics, because a system with the spherical symmetry is easier to deal with and most systems in physics have this approximative symmetry. The group 80(3) is a Lie group, where lts elements can be characterized by a set of continuous parameters, and it can be dealt with by mathematical analysis. T herefore, t.he study on threedimensional rotation gronp is very important bot.h in physics and in mathematics .
4.1
Threedimen sio n a l Rotations
In a real t.hreedimensional space, a spat.ial rotation keeps both the position of the origin and the distance of any two points invaria nt . T here are two viewpoiuts in de;;cribing a rotation in a titret:ldimensional sp ace. [n one viewpoint the system rotates in a· laboratory frame , and in another viewpoint the coordinate frame rotates but the system is fixed . III this book we adopt the first viewpoint. Let K be a perpendicular coordinate frame which is laboratoryfixed. All arbitrary poiu!. P is described by a position vector r which is a veetor pointing· from the origin 0 to P. T he coordinate basis veetors in the laboratory frame are denoted bye", a = 1, 2, 3,
,
r =
L
(4.1)
e"x"
a=l
There is a onetoone {;orrespolldence between the position vector r and
'" I).
pr
~ido
ocr
.08
Chap.
4 Threedimensional Rotatim. Group
three coordinates Xu when e o arc fixed . In a spatial rotation R the point P transforms to pi wbose position vector is denoted by r ' = L:u e ax~ . Since the position of the origin and the dist.ance of any two points keep invariant in R , R can be denoted by a threedimensional real orthogonal matrix
E= (;£') T
l
R:E.,
R* = R .
= ,£T,!.,
(4.2) (4 .3)
T he set of all threedimensional real orthogonal matrices, in the multiplication rule of matrices, satisfies four axioms of a group and forms a group 0(3). Let K' be a bodyfixed frame wit h the coordinate basis vectors e~. The components of r' in the frame f{' are invariant in the rota tion R,
r' =
, L
, eax~ = L
«= 1
e~xa.
(4.4)
b= L
Suhstituting EC]. (4 .2) into Efj. (4.4) , one ohtains
, e~ = L
c aR,,!,_
(4.5)
,.=\
The chirality of Ii coordinate frame is determined by the mixed product of coordinate basis vectors. /( is a righthanderl coordinate frame,
(4 .6) The chirality of /{' depends
Oil
det R
(4.7)
K' is righ t handed if R is a proper rotation, det R = 1. /{' is left handed if R is an improper rotat.ion, det. R = 1. The set of all proper rotations forms an invariant subgroup of 0(3), called the threedimensional rotation group SO(3). The set of all improper rotations forms the only coset of 50(3) in 0 (3). The spatial inversion q is the representl:l.tive element ill the coset. Every improper rotat.ion R' is equal to the product of (1 and a proper rotation R. Study some special rotations . A rotation around the zaxi>; through an angle w is deuoted by R(e3Jw) :
§4.1
Three dimensional Rotations
.09
x~ = XICOSWX2sinw,
 sin,,",'
Xt = Xl sinw I X"l COSW,
cosw
(4.8)
o
x~ = X~, y
P'
Fig. 4 . I A rotation around the zaxis.
In terms of the Pauli matrices. 0,
~
(1a(1b = Oab 1
, I i'L
=
(12
(: :)
(0 i0 )' OF (:~1) ' i
Tr
(ab<;(1c ,
(1a =
0,
Tr ((1a(1b)
(4.9)
= 26"b.
c=l ~xp
. }= L { lW0"2 "
= 1
1 "(  UJ"l . )" w
n!
(1  ~!,,",,2 +
:!",,4 _
..
+.,.)
(COSW Sinw) . . SIIlW COSW
= l COS,,",'  ·lO"ZSlIl W =
(4.10) R ( e 3,w) can be expresseO as a n exponential function of mat rix
R.(e3 ,w) = exp{  iwT3} =
(
COSW
sinw
i
Si~w
co~w
o
o
From t.he cyclic of three axes, one has
R (el,W) = exp{  lwTI } =
(~ ~w o
SlllW
IIII
(:hll7!.
4 Threedimensional Rotation Group
cosw (
s~nw

o
SinW)
10, o cosw
The matrix entries of Ta are
(4
Another useful rotation is S(cp,B), which transforms e3 to the direc n(B, cp), where 8 and cp are the polar angle and the azimuthal angle of
cp cos B  sin cp cos cp sin B) sincpcosB coscp sincpsinB , cosB  sinB 0
COS
S(cp,B)
= R(e3,cp)R(e2,B) = S(cp,B)
(
(~) = (~~~;::~;) = (~~) . cos B
1
(4
n3
It is easy to check (see Prob. 9 of Chap. 1 in [Ma and Gu (2004)])
S(cp,B)T3S(cp,B)1 = nlT! + n2T2 + n3T3 = n· T, T = e1T1 + e2T2
(4
+ e3T3.
Thus, the rotation R( n, w) around the direction n through an angle w be expressed as an exponential function of matrix ,
R(n,w)
= S(cp , B)R(e3,w)S(cp,B)1
= exp{iwST3 S I
= exp
{i t
= w sin
Bsin cp,
= exp{ iwn . T}
waTa} .
}
(4
a= l Wj
=wsinBcoscp,
W2
(4
Note that R(n,w)n = n, TrR(n,w) = 1 + 2cosw, and
R(n,w+27T)
= R(n,w)
= R( n, 27TW) ,
R(n,7T) = R( n, 7T) . (4
On the other hand, any element R in SO(3) is a rotation aroun direction n through an angle w where n is the eigenvector of R to eigenvalue 1 and w is determined by Tr R = 1 + 2 cosw. Thus, the elem R(n, w) E 80(3) can be described by w through its spherical coordin (w, B, cp) or its rectangular coordinates (WI, W2, W3). The variation re of w is a spheroid with radius 7T, where in the sphere two end points diameter describe the same element (rotation). Equation (4.14) shows the elements R(n,w) with the same w form a class in 80(3).
§4·2 Fundamental Concepts of a Lie Group
4.2
4.2.1
111
Fundamental Concept of a Lie Group
The Composition Functions of a Lie Group
A group G is called a continuous group if its element R can be characterized by 9 independent and continuous real parameters rp, 1 :::; P :::; g, varying in a gdimensional region. The region is called the group space of G and 9 is the order of G. It is required that there is a onetoone correspondenc between the group element R and the set of parameters r p, at least in the region where the measure is not vanishing. For example, the group space of 80(3) is a spheroid with radius Jr. Each point inside the spher corresponds to one and only one element of 80(3), but in the sphere wher the measure is vanishing, two end points of a diameter correspond to on element. Hereafter, a point in the group space which corresponds to an element R of G will be said as a point R in the group space for convenience The parameters tp of the product T = RS of two elements are th functions of the parameters of the factors Rand S
(4.17
Jp(r; s) are called the composition functions, which describe the multipli cation rule of elements in G completely. A continuous group is called a Lie group if its composition functions are the analytic functions or at leas piecewise continuously differentiable up to the second order. The mathe matical analysis can be used in studying a Lie group so that the theory o Lie groups has been studied thoroughly. The domain of the definition of Jp(r; s) is a square of the group space and the domain of function is the group space. In order to meet four axiom for a group, the composition functions have to satisfy the conditions: Jp(f(r; s); t)
= Jp(r; J(8; t)),
where Tp are the parameters of the inverse R 1 , and ep are the parameter of the identical element E. e p are usually taken to be zero for convenience As a matter of fact, the composition functions are mainly used in the the oretical analysis, but rarely in the real calculations. Even for a very simpl Lie group, such as 80(3), the composition functions are quite complicated and their forms do not write evidently. Many fundamental concepts of a group are also suitable for a Lie group such as the concepts of an Abelian group, a subgroup, a coset, the conju gate elements, a class, an invariant subgroup, the quotient group, the iso
112
Chap. 4 Threedimensional Rotation Group
morphism, the homomorphism, a linear representation, the character, equivalent representations, an irreducible representation, a selfconju representation, and so on. The matrix entries and the character of a re sentation of a Lie group are the singlevalued analytic functions of the g parameters in the group space, at least in the region where the measu not vanishing. 4.2.2
The Local Property of a Lie Group
The group elements in a Lie group G are said to be adjacent if their rameters differ only slightly from one another . An element A(a) is sai be infinitesimal if it is adjacent to the identical element E . The parame e p of E are chosen to be zero. Thus, the paTameters a p of an infinites element A(a) are infinitesimal. Evidently, RA and AR are both adja to the element R, and the product of R 1 and an element adjacent to an infinitesimal element. To speak roughly, the point R moves continuo in the group space if R is multiplied with infinite number of infinites elements continually. Conversely, if the points Rand E can be conne by a continuous curve embedded completely in the group space, R is product of infinite number of infinitesimal elements. The infinitesimal ments are related to the differential calculus of the group elements so they describe the local property of a Lie group. The precise version of statement will be discussed later. The product of two infinitesimal elements A(a) and B(j3) is an infin imal element. The parameters of AB can be expanded as a Taylor s with respect to the infinitesimal parameters a v and j3v
(4 where e p = 0, AE = A, EB = B , and the infinitesimal quantities of second order have been omitted. Therefore, the product of two infinites elements are commutable, and the parameters of the product are the su those of two elements. It does not m ean that the group is Abelian. Deno by Ci p the parameters of the inverse AJ of an infinitesimal element A one has
(4
§4.2 Fundamental Concepts oj a Lie Group
1.2.3
11
Generators and Differential Operators
There are infinite number of infinitesimal elements in a Lie group C. How t study their property in the transformation group Pc and in a representatio D(G)? PR is the transformation operator corresponding to the element R o G, PR'Ij!(x) = 'Ij!(R1x), where x denotes the coordinates of all degrees o freedom in the system. Let R be an infinitesimal element A(a),
(4.21
The differential operators 1~O), whose number is g, characterize the actio of every infinitesimal element A(a) on a scalar function 'Ij!(x). For the group SO(3), its element R is a rotation in a real three dimensional space. If the system is a mass point and x denotes thre coordinates of the mass point, one obtains from Eq. (4.11), (Ax)"
=
L
{Jab 
i
b
L
ad(Td)ab} Xb
= Xa

L
d
adCdabXb·
(4.22
bd
Substituting Eq. (4.22) into Eq. (4.21), one has 1d(0)
__ . ' " ' 2L CdbaXb ab
8a = L d,
(4.23
Xa
where the natural units are used, Ii = c = 1. The differential operators 1~O of SO(3) are nothing but the orbital angular momentum operators. For a nbody system, 1~O) is the total orbital angular momentum operators (se Eq. (4.220)). If an mdimensional functional space £ spanned by the basis function 1/JJ1.(x) is invariant with respect to the transformation group Pc and corre sponds to a representation D( G) of a Lie group G, m
PR 1/JI'(x) =
L v=l
1/Jv(x)DvJ1.(R).
(4.24
Chap. " Threedimensional Rotation Group
Expand the representation matrix D(A) of an infinitesimal element, 9
D(A)
=1
i
L
I
o.pIp,
_ . oD(A) p 
00.
Z
p
p=l
I
(4
.
0'=0
Ip is nothing but the matrix forms of the differential operators I~O) in basis functions 'l/JJ.L(x) of L, m
I~O)'l/JJ.L(x)
=
L
(4
'l/Jy(x) (Jp)yJ.L .
y=l
I p , 1 :::; P :::; g, are called the generators in a representation D(G) o which characterize the transformation of the basis functions 'l/JJ.L(x) as as all functions in L under the action of every infinitesimal element A If D(G) is a unitary representation, Ip are Hermitian because we introd a factor i in front of Ip in Eq. (4.25).
4.2.4
The Adjoint Representation of a Lie Group
Let RSR 1 = T, where the parameters ofT are the functions of parame of Sand R, (4
For a faithful representation D(G), one calculates the derivative D(R)D(S)D(R)l = D(T) with respect to the parameter Sj, and th takes iij = 0, D(n) oDeS) D(R)  ll = ""' os J· ~ s=O k D(R)IjD(R)1
=
L
h D t1(R),
oD(T) otk
Dt1(R)
I [=0
PRI?) p RI
I '
OSj
s=O
= o'I/J~(s;r) I sJ
k
Similarly,
o'I/Jds; r)
=L
IkO) Dt1(R).
.
(4
s=O
(4
k
Equation (4.28) (or Eq. (4.29)) gives a onetoone or manytoone co spondence between the group element R and the matrix Dad(R), and correspondence is invariant in the multiplication of group elements. Th D"c\ (n) is a representation of a Lie group, called the adjoint representat The dimension of the adjoint representation is the order 9 of G. Every )!;roup has its adjoint representation.
§4.2 Fundamental Concepts of a Lie Group
4.2.5
115
The Global Property of a Lie Group
[n fact, the global property of a Lie group is the topological property of the group space of the Lie group. We will explain the global property of a Lie group with the language familiar to physicists. First is the continuity of the group space. A Lie group is called mixed if its group space falls into several disjoint pieces. A mixed Lie group contains an invariant Lie subgroup G whose group space is a connected piece in which the identical element lies. G is called a connected Lie group The set of elements related to the other connected piece is the coset of G. The property of the mixed Lie group can be characterized completely by G and the representative elements belonging to each coset, respectively. For example, the group space of 0(3) falls into two pieces where det R = 1 and det R = 1, respectively. An invariant subgroup 80(3) of 0(3) has a connected group space with det R = 1. A representative element in the coset with det R = 1 is usually taken to be the spatial inversion a. The 80(3) group and a completely characterize the group 0(3) because any improper rotation in 0(3) is a product of a and a proper rotat.ion belonging to 80(3). Hereafter, we will mainly study the connected Lie groups . 8econd is the degree of continuity of t.he group space. The group space of a connected Lie group G can be simply or multiply connected . The Lie group G is called a multiply connected with the degree of continuity n if the connected curves of any two points in the group space are separated into n classes where any two curves in each class can be changed continuously from one to another in the group spa.ce, but two curves in different cla.sses cannot, where "continuous change of two curves in the group space" means that there exist a set of curves ](t) that depends on a continuous pa.ramete t, 0 ~ t ~ 1, such that each ](t) embeds completely in the group space, and ](0) and ](1) coincide with the two curves, respectively. The Lie group with the degree of continuity n = 1 is called simplyconnected. As proved in mathematics, for a connected Lie group G with the degree of continuity n, there exists a simplyconnected Lie group G' homomorphic onto G with an ntoone correspondence . G' is called the covering group of G . Consider the group 80(2) composed of all rotations R( e3, if!) around the zaxis. Every rotation is characterized by its rotation angle if!. if! varies in a unit circumference, which is the group space of 80(2). Any two points in the circumference can be connected by a curve in the circumference through several loops around the circumference. The loops contained in the curve is enumerated by an integer n: + 1 for each clockwise loop and 1 for each
116
Chap. 4 Threedimensional Rotation Group
counterclockwise loop. The curves connecting two points are sepa into classes signed by n. The degree of continuity of SO(2) is infinite SO(2) is an Abelian Lie group, whose representation has to satisf condition D(rp + 2'll') = D(rp). There are infinite number of inequiv representations denoted by an integer m,
e i2mrr = l.
The covering group of SO(2) is composed of all real numbers wher multiplication rule of elements is defined with the addition of digits. the elements with the parameters rp + 2'll' and rp in the covering grou different, its representation is eir
,..
ll7
§4·3 The Covering Croup of 50(3)
variation region is open because the velocity can tend to the velocity c o light, but cannot be equal to c. In order to generalize the formulas for the finite group given in Chap. 3 to a Lie group, the key is to define the average of a group function over the group elements. For a Lie group the average becomes an integral over the group parameters, instead of a sum over the group elements. As proved in mathematics, the integral can be defined only for a compact Lie group.
4.3
4.3.1
The Covering Group of SO(3)
The Group SU(2)
The set of all twodimensional unimodular (det u = 1) unitary matrices u in the multiplication rule of matrices , forms a Lie group, called SU(2). An arbitrary element u in SU (2)
u
=
(~
!)
E SU(2)
satisfies aa' + ec' = bb* + dd* = ad  be = 1 and abo + cd* = O. The solution is a = d', b = c', and Icl 2 + Idl 2 = 1 (see Prob. 7 of Chap 1 in [Ma and Gu (2004)]). Letting d = cos(w/2) + isin(w/2) cos() and c = sin(w /2) sin () (sin
u(n,w) = 1 cos
(~)
 i (0"' n) sin
(~)
,
(4.31
namely the generators of the selfrepresentation of SU(2) are O"a/2, wher are the Pauli matrices given in Eq. (4.9). Let 0" = La eaO"n.,
O"a
(4.32
Taking the trace of (0" . U) (0" . V) multiplying with the unit matrix 1 o 0" a, one shows
(0" . U) (0" . V) = 1 (U . V) where 0" . (U
X
V) =
Labc EabcO"aUb Vc·
 . I...:;.
.\
_ /
x V),
(4.33
The direct calculation leads to
u(n,wl)11.(n,w2) = u(n,wl 'u(n, 411") = 1,
+ iO" . (U
+ W2),
u(n, 211") = 1,
(4.34
Chap. 4 Threedimensional Rotation Group
ll8
Thus, the element u( it , w) E SU (2) can be characterized by w through sphericaJ coordinates (w, e, ip) or its rectangular coordinates (WI, W2, W The variation region of w, which is the group space of SU(2), is a spher with radius 271", where all points on the sphere (w = 271") represent same element, 1. Although the curve connecting any two points in group space of SU(2) may contain a jump on the sphere, the jump be shrunk in the group space of SU (2) because the jump can be loo as a continuous curve on the sphere. Thus, the group space of SU(2) simplyconnected closed region, and the SU(2) group is a simplyconnec compact Lie group.
4.3.2
Homomorphism of SU(2) onto SO(3)
A real linear combination of three Pauli matrices is traceless and Her tian. Conversely, any twodimensional traceless Hermitian matrix X be expressed as a real linear combination of the Pauli matrices. Taking components x" of the position vector r of an arbitrary point P in the threedimensional space to be the coefficients in X, one obtains a onetocorrespondence between X and r: 3
X
= '\""' o a=1
(laXa
=
(J .
r
=(
X3
XI
+ iX2
Xl  iX2) X3 '
3
det X
=
L
X;
(4.
= _r2
a=l
After a similarity transformation u 1 E SU (2), X becomes another trace Hermitian matrix X' with the same determinant, X'
= uXu 1 ,
det X'
= det
X.
(4
X' corresponds to the position vector r' of another point pI
(4. a
where R is a real orthogonal matrix , R E 0(3) . If u is changed from continuously, R is also changed from 1 continuously. Thus, det R = 1 R E SO(3). Conversely, an arbitrary element R E SO(3) rotates r to r ' , then, X changes to Xl Two traceless Hermitian matrices X and X' w the same determinant can be related by Eq. (4.36), where u is a unit matrix with det u = 1, namely, u E SU(2) . The matrix u relating X and
§4·3 The Covering Group of SO (3)
11
is not unique. If u1Xu 11 = U2XU;1 = X', U;lUl is commutable with an matrix X, so it is a constant matrix, Ul = AU2. Due to their determinants ). = ±1. 8ince r is arbitrary, from Eq. (4.36), one obtains a twotoon correspondence between ±u E 8U(2) and R E 80(3), 3
UO"au  1 =
2:= O"bRba·
(4.38
b=1
Evidently, the correspondence is invariant in the multiplication of grou elements. Hence, 8U(2) is homomorphic onto 80(3): 80(3)
~
8U(2).
(4.39
To show the concrete correspondence between U and R, we calculat u(n,w) (0". r) u(n,w)I. Decompose r into two components parallel an perpendicular to n, respectively, r = na + mb where n . m = O. From Fig 4.2 one sees that R(n, w)r = na + [m cosw + (n x m) sinw] b. it ita
w
m'b mb
itxm
r
o Fig. 4.2
Rotation of a vector r around it through w.
Due to Eq. (4.33), one has
(0" . n) (0" . m)  (0" . m) (0" . n)
= 2iO" . (n
x
m) ,
(0". n) (0". m) (0". n) = i {o·· (n x m)} (0". n) = 0". m. Then,
= 0". n, u(n,w) (0". m) u(n,w)l = 0". [m cosw + (n
u(n,w) (0". n) u(n,w)l
u(n,w) (0". r) u(n,w)1
x m) sinw],
= 0". r' = 0". [R(n,w)r].
(4.40
3
u(n,w)O"aU(n,w)l
= 2:=
O"bRba(n,w).
(4.41
b=1
The elements in 80(3) and 8U(2) are both characterized by the paramete
w. The group space of 80(3), which is doublyconnected, is the spheroi
with radius
7f.
The group space of 8U(2), which is simplyconnected, i
§4.3 Th e Coveri"9 Group oJ SO(3)
119
is not. uniqne. If lL1X1Jil = u2Xu2"1 = X ' , tl2"lVl is <.:onunntable with allY matrix X , so it, is a constant matrix , UI = >'1(2' Due to tlleir det.erminants , >. = ±1. Since r is arbitrary, from Eq. (4.36) , one obtains a twotoone cOr1"cspondence betwccn ±u E SU(2) and R E 80(3),
,
"(Wa u 
I=
L
(1bR/,a.
(4.38)
h= l
Evidently, the correspondence is invaria.nt in the multiplication of group clements. He nce, SU(2) is homomorphic onto 80(3):
80 (3)  SU(2).
(4.39)
To show the concrete correspondence bet wccn U and n, we calculate u( il, w) (0"  r ) u( 11., w) 1. Decompose r into two components parallel and perpendicular to n, respectively, r = fw t 1hb where n· m = O. From F ig. 4.2 onc SI.."es t,hat R( n ,1.I.' )r = rIAl + [m cosw + (n x m ) Sill wIh. Ii
7n'b nxTh
o fi g. 4.2
n
Rotation of a vector
T
around
n
through w.
Due to Eq. (4.33) , one hllS
(0" ' il.) (0" ' m )  (0" ' m )(O" '
f~ ) =
210"' (il. x m ) ,
(0" ' Ti) (0"' ,Tt ) ((1 , il l = i {(1 ' (it x fh)} (0" ' Til =  (1 . m.
Then,
U(1'l ,W) (0" ' it )u(n,w)  l
= 0" ' ii,
U(n ,IA') (0" ' ml 'u(fl, w)l = 0" ' [m cosw + (n x 7h)sinwl , u( n ,w) (0" . r )1J(n ,w)  1 = 0" ' r ' = 0" ' [R(il. ,w)r ].
,
n(n,W)U"lt(ii. ,w) 1 =
L O"hRoo(n , iN'). ,,
(4.'10 )
(4.41)
The elements in 80(3) and 5U (2) are both characterized by the parameter w. The group space of 50(3), which is doublyconnected, is the spheroid with radius 7r. T he group space of 8U(2 ), which is ;;i m piycOJlIJected , U!
120
the spheroid wilh radius 21T. Inside the s pheroid with radius ;rr t here is a onelaone correspondence between elements in 50(3) alld SU(2). rl(ii,w) in the ring with IT < W < 21< is equal to u(  ii, 27rw) owing to Eq. ('1.31). T he pair of ±u(n ,w) in SU(2) Illaps onto one e1emellt. RUI. ,w) in 50 (3). The group SU(2) is t he covering group of SO(3). A faithful representation of 50(3) is simple\1l1ucd and it is an unfaithful represe ntation of SU(2). To spcftk st rictly, a faithful representation of SU(2) is not a re presentation of SO(3). H owc n~r , due to physical reason , it is called a double\'illucd rep
resentation of SO(3). Similar to tile clflSSC8 in SO(3), the clements II( n ,W) with the same w form it class of t.hc SU (2) group (sec Prob. 13 of Chap. 4 in [ll,'la and Gu (2004 )]). Dne to the homomorphism of 8U(2) onto 80 (3), it ils cOllvenient to call n(n, w) E 8U(2) a '"rotation" around the direction n through an angle w. It will be known later that the gTOUp 8U(2) is related to the spinor. T hcreforc, the Jleriod of a s pinor i.'; 471" in "rotation" .
4.3. 3
T ile G roup [u teYTul
MallY propurlie>! of a fillite grollp are bl\Sl.·d on the (.'Ollccpt. of the a\'crHge of a group fUllct ion which is invariant ill t he left and rightllluitiplication with any group clcmcnt. For a Lie group, if the ttverag,e of a group function call be defined as all integral ovel' t he gTOUp space, thooe pl"Operties of It finite grou p will be s uitable for R Lie group
~ I:
1'(//) 
j dnF(R)
~j
(d') W( R)F (Il),
(4 .42)
.7 REG
j dn 1'(/1.)  j dR P(SIl )  j dn 1'(/15).
(4 .43)
The group integral is linear with respect to the gro up fUllct ioll. The weight function lV( R) can be llmler;;tood as the relativt: dens ity of elemcnts in thc neighborhood of R. I f P(R) > 0 and F ( R.) is not equal to zero c\"el)'where. ihe average of F (R) is larger than zero. Thus, lV ( R) has to be single valued, finite , intograble, nOll_ll('j!ati\,c, and 1I0t u
fin ite region . IV (R) is Ilonnalized by
(4.44) Letting T = SR and I\f (S ) = IVo, one ubtaill.:; frolll Eq. (4.43)
§4.3 The Coveri"9 Group oJ SO(3)
j (dt)W(T)F(T) = j dT F(T) = j dR F(R) =
j dR F(SR)
=
j dR F(T)
=
j(d") W (R )F(T) .
(de) W(R) = (dl.) W (T) = (da) Woo
(4.45)
To speak roughly, the "number" of elements in the neighborhood of any gl'OUp clement is the same. ll'(R) can be calculated through the Jacobi determinant ill the replacement of integral variables. Denote by A the infinitesimal element with tIle paramet ers 01" and by R' the element with the parameters r~ in the neighborhood of a given element R. The Jacobi determinants for R' = AR aud A = R'R 1 are, respectively,
Wo = W(R)
det{af~(a;r)} Va"
W(R)
=
Wo
(4.46)
.
(4.47)
,,=0
dct{af~~;r)} ,.
,
r' r
T wo conditioll5 for W (R) are C<juivalcnt. From any of them \V(R) cun be calculated where Wo is determined by the normalization condition (4.44). f.Or the group 5U(2) the weight function is W(h.w) ,
(448) No t e that the elements u(n ,w) in SU(2) with the same ware conjugate to eadl other. The conjugate elemenb are locllted Oil a sphere with a radius w ill the group space of 8U(2) and their parameters can be related by a rotation R. The Jacobi determinant in the replacement of variables of Eq. (4.45) for two conjugate elements is nothing but det R = 1. T hus, the weight fnnction IV (n,w) is inoependent of the direction n, HI (fl , w) = W (w). T his is the direct. result of an isot.ropic space. In Eq. (4.46) the clement R becomes u(e 3,w) and the infinitesimal element A becomes 1J(A)
u(e 3,w) = 1 cos(w/2)  i173sin(w/2),
u(A ) = 1  i (1710'1 + 1720'2 + 1730'3) /2, u(A)IJ(e3,w) = 1 cos (w'/2)  i (u· ii')sin (w'/2) = 1 {cos(w/2)  (0'3/2)sin (w/2))  i171 {cq cos(w/2) i172
f0'2COS(w/2)  0' 1 ~in(w/2)} /2
 i173 (a3cos(w/2)
+ 2 sin(w/2)} /2,
+ 02sin(w/2)} /2
Chap.
T","",,dimens;onai Rotati"" Group

where only the terms up to the first order of OJ is reserved because OJ in Eq. (4.46) is taken to be zero after the derivat.ive. The parameters w' alld n ' are calculated (Ill follo\\TS:
cos(w' / 2) = cos(w/2)  (03/2)s in (w/2) = cos{(w
+ cq)/2} ,
sinew' /2) = sin{(w + (3)/2} = sin(w/2) w'n'J =
+ (Q3/2)cos(w/2), w{sin(w / 2)}  I {aJ cos(w/2) + 02sin(w/2)} /2.
w' n; = w{ sin(w / 2)} w' n3 =
1 {02
w·' {sin(w' / 2)} 
cos(w / 2) 
J {03
cos(w /2)
01
sinew /2)} / 2,
+ 2 sinew / 2)} / 2 =
w' = w
+ 03.
Substituting them into Eq. (4 .:16) , one ha.s (w/2)cot(w/ 2) w/2 0 w / 2 (wj2)cot(!'<"/2) 0 o 0 I
Wo lV (w) _ w 2 {4sin 2 (w,/2)}  I.
T hus, W (w) = W0 4w 2 sin 2 (w/2) . Due to normalization
r )0
21r
1 = 41·Vo
sin 2 (w,/2)dw
r )0
SillOdOJ" dcp = 16rr2Wo,  "If
one oht.ains Wo = (161T 2) I. The gTOllp integral of a fllnction F( 11) = P( w ) of 8U(2) is
J
(du)F(u) =
:r I 411"
J' l' dcp
sinOdO
0
 1f
1"
sin 2 (w/2)F(w)dw.
For the group 80(3) , t.he variatiolll"egion ofw reduces to
J
(dR)F(R) =
 I2
211"
For a cllll» function, first, namely,
J
J' l' dcp
 "If
0
(4.49)
0
sinOdO
l'
11",
sin 2 (w/2)F(w)dw.
(4 .5U)
0
pew) = F(w), the illtegrl;l\
Oil
8
tllld
cp call be done
sin OdOdr..p = ' l 1l".
Now, for a compact Lie group, such as the groups 8U(2) and 80(3), t he average of a group function which is invariant in the left and right1I11litipii(:atioll with any group dement can be defined. Thns, Ii cOl!lpad Lie gTOUp has the following properties like those for a· finite grou p: (a) Any representation is equivalent to a unitary representation , and two equivalent unitary representations ean be related th rough a unitary ~i!llilarity tran;fol"lnatiOIl.
§4.3 Th e Coveri"9 Group oJ SO(3)
123
(b) Any real representation is equivalent to a real orthogonal representation, and two equivalent reHI orthogonal representations call be related through I) real Qrthogonal similarity transformation . (c) A representation is reducible if and only if there exists a nOllconstant matrix commutable with its evcry represcntation matrix. Any reduci blc representation is completely reducible. (el ) The representation m atrix entries and the characters of the two satisfy the orthogonal relainequivalent irreducible uililary repn."8cntations tions:
D~f'(R) * lY,,), (R) =
J J
dR
dll x ' (l1)"><' (II)
Any representation sentations:
_1 J'jOI",Jp>.. , "mj
(4 .51 )
~ 'oj.
be expanded with respect to t.he irreducible repre
Catl
XI R) ~
, aj =
J
L ,
a,K(II), (4.52)
dR XJ( R)*X( R ).
The group integral for the characters reduces to the elllSS integral. For thc gro up 5U(2),
(4.53)
121: dw sin2(w/2)x1(w) . *X(w) .
I
aj = 
a
rr
(4.54)
(e) Two representations are equivalent if and ollly if the characters of every clement in them are equal to each other, respectively, X(R) = X(R). A representation is irreducible if and only if the character X( R) in the representation satisfies
J
dll
Ix(II)I' ~
14.55)
l.
For the gTOUp 8U (2), Eq. (4.55) becomes
1 rr
1" "
2
dw sin 2 (w/2) Ix(w) 1 = I.
(4.56)
Chap.
T" ,"",,dimens;onai Rotati"" Group

(f) T he similarity t ransfonn ation matrix X, which relates t he irreducible unitary ~elfcolljug"'te representatioH and its conjugate o ne, is ~ymrnetric or a ntisym metric. X is symmetric if lind only if t he repre;;elltatiol) is real. satisj}·ing (g) Although the number of the standard irreducible basis Eq. (3.130) becomes infinite for a Lie group, t heir fo rmulas can still be used,
bW
(j)
, (j) 
(4.57)
/'"  e,l .
(j) T . bpl'
Th~ b{l$i$
"~ ,(i) TI'
,
DiTp (T) ,
(j) T  "~ D pr(Tl j (j) bvp b"T·
bW and the idempotent ef. b~l =
Inj
J
be expr~$ed as
dRD{I« R)* PR,
el'(j)  ,(j) 1'1' e(j) =
(:(l.U
(4 .58)
Inj
L eW) =
J
dRDi1<1' IR)' PH,
111.)
!:,
14 .59)
J
n(iUj) " nUl "v "I' = U;j "1'''/''
dR Xj( fl )' PH. ~(;),n(j) _ ... v "'I' 
J ..~(j),n(j) 'J"'V
"'I' ,
where the gro up element R is replaced with the transformation operator PH which acts on the scalar wave fu nctions, and t is a linear combination of g roup elements. Hthe compact Lie gro up G is the symme tric group of a quantum system , the static wave functions 1/l p (X) can be combined int.o that belonging to an irreducible repre;;entation. In fact , bLl 1/J,,(X), if it is not vanishing, is just t hat fu nction belonging to the 11th row of the irreducible represelltation
DiIC) , T {b{",¢,lx)} ~
4.4 4.4.1
L.:
{o:,,¢,(x)j D;" IT) .
14.60)
Irreducible R e presen t ations of 5U(2) E uler A n gl c8
The g(.,,(Jmet,ric meaniug of the parameters w for chanu;tElrizing Ill! element R(n ,w) in 50 (3) are evident . R(h ,w) is a rotation around the di rection n t hrough an angle w. T he more important merit. of the parameters w is that, at least in the neighborhood of t he idelltical element E , there is a o netoone eOlTeHpoudellce between the group element and t he point il]
§4.4
fnu/""ibl e Rel""~tmt"t.;o" .• uf SU (2)
125
t he group space. This merit makes thc parametcrs w more su itable in theoretical study. However, the parameters w are not very cOllvell ient. in the real calculation, because the parameters w are hard to be determined from thc matrix form of R or from the relative positions of t.wo frames before and after the rotation R. On t hc other hand, t hcre is another set of parameters, called the E uler angles, which are easy to be determined from thc matrLx fO I'm of R and from the relative pooitions of two framcs before aud after t he rotation R. There i.<; a ~horkolll iu g for the Euler angle; t hat the Euler angles of an ciemcllt in the neighborhood of the identical element. E are not unique. T his shortcoming makes the E uler angles inconvcnient in theoretical study. For a given matrix R. E 80(3), its third column is a n nit. vector , to the direction of which R rotates the z axis. Deno t.ing by f3 and 0 the polar angle and the azimuthal angle of the unit vector, respectively, o ne has
S(a,,B)
0 ) (Rn ) ( 0 I
=
R 23
(4.61)
R33
where 5(0, ,8) is given in Eq. (4 .12). Hence, 5(0 ,tl)I R preserves the zaxis invariant so that it is a rotation around zaxis, denoted by R(e3,1'). T herefore, eyer)' rot.ation in the threedimensiOllal space can be expressed as a product of three rotations around the coordinate a xes
 Co.C{j8""f 
SoCiJ8""f
Bo.Cy
+ coCy
(4 .62)
8f38""f
where Co = cos 0', So = sin u, etc. Three angles a , /3, Ilnd , afl;~ caHed i be Euler angles. When the matrix form of R. is given, (J and 0 arc respectively the polar angle and the azimu thal angle of a unit vector given in the third colullln of R. Thc third row of R is a lso a unit vector. whose polar anglc is fj and whose azimuthal angle is (r.  y). If the rotation R t.ransforms the coordinate frame f( to [(' , then in the [( coordinate frame , the polar angle
and the azimuthal angle of the z'axis of f{ ' are /3 and Ct, respectively. In the J(' coordinate framc, thc polur angle and thc azimuthal angle of the zaxis of J( are fj a lld (7f  1') . The domain of detillition for t he Euler angles in SO(3) are
(4.63)
Chap.
T" ,"",,dimens;onai Rotati"" Group

\Vhcn (3 = 0 or r., R(c~ ,o+1'),
R(o, O,,) =
R(o·, To,,) = S(a  " To ),
(4.64)
only one paramet.er in 0: and l' is independent. This is the short coming of the parameters of the Euler angles. In Eq. (4.62 ) a I"O tation is expressed as a product of t hree rotations around the coordi nate axe!; of the laborat.ory frame K . If Ollt) choo>;es t he rotations around the coordinate axes of the bodyfixed frame /(' , the proonct order will he changed :
R (a,p, ,)
~ { [R(e" a )R(e"
m[R(e" ,) [R(e" a) R(e2, mr'}
(4.65)
. {R ( C3, 0: )R( cz , j3) R( C3 , 0) 1 } R( C3, 0:). R. i1; a rotation which fi rst rotates around the Z'llXi!; in J{I t hrough 0: a ugle, then rotates around the y'ax.is in the new /(' frame tllrough (3 angle, and at last rotates around t he z'axis in the newer f{' frame through , angle. The group int.egral of SO(:3) with the parameters of the Euler angles can be calculated as follows . Let R. rotate the J( frame to the Je frame. Denote by P and Q the intersections of t.he z'a:"is and the x'a.xis of Ie with t.he uni t sphere ill K, respectively. The position of P on the unit sphere is clJaracterized by the polar a ng·le fJ and the azi.muthal angle a of t he direction OP. For a given P , the position of Q is charactcrized by t hc angle ,. For the rotation in the neighborhood of R, P (:hang~ in the area (sin fJdo:dfJ), and Q changes in the arc (d,) when P is given. Since the unit sphere is isotropic, t he rela tive ~ number" of elements ill the Ileig·hborhood of R is proportional to the area (sin (3dodfJ) and to the arc (d,) . T hus, t he group integral of 80 (3) with the Euler angles as parameters is
f
p(R)dR~
I,]" da 10f"'in fd P ]"_". F(a ,f, o)d"
811"
(4 .66)
_"II"
where the coefficient is determined by the normalization
J
I ]" do: 1"
dR = 8 z 11"
_ or
sill fJdf3
0
]"
d, = 1.
(4.67)
_ or
For the SU(2 ) gTonp,
(4 .68) where thc domain of definition for the Euler angles is cnlarged,
o :s: f3 :s: To ,
 27r
:s: , <
2r..
(4.69)
§4.4
fnu/""ibl e Rel""~tmt"t.;o" .• u[ SU (2)
The group integral for SU(2) is
J
F (u)dn=
4.4.2
I2 lG 1T
J' l' da
_ ,..
sinjJdjJ
0
J"
F (a, jJ , ,)d'Y.
(4.70)
 2".
L inear R epre8entatio n8 of S U(2)
The basic method fo r calculating a repre;entation of a gTOUp G j,; to find a functional space CU ) invariant with respect to C. Applying tIle transformation operator PH to the basis functions .,pt, in e {j), one obtains a representatio n Di(R) of G:
The element 11 E SU(2) is a nnitary transformation in a twodimcns ional complex space,
(''') r/,
="U
, (') 11·
The homogeneous functions of order n with resp(."Ct to ~ a nd II constnlct lin (n + I)dimensional space cC n) which is invariant to the group SU (2). T he basis functiolls in e (n) are ('''Ti''''', m = 0,1 , . .. , n . For cOllvenience one choost!S t,he coeJ!icients and enumeration of the basis functions as follows
. ¢f,(~,11) =
(_ l )ip
J(.)+/t)'(. .J
. .+ ),el'ry1 1',
/1.
j = n/2 = O, 1/ 2, 1, 3 / 2, . .. , /1 =j  1II.=j, (j  I ), .. . ,  (j  I ),  ) .
(471 )
Calculate the matrix form D il'(u) of P" in the basis functions 41,(~,11), p .. .pj, (~,ll) = .j1,( ~11, .,/') =
L
1/{(f"I/)DtIJu) .
" Not e that
II
_,
1/ 1
=
= l cos(w/2)
+ isin{wj2)(u
(COS(W/ 2) + in:c sin(w/2) sin(w/2)( 1L2 + ·ind
· n ),
sin(w / 2) (n2 + ind ) cos(wj2)  in3 5il1(,,'/2) .
(4. .72)
Chop. l
128
When it =
I'll, {" =
'I'/jree.di"'~n,iO"111
RotatiQl' Group
{cxp(iw/ 2) iUlel,/, = 'lcxp(iwj 2). Theil,
= '1/1.({, II)e i ,."" (4.73)
Dt l ,(e3 ,w) = 6""e 'I"<' . cz , {II = f,cos(w/2) I 11sin{w/ 2) and 1/<.:os(w /2). Then ,
When r't
,
~ ~
1)" =
 { sin(w/2)
+
J(j + I' )! {_{sin(w/ 2)}i+,, m {l/ cos(wJ2)}m (j
+ It
m)!m !
I.n order to express the righthand side of the equality to be 8. lincar combination of !/1,(f, . '1), one makes a replacement of the summation indices from 11} to v , v = n + In  j, ;
L di,,(w)
=
V~I,(e2 'w) _ ' " ( 1)" {(j + ",)i(i _ 1I)!(j + p )!{j _ IL) !}1 /2 L (j + v n)!(j /' /I.)!n1(1/ II + II)!
(4.74)
. "{cos(w j 2)}2j+,,p2n {sin(wj 2)} 2n ,,+I' ,
where n runs from the ma.ximWll between 0 and (v  /J) to the minimum between (j + v) aud (j  //). Hence, the rept"L'SentatioIl D i ofSU(2) is
Dill (a,
p,i) =
{1Y (e:!, 0 ) Dj (e2 , p) DJ(e3.,.)}
= e "''' di"I'
(")e  i " ., }J
v,.
(4.75)
,
where DJ( e3,w) is diagonal and DJ(e2 ,w) = di(w) is real. As a cOllvention, the arrangillg order o f row (column) i.., j, (j  I), .. ., (j  I),  j .
fnu/""ible Rel""~tmt"t.;o" .• u[ SU(2)
§4.4
'"
The representation matrix (jJ hI\." some evident SYlllmetry, which can be proved by replacement of summation indices:
Flom Eq. (4 .74), one has
(4 .77)
Now, we analyze t.he property of the reprf'..sentat,ioll Dj of 8 U(2). (a) T hedimensionofO j js(2j+l),j = O, 1/2 , I , 3/2, .. .. OO(u) = 1 and OI /2(u) = '\1. (b) Wben j is an integer, Dj is a singlevalued repre~elltatioll of 80(3) , but an llllfaithful one of 80(2). When j is half of an odd integer, OJ is a doublevalued representation of 80(3) and a faithful Olle of SU(2). (e) d j is a real orthogonal matl"ix and Dj is a unitary rcpr(£cntatioll. (d) The character of the class with angle w can he calculated from Lhe diagonal matrix o j (e3, w) : j
X'(w) ~
j
L ,"'" ~ L I'=  }
sin{(j + 1/2)w} sin(wj2) .
eif'W _
(U 8)
1'= )
Since X'(w) satisfies the orthogonal relation (4 .53), the n~presentations OJ (Ire inequivalent irreducible representations of SU(2). We are going to show by reduction to absurdity that there is no other rcpresentation of 8U(2) with a finite dimension which is inC(IUivalcnt to all OJ. If this representation exi.'>ts, its character X{w) hus to satisfy 0=
!. 12l1" Jw sin2(w/2)Xi(w)*X(w) ,
0
'1 ,
0
2 r.
riw sin {(j + 1 / 2)w} {X(w) sin(w / 2)} .
From the theory of the Fourier ;;erJ(.."li , the ort.hogoual
ba.~i.<;
fllllct.iolls
no
Chap.

Th"",,dimens;onai Rotati"" Group
Sill {(j t l / 2)w), where j arc half of nOllnegative integers, arc complete in the region [0, 211"], namely, any function {x(w)sin(wj2)J orthogonal to all sin {(j + J /2)w} 1ll1L<;t be zero. \Vhcn the group parameters are taken to be the Euler angles , the 01"thogonull'clations (4.51) become
'!'/" 1610
_,,
do 1 1 " dfJ 0
Sin i3 / 211"
111 D;,p(n, fJ"Y DL\(a ,{3,"I)
 21.
1
11"
2"
(4 .79)
 2j + I 0;;0,,,,01').'
..
0 dfJ sin !3~w(f3)d1wU~) = 2j
j
+1
O;j.
(e) The generators in the represclltatioll D j are calculated by expanding the representation matrices of the rotations around the coordinat e axes,
D{,,(C3,W) = tS"J1{ 1 il1w
+ ... } ,
+ (w/2) {JIF(I'_l) r~ 
J"il'+l) r {,}, Dil'Jc l,w) = {Dj(C3J  1l"/2)dj (w)Dj(C3, rrj2)} "I'
di,.(w) = 0"11 =
J"II + (w/2)
{ iO"(I' I)f{.  i8v (I'+ljrt}
,
(11) = ~ (8v(/.+!)rt + 8"(1'_1)r~,,] , (14),,1' = ~ [8 (I,+l) r {  8"(I, _ l)r~I'J (11) ".. = Its''1'' VI'
(4 .80)
v
rt
=
r~V+ l
r;"
((j+l')(j _ V + l)}I/2, I'~ v = rt+l, and = J(2j n)(n + I). In comparison with T" given in Eq. (4 .11 ), the repreS~lltation Dl is eqqivalent. to the selfrepresentation of 50(3) (see Prob. 10 of Chap. 1 in [.Ma and Gn (2004)J): where
=
1 :k0 1) /2 (
a = 1, 2, 3,
M = 
1
~l
~i
. (4.8 1)
Since the generators IJ: including Ta, are the matrix forms ofihe different.ia1 operators La , t hey all satisfy the typical commutative relations
[1~,
It] = iL" =1
(fJ/x; J{
(' .82)
fnu/""ibl e Rel""~tmt"t.;o" .• u[ SU(2)
§4.4
131
!.n physies , the combinations of generators, I ~
,
L a ( I~) ~,
are often used:
(Ii) (I~ )
VI'
=
(If + iJ4) "I' =
=
(Itat)
JV(I'+l) r t =
= bv(I'I) rl, = bv(I' _J) r~",
"I' VI' +r Jj+ + 2 ( [;)21'1' = ~2 {Ii+ Ji
If the basis function
J"(I'+I) r~I" (4 .83)
(p)' } ~ ',.(J. + 1). vI' ",. 3
IPt (~,I}) belongs to t he Ilrow of the. representation Di . "
(sec Eq. (4 .72 )), it is the common eigenfunction of
(fJf
(Ij) 2 v1. (~, 1}) = j(j + 1)1/1, ((, 'I), 151/1,((, rr} = J.ltP~((, ,'1) , '4: 1/1,((,1])
=
and
Ii,
(4.84)
r{1'~J:I ((' 1)).
(f) Since the characte r X7 is real , D ) is selfconj ugate. Due to Eq. (4. 75) , the similarity t rans formation X , transforming D i to D j · , changes the sign of w in Dj (C3, w), but preser ve;; DJ (c z, w) invariant. Thus, X is
d'(") ' (4.85)
e,
Wben j is an illteger tl'(1T) is symmetric [see Eq. (4.76) J and Dl is real. When j is half of an odd integer, di (1T) is antisymmetric and Dj is sclfconjugate, but Hot )·cal. The explicit romls of d[l'(w) with 1/2 > j:::>: 3 are lisk'
4.4.3
Spherical H armonics Fun ctions
Collsider a singlebody system moving ill a spherically symmetric potential. Its Hamiltonian H (r ) is invariant in a ny rotation R around the center, H (r) P/i = P/iH(r). The symmetry group of the system is 50(3). If the energy level E is degenerate with multiplicity n, there is n linearly independent eigenfullctions tPl'(x), H (J' )WI'(r ) = Ewl'(x), where 11. = 1, 2, "' , II. Pnwl,( r ) is also a n eigc nfunction of fl (r ) with the same cnergy E so that Pjfl/JI,(r ) can be e xpanded with respect to the basis functions 1/!,,(r )
PR1/J,,(r ) =
L v=1
1fv(r )D VI ,(R).
>3,
Chap.

Th"",,dimens;onai Rotati"" Group
Since 1f!1,(r) rcmums invariant III a rotation through a ll angle 21T, D (R) is a singlevl!.lued representation of 50(3) conespolldillg to the cncrg..y E. Generally, D(R) is reducible ,
X  I D(R)X
E9 ,
=
at Dl(Rl ,
X  1J"X =
E£1
,
(ll l~,
(4 .86)
where fa are the generators in the representation D (R) and e is an integer. at can be calculated by the characters, X( R ) = LtU1Xi(R),
21' , l'

IT
.:: IT
dw
0
(4 .87)
dw siJl(w /2) sin{ (e + 1/2)w h(w).
0
Sui::>stitutillg af into Eq. (4.86), where I a is taken to be h first , and then to he i ±1 olle obtains the s imilarity transformation matrLx X. The ww of X is eHnllJefat~1 by 1', and t he eoll1Ulll by linT, where T is IlCtxlcd whell at> I. The new busis function ¢;'T( r ) is
tJ;f.,r(r ) =
L ,
PR"~ c(r) ~
t/ill(r)XI'.lm r
I:
"~'c(r )D~'m(R).
(4 .88)
m'
lience, the static wave function of the singlebody system with spherical symmetry can be chosen as ¢~".,.( r) which belongs to the 7tH'OW of the irr<'du{;ible representation f)t of 8 0 (3) . ,1jJ!nr(r) is the comnlOll eigenfunc tion of, in addition t.o the Hamilt.onian, the orbital angular momentulll £2 and L3 (sec Eqs. (4 .23), (4.26), and (4.84)) £2 l/J~,.,.(r) = f(R
+ 1)¢o!Il.,. (r ), (4.89)
£ 3l/J:"rCr ) = m¢':"r(r ),
£±'tP:".,.(r ) = r~"'~!(m±I)T(r), where £± = £1 ± iL2 and £2 = 2:a L~. Now, neglect the index T in l/J~ T(r) for simplicity. Let r = Cr, 0, IP) aud r o = (I", 0, 0). The rotation T = H(IP , 0, y) changes the point r o at the ..axis t o r , J' = Tr o. From Eq. (4.88), one has
l/J~,(r ) = l/J~( Tro) = Pr,l/J!.,(r o) = ~
L m'
"L m'
.,,, 'I'm' (r o ) e'mOd'mm' (0) e ,," .
¢!~,,(ro)D!nm,(Tt
(4.90)
fn u/""ible Rel""~tmt"t.;o" .• u[ SU (2)
§4.4
Since the lefthand side of Eq. (4 .90) ill independent of ,,{, the righthand J;ide of Eq . (4.90) hru; to be independent of ,,{, namely, t he tennJ; with m.' i 0 011 the right hand side of Eq. (4.90) are vanishing. Letting
¢!,,(r o)
2l+ 1) ' 1'
= 6",0 (
411"
(4 .91 )
.pt(r) ,
from Eq. (4 .90), one has
¢~,(r )
(
.p£(r )
=
2l+ 1)'1' D!"o(IO,O,Or ·
(4 .92)
1 11"
Hence, the static wave fu nction tb~, ( r ) of the singlebody system with the spherical symmetry can be decomposed ru; a product o[ a radial fuuction .p£( r) and all angula.r function Y;' (0,10),
i(O, 10 ) ~ Vm
(2'+• 1)
' I' D'
4
(
mOVl"
0
0)"
~
(2'+1) 4
1/ 2 im'i'di (") V e
·mO
'
••
(4 .93) Since the radial function remairu; invariant in any ro tation, the angular fllllct,ioll lwlongs to the m th row of lhe irreducible representatioll f) f of SO(.3) so that it ill the common eigenfunction of L2 and L3 as given in Eq. (4.89). In quantum mechanics, Y},(O,IO) is called the spherical harmonic fUllction . Due to Eq . (4.79) Y,~( O ,IO) is orthonormal
j _" dlO Jro dO sinOY.~,(O"p) · Y,~,(O,IO) {(2l + 1)(2l' + I n 1/2 j " 1" '. C • d'P dO sIHODmO(rp, O, O) Dm,o(rp, O,O) 1f
f'
=
4r.

0
T,"
(4 .94) From the symmetry (4.76) of di(O) , one has
Y'(,.o)"~ (2C+l) 1/2ei"''i'dlmO (') ~ (_ I )"' yf_ m (0 ,.,... m , .,.. 4 11"
0 )
The Legendre functioll is defined
Pt(cosO)
=
(
.
(4.95)
a;;
4;r
2l + 1
)
'1' Yo (0,0) = l
t
doo(O) ,
(4.96)
which satisfies
£ 'Ir
. ()
•
2bu'
d8 sm 8Pt(cos O)Pt{cosO) = 2l
+1
.
(4.97)
Chop. l Threedim e",'''''''/ RoIMiQt. Group
134
In t he spatial inversion , r
Y,~(1r 
9, To + 'P) =
_
>
 r , (}   r.  0, a nd !p
Cf
: 1) 1/'1 e 4
im ("+v»
+ 1r
Jl'mO(7r _ 0)
! 'P,
(4.98)
(l)ty,~(O ,'P).
The parity of the spherical harmonic function is ( 1)1. i\'lu\lipiying Y,~(O.'P) witll rt , olle ohtain..:; the harmo nic polynomial Y:,(r ) which i~ /I. homogeneous polynomial of order (' with l"C!:Ipeo;:t to t he
rectangular coordinates
(XI, X2, X 3)
and satisfi es fhe Lnplu.cc equation
Y,~( r ) = ly,~( r ),
Y :,(r ) wi t h f ~ 3 arc listed
4.5
IllS
(4.99)
follows.
T he Li e T heOl·ems
The Lie theorems tire the fundamental theorems in the theory of Lie groups which characterize the important role of gcucr&lors.
Theo r em 4. 1 (The First Lie T h eore m ) T he rcpImjclltation of n Lie group G with a conllccte
§4.5
>35
The Lie Theorems
equation with respec t to the parameters /'8 for the fixed S,
T hen, taking 5 1 = R, one has T = E and obtains from Eq. (4.25) (4 .100) where
8fAer';'i') 8,Je
r'=,·
5,w(r) is indeptmdcnt of the representation D(C) , hut depclJds 011 the choice of parameters of the Lie gro up C. The determinant of S( 1') is nothing but t.he Jacobi determinant ill the int.egral transformation (see Eq. (4.47)) so t hat 5(1') is nonsingular . Dellote by S(r) the illverse matrix of S(r):
L
SAD(rl5DB(r) = "AB.
(4.102)
o
Letting R = E ill Eq. (4 .101 ), one has
(4 .103) lind Eq. (4 .100) redl\ce~ to Eq. (4 .25). Equatioll (4.1 00) is solved under the m 2 boundary conditions (4 .104) where m 1;; the dimension of the represelltation. Since D( R) exists, one may choose a convenient path to integn.t.e Eq. (4.100) snch that only one variable changes in each segment. of the path. F irst, let all r A = 0 except for 1'1 which changes from O. Equation (4.100) with the boundary conditioll (4.104) is a differential equation of first order and its solution D(r I, 0, ... , 0) is an exponential function . Theil fix ·r l and let thc rclIlainiug I' A = 0 except for 1'2 which changes from O. Eqllation (4.100) with the bounda.ry condition D(rl' 0, ... , 0) is a differential equat.ion of first order again and its solution D(r\, rz , 0, . .. , 0) is the product of two exponential functions . In this way olle cali obtain the solution D(R), which i!:l a product of a few expollelltial f\lncti()II~ . o
136
Chap.

T","",,dimens;onai Rotati"" Group
Corollary 4.1.1 T wo representations of a Lie group with a connected gro up space are equivalent if and only if t.heir geul;!rators arc related by a comlllon similarity transforma t ion
Corollary 4 . 1.2 A representation of a Lie group wit h a connected group spar;e is reducible if ami only if its rep rescntation Spll.CC eontaills a nontrivial invariant subspace with respect. to its all generators.
11
lf 1 ) and l~) arc t.he generators of two inequivalcnt Corollary 4.1.3 irreducible representaLiom; of I;). Lie group with a cOllllect.ed group ~pace and X satisfies
for all generators , then X =
o.
Co r o ll a r y 4. 1.4 If X is commutllble with all generlltors fA of 1111 irreducible representat ioll of a Lie gTOUp with a connected group space, then X is a constant matrix. Oue to Theorem 4.1, the coro llaries are obvious. For the mixed Lie group, one can choose one element in ellch cOllnected piece in t.he gTOUp space. The corollaries hold if, in additloll to the generat ors, the representat ion matrices of those chosen elements also satisfy the condition. Theorem 4.2 (The Second Lie Theorem) The ge nerators III any rcprc;;entation of a Lic gTOlJP C satisfy the common COllllIIutative relations
lAIn  I n l A =
D C A8 I D,
(4.105)
D
where C AnD are cll.lk'
8'D( R ) fJrBfJr II .
(4.106)
§4.5
The Lie Theorems
Differentiating Eq. (4.100) with respect to 7'A, one has
[)2 D(R ) B,'' ,8"8
=
i ~ LV
To DSvB(r) D (R) _ ~ To TI'Sv B(r) SpA(r) D ( R). Or A LVI'
By interchanging A and B , it becomes ()2D (R ) =
fJrBlJrA
i ~
L...J D
10 8S DA (r) D ( R) _ 8"1"8
~
L...J DP
l DIpSOA(r)Spn(r)D(R) .
Tlnm , Eq. (4.106) becomes
L
{I o l p  l p I D }SD,dr)SpB (1')
DP
=iL V
Rightm ultiplying it with S(r) (see Eq. (4.102)), one has
[..1.18  [81..1. =
iL {L o
10
PQ
(fJS;Q("I") _ 8S;p(r)) SPA(r)S'Q8(r) } . rp rQ
(4. 107) T he lefthand side of Eq. (4.lO7) is independent of R , so is its t'ight hand side, nall!d y, t he qnanti ty ill t he cnrve bracket of Eq . (4. 107) is COllstant. Letting R = E , oue obt.ains
(<1.108) T hus, t he structure constants are real and independent of the reprcscntation. COlwerseiy, if 9 matrices f A satisfy the commutative relations (4 .105)' Eqs. (4.1 07) aud (4 .106) also hold, s o tlmt t he solution D ( R ) is independent of the paths . Due to Eq. (4.103), 1A satisfy Eq . (4 .25) obviously. The next problem is whether t he solutio n D(R ) constitute a representation of G , namely, whether D ( R )D (S) = D (T) if RS = T . !nteg rat.illg Eq. (4.100) with the boulldary conditio n (4 .10,1), first from E ~o S, olle ou~uilll:; D (S), T iltH, iulegmlillg, Eq. (,1. 100) wi l li L1 le UOlUl l!U fY condition D (S) b:om S to T , one has
DO(T) B ~ i Iv
'filII";,
L A
JAS,w(t)O(T ) ,
O (T) I T~S ~ O(S).
>38
Chap.

Th"",,dimens;onai Rotati"" Group
= SABlr). J(.;u)=f'
On the other halld, rightmultiply ing Eqs. (4 .100) and (4.104) with 0(8), one hILS
&D(R)D(S) =  i " lASAB(") D (R)D(S) • OI'B
L
A
D(R)D(S)IRS ~S = D(S).
Since both D(T) and O(R)D(S) satisfy the same IXlnation with the same boundary conditioll, they equal each other. 0 Because the generator I A is the matrix form of a diffcreutial operator in the representation space, 1.~O) satisfies the same commutative relation
[1°)
(0) 1(0)] " '"" C 0 1 (0) [1A ' 11  I L.. AB D'
(<1.109)
D
Usually, the stl'udure constants C ABD arc not calculated from Eq.
(4.108) . Instead, for a given Lie group and its parameters, CABO are calculated from the commutative relations of the generat.ors in a known faitllful representation, say the selfrepresentation. The difIerential operators of the 80(3) group are the orbital angular momentum operators,
[La> LbJ
= i
L
EaWLJ,
(4.IlO)
< From Eq. (,1.110) one obt,ains that the structure constants C abd of both the 80(3) group and the SU(2) group are the totally antisymmctric tcnsor (aM. Thc g'cnerators of every representation of the SO(3) group and the SU(2) gro up, iucluding their selfrepresentations (Sl,.>€ Eqs. (4. 11 ) Ilud (4 .3 1)) , have to satisfy the commutative relations (4.110). In qu ant um mechauics, the matrix forms (4 .80) of the angular momentum operat ors are calculatcd from the commutative relatioJ1>l (see Prob. 14 of Chap. 4 ill IMa and Gu (2004)1) .
AlP of a
What conditions the s tructure constants C satisfy? From Eq. (4.105) and the Jacobi identity,
Lie group s hould
Ili A, 181, 10 1+1118 , 101, IA I + 1110, IAr. 181
= fAI BID IBIAID  IDfAIB+1D IB IA+IB1VIA1vIBJA [AI 8 ~ + [AI V lll+lv~/H  JA[V I B JB ID~ + 1 8 ~JD
 0. one obtains
CABD = COAD,
E
Q {CABPCPDQ I CBriC PA t CVl'CPBQ} =
o.
(".11 1)
p
Theor em 4. 3 (T he Th ird Lie Theore m ) A set of constants C ABD CIUl he the structure COllst •• ub; of fI Lie group if and ollly if they satisfy Eq. (4.111 ). We will not prove tbis tht'Orem. The Lie grOllps Cftll be clf\SSifieri based 011 this theorem (see Chap. 7). 1'wo Lic groups with the same structure constants arc sftid to be locally isomorphic. Two locally isomorphic Lie gronps are lIot isomorphic generally. There are t.wo typical counterexamples. SU(2) and 50 (3) have the same structure cons t ants , but globally they are only homomorphic. The twodimensional ullitary STOUp U(2) contains a s ubgroup 5U(2) as well as a subgrou p UtI) eomposed by the determinants of the clements in U(2). However , two s ubgroups contain two COUlIllOI) clements ± 1. TILlIS , U(2) is IIOt isomorphic onto the group SU(2)0 Ut I), but they are locol1y isomorphic. The adjoint rc presentation of 8 Lie group i.~ defincd ill & 1. (4.28).
D(R)IAD( R)1 =
E
InDB~(R ).
B
Whell R is an infinitesimal element.. from Eq. (4 .25) one has
D (R ) = l  i L
ro JD,
DitA (R) = dlJA  i
o
D
i L: f)
CVl TO = ll o ,
L
IIII = L:
ro (Il?) BII '
10( / /'51)8/1'
(4.1 [2)
11
TIIIlS, t he gellerllton; of the adjoi nt. representation 1)""' (R) of a Lie group
''0
Chap.

Th"",,dimens;onai Rotati"" Group
arc directly related with the st.ructure constants,
(
It is ~ltsy to check that
(fA") BD
sati~fy the commutl\tive relalion (4.105),
L {(I;;')RP(I~d)pS (I~d)RP(I~d)pS}  L {CApRCnsP  Cn/CAs P }   L {CApRCnsP +cnlcsA P } p Lp CSpRcAl =i Lp cAlUP')RS'
[I:\" , Iwl Rs ~
P
p
Botl] Eqs. (4.28) and (4.l 1.3) can be used as the definitions of the adjoint. representation of a Lie group. Usually, the adjoint rcprelClltatioll of a Lie !!,l'OUp is calcuhued neither by derivative ill Eq. (4 .28), nor by solving the differential eqlllltion (4. J 00) with the generatorli (4. 113). III fact , the adjoint representation is determined by comparing the known representation and its generato rs with Eqs. (4 .28) and (4.113). Since
(
, 1 PRL"Pii = L
LbRba .
(4,115)
b=l
Let R = R(
Since P R {<'3.,) is commutable with L 3 , R in Eq. (4 .116) can be rcplaC(.d with R(rp,O,")'). Bnt it is convenient. to take")' = o. If r,l;f,,(x) belongs to the mth row of the irreducible representation Dt( SO(3)), it is the eigenfunction of L 3 with t he eigenvalue m , and PR '~~,,(X) is the eigenfullction of L · h with the eigenvalne m where R = R (rp,O,"),) and it = n(O, 11'). Obvioll~ly, PR"IjJ!,,(x) ~ also the eigenfunctioll of 1.,2 with the eigenvalue e(C + 1).
§.j'.6
4.6 4.6.1
CI~b~ch Gonl",' Coelfic,~,,~
of SU(2)
141
C lebschGordan Coefficients of SU(2)
of Representations
Direct Product
For a compound system composed by two subsystems with the spherical symmetry, the wave fUllctions of the subsystems belong to the given irreducible re presentations of SU(2), respectively
_I" (x(t))D'1"1" (u) '
L"
(4.117)
1/J~,(x (2))DZ,"(u) .
T hey are the eig"enfuuctiolls of the angular momentulll operators, respectively. Here, we generalize the l'ot
PI' { W{,(;Z;(I))WZ(X(Z)) }
L { ¢{,,(x(t))1'Z'(X(2)) } [Dj(u)
=
X
D ):(u)]I"VI,I'l"
I" v'
(4.118) Geuerally, the represellt.ation I)i (tI) X DI.' (tI) i<; reducible and call be reduced J of SU(2) by a similarity w; u d irect lSum uf irreducible representations D tnlllsfoTlllation P',
e
(CJ):)  I {Di(u)
X
I)k(u)}
eik =
Ee
aJDJ(u).
(4.119)
.1
The series in Eq. (4.119) is called the ClcbswGol'dan series. The ehara<:ter of the direct product of representations is 0(w)x /'" ("") =
L
aJxJ(w) ,
(4 .12U)
J
where the multiplicity aJ can be calculated by the formula (4 .87) for characters. Here, we apply the formula (4 .78) directly. When j > k,
,"(wi ~
, I:
e;/'W =
, L I'=j
>4,
Chap.
Xi(w)X'(w) ~
Th "",,dimensionai Rotati"" Group

, L
10'(;+1'+1 ),,' _ e;(j+I< )w
e iw
/.= 1;
1
(4.121)
j+ k
~ L J =j k
When j
J = j  k'
< k, .i  k is replaced with k  j. Generally, one has when ) = j+k , j+k  l , ... , Ij  l"l, the remaillillg cases.
(4.122)
In the coupling of the angular momentums j and k of two subsystems , the total angular momentum J of the compound system can take (2k + 1) values (when j > k) or (2j + 1) values (when j ~ k) given in Eq. (4 .122), lind each value of the total angular momentum .J ap pear:s once. Three angular momentums,7 , k , and J satisfy the r ule that each one in three is not larger than the sum of the other two and not less than their difference. In addition, the three IlUguiar momentums j , h, and J a re all integers, or one is int.eger and the other two are half of odd integers. T his nIle is called the rule of a triangle, denoted by 6(j, k , J). The row of the matrLx jk is elJumerated by /lV, and its column by J AI , The !lew basis functions arc (:ombint,."tl by the m[l.trix cutries of Cj l ,
e
... (x(l) ;(;(2») lH'
~ L... " { 'Pi' .'J(:lP l)_'~ lv(x{21) } Gjl: I'v,JM'
P" iJil,(x(l),aP l)
'w
=
L
iJild x (1),x(2»)Dif'M(U),
(4. 123)
M'
where I /~I < j, Ivi ::; k, and IMI ::; J. The entries C:'~ ..IM are called the Cleru;cbGordan coefficients, or br'ieHy, CO coefficients. Since the coefficicut" arc relat.ed to tile addition of auglllar 1ll01llCllt11ll1S, tile CC coefficients are also eaBed the vector coupling coefficients. Sometimes, the CG coefficients Cf,:.JM are denoted by the Dirac symbols (jkj1VljkJJ\I ) or (JIL, kJ/I1.M). Now, let
' IS
disc \l!;!; t l..., prOPQrtiffi of thQ CG coefficient.>; . Rflwritfl Eq.
(4.119) in the form of generat.ors. Since t.he generators of the direct product Di{u) x Dk (u) of two rcprcscntations are (4. 124) one
IlIlS
CI~b~ch Gonl",' Coelfic,~,,~
§.j'.6
of SU(2)
When a = 3, h is diagonal an d Eq. (4 .125) becomes jk (11 + IJ )GJ"",,.lM
. I ,,' I = /1 + 11 In namey,
jk G 1''',JAt '
jk
_

'IC,j!:
"
1''',.1,\/,
or
GJ"",.1 AI  0,
when AI ! I'
+ IJ.
(4. 12u)
In t he sum of two angular momentums , the com ponent along the zaxis is summL>cl like a scaiolI'. Replacing fa in Eq. (4.125) with I± = 11 ±ifz, one hus from Eq. (4.83)
r' Gjk r '±I' Gik (J.'fI) v ..lM + ± v l'(v'fI).JM
 G"I'v.l(Al±I) r ::;J Al'
(tl.127)
T his is the reeunellce relat.ion for calculating the CG coefficients of8U(2). Since 8U(2) is a compact Lie group, D j is unitary and Gik can be chosen to be unitary. As pointed out ill §3.fi.2 , there is Olle undetermined phase angle for each .J in the CG codfieicnts. Choo~ e the pha..o.;e angle such that C]tk) ,.I(jk) is real and positive. Taking the lower signs in Eq. (,1.127) with II = j fixed and IJ =  k , (k+l), . .. , (1.: 1) one by one, one obtains that the first terlll on the Jeft,lmlld side of Eq. (4.12 7) is vanishing and C;:.Ji]+ vl are all real positi\'c. In the same way, taking the npper sig'n in Eq. (4 .12 7) with IJ =  /.: fixed anJ p. = j, (j  1), ... ,  j + lone by one, one ohtains that the second t.erm on the lefthand side of Eq. (4.127) is vanishing and G;'~ k),J(J,j:) are all real positive. Again from Eq. (4. 127)' the CG coeflicients for SU(2) are s hown all to be real and Gik is a real orthogonal matrix:
(4 .128) .1M
Due to Eq. (4 .12fi) one has
k '" L... c3I'(AlI,).JM Gjk I' ( Al  I'}.J'M 
J JJ', (4 .129)
14.\
Chap.
4. 6 .2
4 T","",,dimens;onai Rotati"" Group
C alculation of Cl ebsch  Gordan C oeffi ci ent s
There arc two equivalent ll1ethod~ fOT the calculation of the CG coefficients of SU(2). Olle is based 011 the recurrence relation (4.127), and the other is based on the group integra.! for Eq. (4 . 119). Tn the first method , one takes the upper Sigl1 in Eq. (4.127) with M = J fixed and Jt = j, (j  I) , . . .,  j + lone by one, and obtains that the term on the righthand side of Eq. (4 . 127) is vanishingalld Cf,r'J "j,JJ are expressed in C:t~j),JJ' which is real po.sitive and can be calculat.ed t.hrough the normalized COlldition (4.129). T he result is jk
'_
(I)J I' (2J + 1)!(j 1 k  J )!(J + I.:  Ji)!{j + IL)! { (J + k j)!(J + j k)!(J + j + k + I)!{k J + p)!(j
C,..(J _I'),JJ =
1/2
p) ! }
(4 .130) Then , (.1  M) till!e~ appliclttioll of Eq. (4.1 27) with the lower ~igll, jk . expresse d as a senes . 0 rcim(Jm).J./" l.· 'I CI'{MI'j ,JM IS !\ . a k·lllg a rep 1acelllcnt of summation index 111 with n = m  11, one obtains the R acah fonn of t he CG coefiicients of 3 U(2) cjJ,;
,,(M  I·j,JM
~ { (2J + J)(j + k  J)!(.l + M)l(.1  M) !(j  fL)!(k  M + IL)! (.I+j+ k + \ )l(.I+j
L
k)!(J
j+k)!(j+fL)!(I.:+M
}'I'
It)!
(_ l)n+i  I'(.I+I.:  Il  n)!(j+lt+n)! (j p. n)!(.1 AI n)!nl(n + /l+k .1)1'
"
(4. 131)
°
where n runs fr 01I1 lhe ma.ximu m between a.nd (J  k  IJ) to the minimum between (j  fl.) and (J  M). T he more symmetric form , caned the Van der \Vaerden form , of the CG coefficients can be obtained by making use of the identitiL'S on eombinatorics (sec Appendix A) Cf,{MI'j,JM
= (2J + 1)1/2 ~(j, 1.:, J)f(j
(k + I'd  /l) l(k  M
L
+ JL)!(j 
+ V)!(.1 + Af}!(J
fL)!
_ M)!} 1/2
(_ 1)" {(n)!(.l  j  M + /l +n) !(J  k+IL+n)!
(4.132)
(j"  p.  n)!(1.: + M  J1  n)!(j + k  J  n) lr l ,
where n rUllS from the ma.ximum alllong 0, (j + AI  fL  J), and (k  /l .I) Ul t. lle minilllum 61l10llg (j  It) , (I.: + M  11) , and (j + k  J), and
CI~b~ch Gonl",' Coelfic,~,,~
§.j'.6
~ (a b C)
= { ( (t
+b
, ,
of SU(2)
e)!( b + c  a)!( c + a  b)! } 1/2 (a +b+ c + I)!
145
(,Ll 33)
T he detailed calculation can be found in §5 .8 of [Ma (1993)], where t he qUfillt nm parameter q is taken to he 1. I,n the sc<:ond method (see Chap. 17 ill [Wigner (1959)]), one first rewrites Eq . (4 .1 19) in the parameters of Euler angles. Dtle to Eq. (4. 126) the factors a ll 0: and l' arc cancelled to each other,
dl,p(f3)d~A(f3)
E,
=
C;'~"J(I,+,,)d(I'+,,)(p+>.)(,B)a~~. J(p+A)·
In t erms of the orthogollal relation (4.79)
e;,~,J{I'+") e;~.J{p+A) =
2J 2+ 1
0 11
the tl i fU liction, olle has
10"d,B sin fJd(I'+")(P+>') (fJ)dtp(,B)d~)" (fJ)·
Second , let p = j and ). =  k in Eg. (4. 77), and t he integral formula
(4 . t:~4 ).
(4 .134) In terms of Egs. (,1. 7,1),
(4 .m) one obtains eik' e jk _ {2J + 1){(2j)!(2k)!}1/2 I",·J(I' +" ) j( " )J{jk) (J + j + k + I)!
(J
+j
+ k )!(J + 1l + /I) !(J  p.  v)! } 'I' (j + /t)!(j /t)!(k + /I) !(k I/)! (_l)",H '"" (J + k + IJ.  m)!(m + j  IJ.)!
 k) !{J  j
{
L
'(J'J'+~k~"")"'('J,+~,~,+~"~~"")~,,,~,7'(~m~+~J~'~k~,~,'"" )! '
m
(4 .136) Since aft "lJU kj is real and positive, it ca n be calculated from Eq. (4 .136) with /1 = j and /I = k. In the calculation Eq. (;\ 1.3) with It = J  j + k , v = J + j + k, l' = J + j  k, alld p = 'TrI. is used for simplificatiol!
aj "
,
_
j("),.I(J") 
{
+ 1)(2j)!(2k )! (J + j + k + l )!(j + k (2J
}'I' J)!
Suh;tjtutiug Eq. (4. 1:H ) into Eq. (4 .136) oue obtains
(4. 137)
Chap. I
146
c J"
=
{(j +
!"',J (", +,,)
L
Rowtion Gro up
k  J )!(J  j + 1. )l( J (1 + j + k + I )!
+
(21 + 1)( 1 { (j + 11,)!(j
Tlp"'udi"'~I\,iotl .. J
+j

1.: )1} 1/2
,.++
II )!(J  11  v )! } ' /' It)!(/.: v) !(k v)!
(_ I )" ,+k+"( J
+ k + J1 
+ J! +
m )!(m
+j
 J1) !
,,) ! (4 .138) T he Wigllcr form of Hw CC codTicicllls of SU(2) is obtui ucd by I'cpluciug the s nmnlat ion inrlex 7/l wi t h n = 111  ~.  1/; m
(1  j + k  m )!(J
,
c:..
II 
m )!m!(m +j  k
I'
{ (2J + I HJ + J1 + V) !(J IJ II )! } ' /' = t::.() , k , J ) (j + /Il !(j Jl l!(/.: + 1/ )1(k /I )! .
,J (p.+,, )
"" . L
( ! )" (J
+ Ji 
" 
11 ) l{n
+j + k 
II
+ v )!
'(.'/J'·""~~"~)!~(~J"~k~+,~'~n~)~!7('~'+ ~k~·C+ v~)i!("~+~", ",)", ,
"
(4 .139) where I l runs from the maximum bet ween ( I.:  v ) and ( j + II ) to the minimum between (J  j  /I) and (1  k t J.l), aud t::. (j, k. J ) is given in 8q . (4.133). T hree fonus o f t he C C coe fficients of SU(2) a re equiva le nt. From the m one obtains t he follow ing symmetry o r t he Clehsch Gordun coefficients of SU(2), where /If = p. + v ,
cil:
_ Cl:i
_ ( I )HI: J C ki ( ")( pIJ{ M I  I'pJ !It
I"',J M 
 ( , )1+1: J

Ci ~'
( I' )(  ,, ).I (  M )
?J' ' ) ' /' CJ1 ( 2k + 1 (  M l l'k(  v) 21 + 1) 1/1 C , . (_I )i  J +v ( 2j + 1 v ( ,\I )j (  I' )"
:: (_ I )I: J I' ~
(<1.1 40)
J
Wig"ner int!'Oduced the 3j symbols which are morc symmetric j k ' ) ( II V P
where
Ii 
~~ I
< f < i +k
(4 .14 1)
a nd /1 '/
II ..
P = O.
(_ I)Hk+t ( j k f ) ~ (k j e ) ~ (jtk) ~ JI VP
V J.i.P
p. p v
(j
k e) , (4.142)
 p.  v  p
CI~b~ch Gonl",' Coelfic,~,,~
§.j'.6
Ltp L
(ike) i kf) (J (
121+ 1)
I"
'W
1/
P
I"
1/
p
(
i
/1'
kPI)
1/'
,4'
of SU(2)
= <>1'1"
<5"""
k ") t =(2f+ 1) _, <>u·<>pp'.
1111 P
Some special CG coefficients are calculated in Prohl>. 16 18 of Chap. 4 of [!l"la and Ou (2004)] .
4.6.3
Applications
(a) The 1/CNTllltation symmetnJ
of wave junctiorls
in a I.wobody system
The wave function of a twohody system wit.h a givcll angular momentum is
In the permutation of two particles, one has
If C1 = C2 =
e, then IP
f,( x (2), :zPJ) =
( 1) L 2l l/t k,( x(l ) , x (2») .
14 . 144 )
For example, in a system comp06ed of two electrons with P wave (C = 1), the wave function with the total angular momentum L = 2 and 0 is symmetric ill permutation, lind that with L = 1 is antisymmeiric.
(b) Th e cI:pallsion of the Lf'.gemire fimction The product of two spherical hamlOnic functions of two subsystems with the same e can he combined to be invariant in rotation (L = 0)
m
m
Let R rotate
to the z..a..xis and ;12 to the half x z plane with positive x {;OllJponCIlt. Denoting by 0 the angle betwetm nl aud f~ 2, one has itl
148
Helice ,
~.(
 ) = ( I)'(21 + 1)"'n ( 0) , r(COS 4,
'Yo U ], n 2
(4.145)
In the method of partial waves in qU811Lum mechanics , 1\ plaHe wave ex!' (i k . r ), which is a sohltion of the d'Alemberl e
.. 00
exp (i k . r ) = exp (ik,"cos 8) = 00
= 41T L
L
,
,l(2C + I )je(kr)Pt(c06 8) (4.146)
,
'i fjdkr )
i_ O
L
Y~ ( k )· y.~( r ),
In ~_ t
where it is the spherical Besscl fuu ction, and k and along the di rections, respectivciy.
r arc the unit \'cctors
(e) The Explmsioll of the product of lwo spherical harmOllie/unctions Replacing Di {u) in Eq . (4.119) with the spheric!:!.] harmonic funct ion (see Eq. (4.93)) , Olle has
V"'"() , n l"'(") '''. n  "L...., {
(2f J
+ 1)(2£2 + 1)
'hr(2L + I)
}'I' (4. 147)
Since C~iJo = 0 whcn L  tJ  tz is lITl odd int.cger , t he pflritics of the wave fuuct iOlls 011 both sides of Eq. (4.147) arc t.he sallic. In terms of the orthogollal relatiQIlS (4 . 128) aud (4.9.1). oue obtai lls
CI~b~ch Gonl",' Coelfic, ~,,~
§.j'.6
of SU(2)
14'
(4.148)
f YA~
'"
(0, ~) * Y;,', (0 , cp) y,~.~ (0,
_ { ( 2fl

4.6 .4
II'
+ 1)(2e2 + 1)
471"(2L+])
(' .149)
e M o CM,
}
OO.Lll
" " rrl2 , U, ' ·
Sum of T hree Angu lar Momentums
If a compound system consists of three suhsystems with the spherical symmetry, the wave function of the compound system with a given angular momentum can be obt,ained by combining first, the wave functions of two subsystems, and then combining t.he result with the wave function of the thi rd one. This is a reduction plOblem of the d il'ect pl"OduCL of three irredncible reprcselltld,iolll; of SU (2). III the reduction, the Illultiplicity OJ of an irreducible representation may be larger than one, and the result will depend upon the order of combination. For example, D! x D] X Dl :::: {D2 ffi Dl ffi DO} X Dl
:::: {D3
total angular momentuUl J , One is to combine the angular momentullls of the first two sllbsystellls into '/12, then to combillc it with the allgular momentum of the third one, Tlle other is to combine the angular momelltulils of the last two subsystems into h3 , then to combine the angular momentum of the first one with h3.
iJlll (J ]2) =
LC(l.l~p)p.J {L C~~,J'2(M_p) 47.( )1jJ~ tP~(x(3)), Lc~f~~_p).JM1}~(x(1)) {L c~,'!~,(M_lj)1jJ~(x{2))'I/)!(x(3) ) } . M
P
1/1,(1 (J23) =
}i
x (1)
(x (2)) }
I'"
~p
('.150)
>50
Chap.

Th"",,di m ens;onai Rotati"" Group
T"",o sets of wave functions can be rciak'{l by a unitary transformation X, which depends 011 J IZ and .h3, in addition to J, j , k, and f , but is independent of the magllctic quantum numbers ,
tJlif (.Jn ) =
L '"
(,1.151 )
Ijfl1(J1'l ) XJ"J03 .
Ext racting a factor from X J 12 J 03 , one llliCS t he more symmetl"ic coefficients, called the R,aOih coefficients W [j kJfj J l 2ha], or the 6j:;;ymbom ,
XJlOh' = {(VI2
+ 1)(2123 + J)}I/2 W [j/.:Jej J 1z 1z3]
= (_ l )i+kH +J {(2112
+ 1)( 2123 + I)}1 /2 {j k lLZ}. e J 1z3
(4.152)
Substituting Eq. (4 .150) into Eq. (4 .151), one obtains
c l'jJ( lIf., .
l' j .JM
Ck! "p,Jo,( lIf  l') 
~ CJ1~ i cj~· ~ ( .I1 p)p,J'\/ IW.J 12 ( M p)
'"W[jkJe; 112123J. {(V12 + 1)(2JZ3 + I)}l/2
(4.153)
In t erms of the orthogonal relations (4 .129), the R..'icab coefficiellts can be exprc!>Scd
ru;
a product of four CO coefficients:
(4. 154)
(,1.1 55)
) ( "/2j, mz m]
/J.3
(4 .156) Obviously, the {lacah coefficients and t he ojsymbols are all real, mid X is a real orthogonal matrix. Hence , the Racah coeHicients satisfy the orthogonal rc!tltiOlll;
§.j'.6
CI~b~ch Gonl",' Coelfic,~,,~
of SU(2)
)5)
(4.157)
In ter ms of the identities on combinatorics given ill A ppelldix A, one is able to calculate the analytic form of the Racah coefficients fI'om Eq. (4 .1 55):
W[abcdj
efl =
L ,
( 1 )a+b+c+d il( G, 1.0, e)il(d, e, c)il (b, d, f)il(a,
( l)Z(z+l) !{(z  a  b  e) !(z  c  d  e)!
(z  1.0 d  f) !(z
/l
c  f)!(a +b+ c +d  ;:;)!
(a +d+e+!  z) !(b+ c + e + !
z=
max
!, c)
a+b+e c +d +e Q,+ c + ! b+ d + !
 z) !r',
(1.l58)
a+b + C+ d } .. . , min
{
a + d+e+! b + c + e +!
,
where Il, 1.0, c, d, e, and f ha\'e to satisfY four conditions of the triang'le rules give n in Fig. ,1.3. The detailed calculation is givcn in §5.4 .2 of [Ma (1993)], where the quantulIl parameter q i.<; taken to be 1.
b
F ig. <1.3
Four triangle rules of the Racah coefficients
T hc formulas Oil the Il.acah coefficients arc quite complicated. A gTaphic method may be helpful to understlllld them . Denote a CG coefficient by three orienten lines intersected at one poiut as shown ill Fig. 4. 4 (a). Remind t.hat (Cj~· }l is the transposeofGjk. \Vhen J = O (or j=O, k = 0) the corresponding oriented line can be omitted . The solid line denotes the magnetic quantum number to be summed, and the doth'.."(] line denotCl; t he magnetic quantuHl Ilumber to be equal to each otber. but not ~ulllmed.
>52
Chap.

Th"",,dimens;onai Rotati"" Group
JM
kv
kv
lit
Fp
 L: (c) T he definition (4. 153) for the R..1.cah coefficients
JM
(d) The expansion (4.155) o f the Racah coefficients . Fig. 4.4
Diagram for the Racah coefficients .
The OJsymbols illt.roducoo Ly Wigne\" are more sy mmetric. T here are 144 ojsymbol<; to be equal to each othe r:
{::;} {:~;}  {~ ;: }  {::;} (a+b  C+ d)/2 (a+b+c  d)j2 { (u  b+c +dlJ 2 ( a+b + c + dlJ 2
e} f
(4. 159)
§4.1
4.7
Tensors and
Spi"or~
153
Tensors and Spinors
\Ve have discussed the concepts of 11. scalar and a vecto\". Now, we al·e going to introduce tllC concepts of 1\ tensor and II spinor. The definitions for Ii tensor and a spinor, as well as those for a scalar a nd a vector, are based 011 their transformation properties in a matrix grou p. In physics , if without spt.'Cialnotification, the matrix gro np is the group 50 (3) , or sometimcs t he Lorentz group. 4.7.1
Vector Fie.ld$
Denote the pOSition \'ector of an arbitra ry point P in a real threedime nsional space by r = La e "x", where c" is the b asis vector in t he labora tory frt\.me J{ and Xa is the coordina t.e in J{. Under a rotation R in the threed imensional s pace, t.he point P rotates to the point P' , and the position vector r becomes r' with t he component x~ in X ,
r =
L
ea Xa
...!!..... r '
=
L
"
(4.160)
"  'L R I
"'I< 
h=1
A q uantity 1/) is called a scalar if it reserves invariant in the rotation R. T he distribution of a sca.lar is called a scalar field . A scalar field is described hy a scala r functio n rJ;(x). In a rotation R, t he value rJ;'( Rx) of a scalar fiel d at the point pi aft.er t he rotation i~ equal t.o the value v.o(x) at the point. P before the rot a t.ion. The t r ansformation of a scalar fUllction in a rotation fl. is characterized by the opera.tor PR,
(4. 161) A quantity V is called a vcctor if its componcnts v" transform in a rota tion R like Xa ,
V~ ..!!...
, v,: = L R ab V b. ,
.
(4.162)
A vector has three eOlllponellt~ , which as a whole characterize t he ~tate of t.he ~y,;tem. Denote by QR the t ransformatiou operator f() r t he components ,
Chap.
T" ,"",,dimens;onai Rotati"" Group

V: 
, (Qn V )p = L R"b Vb. ,,
«1.163)
In fact , Q R is the R matrix. The distribution of a vectOl" is called a vector field. A vecto\" field is described by t hree f\lndiom; V (x)a for its cOlllponeJlt~. In t he rotation R, the vector V '(Rx)" at t he point pi after t he rotation is rot.ated from the vector V (x)" at thc point P before the rotation according to Eq. (4.162), V '(Rx)" = Lb Rab V (X)b . Introduce the traIl!;fol"Illatioll operator On, for the rota tion R,
,
V '(x)"
=[OR V (X)]" L
R"bV (Rlxh,
=
OR = PHQH. = QI1PU ,
(4. '64)
,
[QR V (x)L ~
L
R o' V (x)"
b=l
Rewrit.e the formula:; in t he vedor eqnalitiffi , On V (x) = n V ( R.I x ), QR V (X) = RV (x ),
PR V (x) = V (J~Ix).
(4.'65)
The bodyfixed frame K' and its basis vectors e~ are fi xed with the systcm a nd rotate togethcr, so that the components x" of the position vector r ' with respt'(:t to the basis vt'(:tors e~ reserve illVariant,
r' =
L o
eax~
=
,
L
ed xd ,
ed = Q Red =
,
L
.,
ebRw.
(U 66)
For t he general vector field V (x), one has
V (x) ~ OR V (X) ~
3
L
eo
[OR V (x) l" ~
3:1
L
,
a m ]
e~V(R  1X)d =
L
eo
L
Ro, V (W'x),
w l
[QRed] PnV (X).J.
d=l
(4.1G7) Before the rotatio n R, t wo u·ames f{ and K' coincide with each other,
e a = e~ , aud V (x)a = V (x )",
Tensors and
§4.1
, V (x) ~ I:
Spi"or~
, e,.v(x)o ~ I: e"V (xio.
'"
In the formula V (x) = La e " V (x)", ea i.'l fixed in the rotation and t he COJllPonellt V (X)a, denoted by the ~Ylllbob in the bold form and called the vector field , transforms according to Eq. (4. 164) . In the formula V ex) = Ld C d V( X)d, C d is a basis vector in 1(' which coincides with the basis vector e a ill I( before rotation and transforms according to Eq . (4.1 66), and "(xl,; is only a coefficien t am1 l.r1'1nsfor1l1S li ke a scalar. In summary, under a rotation n, a vector field V (x) changes as follo ws
• L
C"OR V (X)a =
On V (:C)a =
L•
L
=
=
•
L
CdPRV (X)d,
R "d V (R 1x)J,
';=1
ed = ORe.;
c"R",N (R 1xld
QRe.;
L•
=
(<1.1 68)
ebRf,d ,
b= l
ORV(X)d = Pnl' (X)d = l' (R lx)J. The position vector field r (x)" is a special vector fidd which reserves invariant in a rotatio n. In fad, r (x)" = Xa , and r (R lx)a = (R  l x)a,
[ORr (x)]...
=
L,
Rabr {R  1x)b
= Xa =
r {x)" .
Namely,
ORr( X) = r ex),
[PR r {x) l"
=
[QR r (x) l" =
r (R  lx)a
=
L
L,
Ra/>Xb,
(RI)abXb .
(4. 169)
b
The operators OR and QR , 11.05 well 1),5 the operator PR , fi re a1\ t he linear unitary operators. The operator L(x), describing a mechanics quan tity, transforms in a rotatio n R as follows: (4. 170)
4.7.2
Tensor Fields
A tensor of order n contains n !;uoocl"ipt& and 3" componentlS, which as a whole chanl{;teri;.::e the ,;tate of t he !;y~telll . [11 ihe rota t iou, each ,;u b,;cript
>56
Chap.
Th"""dimens;on,,i Rotati"" Group

transforms like a vector subscript,
T ,,,,, •... ,,,, .!!..... (OuT )"]", ...",,
L
=
R" ,b, . . . R""b" Tb,b, ... b,,. (4. 171)
b, bz ...""
A distribution of a tensor is called a tensor field. A tensor field of order n is described by 3" functions for its components transforming in a rotation
HAS
Qn = RxR x · ··xR,
OR = QnP R.
[p//T (:c)lu,a, .. a"  T (R  Ix) " '''2···(1",
L
[QR T (x) la,a,., .a" =
R"jb, fl" ,bo '" R""b" T (xhb 2 •• • b.. .
b Ib • . ,/> "
The basis tensor Od,d, __ .dn component which is equal to ]
(4.172) is a tensor containing' only olle nOllvanishing
(4 .173)
A tensor field can be expanded with respect to the basis tensor: T (x)", ... a"
L
=
(Odl ... J").., ... ",, T(X)d 1... d" = T(x}" •... "",
d, .. ,d"
[OR T (x) I",01 ...
Q"
L
=
(QR (}d, ...d"),,, ...an PRT(x)a, ... a",
d, ... J"
(Q n lJd,d, ...d" )",,,, ..
G
n
=
L
R u , b, R a,b2 ... Ra"b"
Il, .. .b n
= Ra ,d, R"2d2 .. . Ra~d" =
L
(lJd, d, .. .d" )b, b, . .b"
(lJb, ...b" )a, .. G" Ro ,d, Rb 2d2 . . . R b"d",
b, ...bn
p H T (x) d, ...d"
~
T (n Ix) d, ...d,,·
(4.174) 'fhe scalar field is a tensor field of order 0, and a vector field is a tensor of order 1. 4.7.3
SpinOf' Field"
A spinor of rank 8 contains (28 + I) components which as a whole characterize the state of the system. In a rotation, the spinor transforms as ", (8) '0
R 
(OR~(")" ~
L,
D;,( R)~i")
(4.175)
§4.1
Tensors and
Spi"or~
>57
A distribution of a spinor is called a spinor field. A spinor field of rank s is described by (2s + 1) functions fOf its components transforming ill a rot.ation as
[OR,p($)(X)L,. = OR = QRPR ,
L,
D;).(R),p (sl (R  1x)).j
QR = DS(R) ,
14 . 176 )
[PH ~ ")lx)L ~ "')InIx)" [QR ~ ")lx)L ~
L,
D~,IRW ") lxk
Usually, a spinol" is denoted by a (28 + 1) x 1 column mah·ix. The spinor of rl:luk 1/2 is called 1:1 fundamental spinor , o!" briefly a spino!" , where the superscript 1/2 is often neglected. A fundamental spillor for the group 50(3) has two components an d is denoted by a 2 x 1 coluIlln matrix. The basis spinor e (sl( p) is a spinor containing only one nonvanishing componcnt which is equal to 1,
(4.177) where the ordinal index p is indicated inside the rou nd brackets. Tn a rotation fl,
, ,
~ D:,IR) ~
L
14 . 178 )
).=8
, "'C\),Dl,IR)
).=s
A .;pinor field call be expanded with rcsp(.'Ct to the basis
~pinor ,
,
. "'Ix) ~
L
'(')lp)~(') lx)"
~,
OR ~ (')I·x) ~
,
L
14.170)
pc: _ .
,
QRe(· l (p) =
E
e (')(>, ) D~ p( R) ,
).=  B
A scalar is a spino]" of rank O. A vector is a spinor of rank I because the selfrepresentation of 80(3) is equivalent to D l (see Eq. (4.81) ). T he basis spinor e (1l( p) of Tank one is also called the spherical harmonic basis vector,
Chap . 4 Threedimensional Rotation Group
158
3
V(X) =
L
1
e,Y(X)a =
=L
e(i)(p)'l//!) (x)p,
p=I
a.=1 3
e(I)(p)
L
3
eaMa.p,
'l/J(i)(x)p
= L (M1)pa V(X)a,
0.=1
e(i)(l) =  (el + ie2) /.../2, { e(I)(O) = e3, e(I)(l) = (el  ie2) /.../2,
0.=1
(4.18
'l/J(I)(xh =  [V(xh  iV(xhl /.../2, { 'l/J(1)(x)o = V(xh, 'l/J(1)(X)_1 = [V(X)I + iV(xhl /.../2. 4.7.4
Total Angular Momentum Operator
Discuss a system characterized by a spinor field. The Hamiltonian of t system is isotropic so that the group 80(3) is the symmetric group of t system,
(4.18
where the transformation operator OR for a spinor field is divided into t operators, OR = PRQR· For the infinitesimal elements, 3
PA
= 1 iL aaLa., 0.=1 3
(4.18
QA =liL aaSa, 0.=1 3
3
OA=liL aa(La +Sa)=l i L a.=1
aaJa,
0.=1
where So. is the generator of DS, and La. is the differential operator of P called the orbital angular momentum operator in physics. Their sum denoted by J a
(4.18
J a, Sa, and La all satisfy the typical commutative relations of angu momentums. The static wave functions with energy E construct an invariant fun tional space. Its basis function !lip(x) is a spinor field, transforming in
§4. 'l
Tensors and Spinors
159
rotation R according to Eq. (4.176). But, after the transformation, it has to be a combination of the basis functions,
OR ljip(x)
= DS(R) Ijip(R1x) = L
1ji>.(X)DAP(R).
(4.184)
A
The set of the combinative coefficients D>.p(R) forms a representation of SO(3). Reducing the representation D(R) by the method of group theory to be the direct sum of the irreducible representations of SO(3), the static wave function is combined to be Iji~(x) belonging to the J..l row of the irreducible representation Dj,
(4.185) 11
Namely, ljit(x) is the common eigenfunction of the generators J2 and h, j21ji~(x) = j(j
+ 1) Iji~(x),
h Iji~(x) = J..llji~(x), .!± ljit(x) = r~1' Iji~±l (x),
(4.186)
Now, the system is characterized by a spinor field, and its conserved angular momentum is not the orbital angular momentum, but other mechanical quantities J2 and J 3. J a is the sum of the orbital angular momentum La and another quantity 5 a related to the spinor. Both J a and 5 a satisfy the typical commutative relations of angular momentum. 5 a should be a mathematical description of the spinor angular momentum, discovered and measured in experiments. Therefore, 5 a is called the operator of the spinor angular momentum and J a the operator of the total angular momentum. The total angular momentum is conserved in a spherically symmetric system characterized by a spinor field. From Eq. (4.178) the basis spinor e(s)(p) is the common eigenfunction of the operators of the total angular momentum and the spinor angular momentum, he(s)(p) = 5 3 e(s)(p) = pe(s)(p), .!±e(s)(p)
= 5±e(sl(p) =
r~pe(sl(p
J2 e(sl(p) = 5 2 e(s)(p) = s(s
± 1),
+ l)e(sl(p),
(4.187)
There are three sets of the mutual commutable angular momentum operators, one is L2, L 3 , 52, and 53, the other is J2, h, L2, and 52, and the
Chap. ;, Threedimensional Rotation Group
160
third set is j2, h, S2, and L . S = L,a LaSa· For the fundamental spin s = 1/2, the common eigenfunctions of the first set are the product of spherical harmonic functions (it) and the basis spinor e(s) (p). Co bining them by CG coefficients, one obtains the spherical spinor functio which is the common eigenfunction of the second set with the eigenvalu j(j + 1), e(e + 1), and s(s + 1),
y,;,
Jl,
yjeS(') yt () (5)( p. ) I'n _0" Cse p(l'p)jl' I'p n e
(4.18
p
\Vhen s
= 1/2 and e= j
yJUl/2)(1/2) (it)
'f 1/2, one has
=
( j (
(
},j(j+l/2)(1/2) (') _ I'
n  (
+ Jl) 1/2 y j 
2j 2j
.,+1/2
( j p. + 2J + 2
1)
.
, .
1/2 yHl/2(it
_(J+Jl+l)
.,  1 / 2 )
J
(4.18
1/2
+2
x is selfinverse and commutable with
(0"' x) Yj(jl/2)(1/2) (it)
J
Jl) 1/2 yj  I/2(it)
j 
2j
Since 0" .
1 / 2 (it)
1'1/2
yHl/2(it) 1'+1/2
J3 and J2,
= C1 Yj(jl/2)(1/2) (it) + C 2 Yj(j+1/2)(l/2) (it),
where C 1 and Cz are coefficients independent of Jl. Letting Jl = j) o obtains C] = 0 and C 2 = l. As calculated in Prob. 22 of Chap. 4 of [ and Gu (2004)]' the common eigenfunction of the third set of the angu momentum operators with the eigenvalues j(j + 1), S(8 + 1), and v i
Jl,
L
e(s) (p)ei(l'p)
(4.1
P= 5
4.8
4.8.1
Irreducible Tensor Operators and Their Application
Irreducible Tensor Operators
In quantum mechanics a physical quantity is characterized by a linear erator L(x) which consists of the coordinate operators X a , the differen operators a/ax a , and the matrix operators (Ja. In a rotation R, L(x) tra forms as
§4.8
Irreducible Tensor Operators and Their Application
161
A set of (2k + 1) operators L~(x), k :::; p :::; k, is called the irreducible Lensor operators of rank k if those operators transform in the rotation R o SO(3) as k
L
L~(x)D~p(R).
(4.191)
A=k
Each operator in the set of the irreducible tensor operators characterizes a physical quantity independently, but in a rotation it relates to the other operators in the set as given in Eq. (4.191). Rewriting Eq. (4.191) in the Form of generators, one has
[h, L~(x)l = pL;(x), [h, L~(x)l = {(k 'f p)(k ± P + 1)}1/2 L~±l(X), 3
L
[la, [.la, L~(x)]]
= k(k
(4.192)
+ l)L~(x).
a=l
This is an equivalent definition for the irreducible tensor operators. The irreducible tensor operator of rank 0 is called a symmetric operator or a scalar operator which reserves invariant in any rotation R. The irreducible tensor operators of rank 1 are called the vector operators. L~(x) are called t.he irreducible tensor operators of rank k with resp ect to the orbital space or the spin or space if replacing OR with PR or QR, respectively. A typical example for the irreducible tensor operators is the electric multipole operators which are proportional to Ypk (it), OnYpk(it)O[/
= PRYpk(it)p;l = L
Y:(it)D~p(R).
(4.193)
A
They are the irreducible tensor operators of rank k with respect to the whole space and to the orbital space, but the scalar operators with respec to the spinor space. The electric dipole operators (it) are the vector operators with respect to the whole space and to the orbital space, and the scalar operators with respect to the spinor space. In fact, they are the combinations of the coordinate operators,
Y;
Chap. 4 Threedimensional Rolalion Group
152
(4.1
ORXaO//
.x xbRba,
= PRXo.p;;l = L b
Rewriting Eq. (4.194) in the form of generators, one has 3
[la, Xb]
= [La, Xb] = i L
(4.1
tabdXd,
d=!
In quantum mechanics there are some familiar operators having sim commutative relations, 3
[la,Pb]
L
= [La,Pb] = i
[Sa,Pb]
tabdPd,
= 0,
d=! 3
[la, L b] = [La, L b]
=i
=i L
t.abdLd,
(4.1
tabdld,
d=l
3
L tabdSd, d=1 ORPaOi/ = PRPaP;;l = L PbRba, b ORLaO R = PRLaPRI = L LbRba , b ORla0"i/ = L lbRba, ORSaO'R = QRSaQ'RI = L SbRba, [la, Sb]
[Sa, L b] = 0,
d=!
3
[la, lb]
L
= [Sa , Sb] = i
I
[La, Sb]
= O.
QRPaQ'F/
= Pa,
QRLaQJ/
= La,
(4.1
b
J
PRSaP;;1
= Sa.
b
They all are the vector operators with respect to the whole space. For vector operators, their components along a given direction n are
n . r
= 0 RX3 01 R = P RX3 pR1'
n .p A
= a RP3 aI R = p RP3 pR1'
§4·8
4.8.2
Irreducible Tensor Operators and Their Application
n . L = 0 RL30r/ = PRL3P'R1,
n· J
n· S = ORS30r/ = QRS3Q R1 ,
R
16
= ORJ30R1,
= R(tp, e, 0) .
(4.198
WignerEckart Theorem
The static wave functions of a spherically symmetric system belong to an irreducible representation of SO(3),
v
=
OR45{,(x)
L
45~:(x)D~:I",(R).
v'
The calculation on the expectation values of the irreducible tensor operator in the static wave functions can be greatly simplified by making use of th symmetry. Since
=
OR {L;(x) tJlt(x)}
{ORL;(x)OR 1 } {OR tJlt(x)}
=L
{L~(x)tJI~(x)} {D~p(R)DL.(R)},
AV
L; (x) tJlt (x) transform according to the direct product of representations Combining them by the ClebschGordan coefficients, one obtains Fit(x belonging to the irreducible representation: Fit (x)
= L L~(x) tJliJ_p(X)C:(M_P)JM' p
L;(x) tJlt(x) ORFit(x)
= L F;+I"(x)C:~J(P+I")
=L
,
(4.199
J
Fit,(x)D'/w'M(R).
!VI'
From the WignerEckart theorem (see Theorem 3.6),
(4.200
where the constant (45i'IIL k ll tJlj), called the reduced matrix entry, is in dependent of the subscripts fl', P, fl, and M, but is related to the explici forms of 45, L, and tJI and depends on the indices j', k, and j. Hence,
(45~: (x) ILZ (x) I tJI~ (x)) =
L J
C:~J(P+I") (45{, (x) IF;+I" (x))
C~~j' 1"' (45/ IILkl1 tJlj).
(4.201
164
Chap. 4 Threedimensional Rotation Group
In quantum mechanics, the expectation values of the irreducible tensor o erators in the static wave functions are characterized in the calculation matric entries. There are (2j' + 1)(2k + 1)(2j + 1) matrix entries in t form of (P~',(x)IL~(x)1 tJli(x)). Equation (4.201) shows that the rotation properties of the expectation values are fully demonstrated in the CG c efficients, and the detailed properties of the system and the operators a left in the reduced matrix entry. The WignerEckart theorem simplifi the calculation of (2j' + 1)(2k + 1)(2j + 1) matrix entries to the calcul tion of only one parameter (pl' II Lk II tJlj). If there is one matrix entry wi given subscripts fL', p , and fL which can be calculated, the remaining matr entries all are calculable. In most cases, even one matrix entry is hard be calculated. For example, the explicit forms of the wave functions tJI~(
and P~',(x) are unknown. However, through eliminating the paramete Eq. (4.201) gives some relations among the matrix entries which som times can be observed in experiments. FUrthermore, the CG coefficients Eq. (4.201) have to be vanishing when some conditions are not satisfied
IfL'I = Ip + fLI 5, j',
Ik  jl
5, j' 5, k
/pi 5, k ,
+ j.
(4.20
Those conditions are called the selection rules in quantum mechanics. T above method holds when 0 R is replaced with PR or Q R.
4.8.3
Selection Rule and Relative Intensity of Radiation
An isolated atom is isotropic and its symmetric group is 80(3), no matt how complicated its internal construction is. Its static wave function b longs to the irreduci ble representation of SO(3) . In the language of quantu mechanics, the static wave function tJlit (x) is the common eigenfunction H(x), ]2, and h. The intensity of the electric dipole radiation betwe two states is proportional to the square of the matrix entry of Xa
because the electric dipole operator is proportional to the distance of t pair of electric charges. From Eq. (4.194) the operators Xa are the com binations of the spherical harmonic functions Y/ (it), which are the vect operators with respect to the whole space and to the orbital space, and t scalar operators with respect to the spinor space. The square of the entri I( tJlt;, IYpl1 tJlit) 12 with p = 0 demonstrates the intensity of the plane light
§4.8
Irreducible Tensor Operators and Their Application
165
the electric dipole radiation polarized in the zaxis, and that with p = ±1 demonstrates the intensity of the circularly polarized light in the x y plane. The WignerEckart theorem shows
(4.203)
The selection rules are the conditions where the CG coefficients are not vanishing, tlM = M'  M = p = 0 or
± 1,
IJ 
11 :::;
J' :::; J
+ 1.
(4.204)
'When J = 0, J' cannot be equal to 0, so the section rule for J' is usually written as
tlJ
= J'
 J
= ±1
or 0,
0++0.
Read as "the transition from 0 to 0 is forbidden". Considering the spatial inversion, the parities of the initial state and the final state have to be opposite because Xa has odd parity. If the spinorbital interaction is weak, the wave functions are first coupled into the product of two parts, the orbital part with a total orbita angular momentum L and the spinor part with a total spinor angular momentum S. Then, the static wave function is the combination of those functions,
iJl JLS M
=~ ~
LS C(MI7)I7,JM iJlLM17 W1S7 '
(4 .205)
17
In physics, this case is called the LS coupling. iJli.fs is the common eigenfunction of ]2, h, L2, and S2. In this case, there are additional selection rules obtained from the rotational properties of the wave functions and the operators in the orbital space (the action of PRJ and the spinor space (the action of QR) [see Eq. (4.194)]'
tlL
= L' 
L
= ± 1 or 0,
tlS
= S' 
S = O.
o/t 0,
(4.206)
Remind that in the complicated system of n electrons, the parity is no equal to (1) L . The transition with tlL = 0 does not violate the conser vation law of parity.
4 Threedimensional Rotation Group
Chap.
166
The relative intensity of the electric dipole radiation is proportional to the square of the CG coefficients in Eq. (4.203),
I(W;{;+1IY/l wt)1 2
:
2
l(w;{;IYd Iwt)1 2
:
I(W~~J IY~J IwtW
 (ClJ ) .. (ClJ ) IM,J'(M+l) OM,J'M )2.. (CJJ (l)M,J'(Ml)
2
For instance, when J' = J  I, from Eq. (4.132) (or see Prob. 17), the ratio of the relative intensity is
(J  M)(J  M  1) : 2(J  M)(J
+ M) : (J + M)(J + M 
1).
The total intensity of radiation with the given .I' and J lJ
1
L
2
(CpM,J'(M+P))
=
2J' + 1 2J + 1
p=l
JJ'
1
L
2
(Cp(MP),J(Ml)
=
2J' + 1 2J + 1 .
p=l
The total intensity is independent of M. Namely, the total transition probability is independent of the direction of the total angular momentum.
4.8.4
Lande Factor and Zeeman Effects
The SternGerlach experiment discovered that a hydrogen atom in Swave deflects as it passes through an asymmetric magnetic field. The two spectral lines on the film shows that the atom has magnetic momentum. This is the earliest experiment observing the spin of an electron. It was measured that the geromagnetic ratio for the spin angular momentum is double of that for the orbital angular momentum. Namely, the operator of total magnetic momentum of an atom Ma is e
Ma. = 2 (La. me
+ 2Sa )
e
= 2
me
~~
(Ja + So.) = 2 , me
where me is the mass of the electron, 9 is the Lande factor, and the natural units are used, c = Ii = 1. The expectation value of M3 in the state 1jI'k is
M = 3
(IjIJ M
1M 3 IIjIJ) = M
egM 2me
= _2 eJVJ _ _ e_( 1jI.l IS IIjIJ ) 2 M 3 M, me
me
(4.207)
The matrix entries of two vector operators .10. and So. in the same states are proportional,
(4.208)
§4·8
Irreducible Tensor Operators and Their Application
167
where the proportional coefficient AJ is independent of the subscripts a, M, and M'. Since J a wfw is a linear combination of w;&" Eq. (4.208) holds if if!£r, is replaced with J a I[Jfw, 3
3
~ (if!£rISaJalif!£r)
= AJ ~
a=]
(if!:{,lJaJalif!£r)
= AJJ(J + 1).
a=1
If if!£r is a state by LS coupling [see Eq. (4.205)]' it is the common eigenfunction of J2, h, L2, and S2 Since 3
3
~ L~
=~
a =l
3
(Ja  Sa)2
a=l
3
=~
J;
3
+ ~ S; 
a=1
2 ~ SaJa,
a=1
a=1
one has 3
~ (if!t, ISaJalif!£r)
= {.J(J + 1) + S(S + 1) 
L(L + I)} /2,
a=l
and
=
A
J(J
+ 1) + S(S + 1)  L(L + 1) 2J(J+l)
J
Taking a = 3 and lvI' (4 .207) , one obtains
=1 9
+
=
M in Eq . (4.208) and comparing it with Eq
= 3J(J + 1) + S(S + 1) 
A
For the system of a single electron, S
=
L(L + 1)
2J(J+l)
J
9
.
= 1/2
2J + 1 2J { 2J + 1 2(J + 1)
.
and L
= J'f 1/2.
( 4.209) Thus,
1  
when L
=J
when L
= J +2·
2' 1
(4.210)
If the atom is in a strong magnetic field such that the spinorbital interaction can be neglected, the wave function can be described by in which the energy E is
1/J;,W;,
E
e
= Eo +  ( m + 2(J)H. 2me
(4.211)
Eo is the energy when the magnetic field does not exist. Equation (4.211) shows that the change of energy of the system in a magnetic field is independent of Land S.
168
Chap. 4 Threedimensional Rotation Group
the leftcircularly polarized light I (a) Observation parallel to the magnetic field (b) Observation perpendicular to the magnetic field
the rightcircularly polarized light
I
.
v
the plane polarized light
Fig. 4.5
The normal Zeeman effect
The selection rule for the electric dipole radiation is
= ±1 or 0, 6m = ±1, 0,
6L
0++ O. 6S
= 0,
6(J
= O.
(4.21
The energy of the radiative photon in the electric dipole transition is
= 0, 6m = ±l.
when 6m when
(4.21
The radiative light is the plane light polarized in the zaxis when 6m = and is the circularly polarized light about zaxis when 6m = ±l. Wh the observation is made in the direction parallel to the magnetic field (t zaxis), the plane polarized light cannot be seen. Two spectral lines the circularly polarized light deflect from the original position (6Eo) equidistance. The redmoved line is th e leftcircularly polarized light (m m' = 6m = I) , and the other is the rightcircularly polarized lig (mm' = 1). When the observation is made in the direction perpendicul to the magnetic field, three lines can be seen where the middle line is t plane light polarized parallel to the magnetic field. This splitting of electron in a strong magnetic field is called the normal Zeeman effect.. For an electron in a weak magnetic fi eld, the spinorbital interacti cannot be neglected. The wave function has to be characterized by lPiA given in Eq. (4.205). The energy of the electron in the magnetic field is e 2me
E=Eo+gMH.
The observed splitting of spect.rum is rel a ted with the quantum numbe
§4·9
An Isolated Quantum nobody System
169
.I, L, S, and M of both the initial state and the final state:
6..E
= 6..Eo + eH
2me
I
I
(gM  9 M) .
(4.214)
This splitting for an electron in a weak magnetic field is called the anomalous Zeeman effect.
1.9
An Isolated Quantum nbody System
An isolated quantum nobody system is invariant in the translation of spacetime and the spatial rotation so that the energy, the total moment.um, and total orbital angular momentum of the system are conserved. The Illotion of the centerofmass and the global rotation of the system can be se parated from its internal motion, and its Schrodinger equation can be leduced to the radial equation, depending only on the internal degrees of [leedom.
4.9.1
Separation oj the Motion oj CenteroJMa88
Denote by rk and by mk, k = 1,2, ... , n, the position vectors and the masses of n particles, respectively. The total mass of the nobody quantum system is M = I:k mk and its Schrodingcr equation is 1
n
1 2 ""' L m k.
n 2
Vrk
\]!
+ 1I1l! = Ell!
'
(4.215)
k=l
where \l;k is the Laplace operator with respect to rk, and V is a pair potentia.], depending upon the distance Irj  rkl of each pair of particles . The natural units are used for convenience, Ii = c = l. Replace the position vectors rk by the .Jacobi coordinate vectors R j : n
Ro
= M
1/ 2
L
mkrk,
k= !
(4.216)
n
Mj
=L
mk,
l~j~(nl),
k=j
where Ro describes the position of the center of mass, Rl describes the massweighted separation from the first particle to the center of mass of th e remaining particles, R2 describes the massweighted separat,ion from
Chap. 4 Threedimensional Rotation Group
170
the second particle to the center of mass of the remaining (n  2) particl and so on. An additional factor VM is included in the Jacobi coordin vectors for convenience. The massweighted factors J.Lj in the formulas R j are determined by the condition n
n)
k=)
j=O
(4.21
J.Lj can be calculated one by one from the following schemes. In the cent ofmass frame, if the first (j  1) particles are located at the origin and t last (n  j) particles coincide with each other,
1 :s:: k
< j,
k oj: j,
then, n
k=j
=
R; = J.Lj [Tj + mjTj/MH d
2
= J.Lj [Mj Tj/M j +d 2
,
and
(4.21
The linear relations between Tk and R j are written in a symbolic for
(4.21
Substituting Eq. (4.219) into Eq. (4.217), one obtains that X is a r orthogonal matrix , nI
L
j=O
R;
=
L kt
nl
(mkmd
l
/
2
TkTt
L
j=O
n
XjkX jt
=
L
mkTr
k=1
Then, Tk = m~I/2 Lj RjXjk' and the Laplace operator in Eq. (4.21 and the orbital angular momentum operator L are directly expressed Rj :
§4·9
An Isolated Quantum nbody System n\
n
l:=
mi:
1
\1;k =
k=l
l:=
\1~j'
j=O
n
L
17 1
=
l:= Tk
i
(4.2:20)
n\
X
\1 rk = i
k=l
l:=
Rj
X
\1 Rj
.
)=0
By the separation of variables, one has
(4.221)
The inverse transformation of Eq. (4.216) is _ [Mk+l ] Tk  m M k
1/2
k
Tk  r j = [1Vh+1 mkMk
R
k
_
~
[ mj ] ~ MM j = l ) )+1
]1/2 Rk _
~
i~1
[
1/2
R. J
mi
Mi M HI
+
M 1/ 2 Ro
]1/2 R; _
[
,
Mj mjMj+l
]1/2 R
j .
(4.222) Irk  rjl2 are the functions of R; . R t which are rotational invariant and
independent of Ro. The last equality in Eq. (4.221) is the Schrodinger equation in the centerofmass frame, which is independent of Ro and spherically symmetric. Hereafter, we will neglect the motion of centerofmass by simply assuming Ro = 0 and Pc = 0 in Eq. (4.221) for convenience.
4.9.2
Quantum Twobody System
For a quantum twobody system, there is only one Jacobi coordinate vector R. The eigenfunction of the angular momentum is the spherical harmonic function Y,;(R) where R = Rjr and r = IRI. Since the angular momentum is conserved, the wave function can be separated (see §4.4.3), (4.223) where () and tp are the polar angle and the azimuthal angle of R, respectively. Substituting Eq. (4.223) into the Schrodinger equation (4.221), one obtains the radial equation
172
Chap. 4 Threedimensional Rotation Group
( 4.2
where the property of the spherical harmonic function Y~ (R) is used.
Y,;,(R) is replaced with the harmonic polynomial Y;,(R), the action of
Laplace operator is divided into three parts. The first part is its act on the radial functions rtcPe(r). The second part is its action on Y;,( which is vanishing. The third part is its mixed action
The results are the same, but the spherical coordinates appear in the calculation.
4.9.3
e and
do
tp
Quantum Threebody System
What is the generalization of the basis eigenfunctions of angular mom tum to a quantum threebody system or a quantum multiplebody syste A naive idea is to generalize the spherical harmonic function (x) D;'m,(n,(3,,)*, as was done by Wigner [Wigner (1959)] . However, derivative of the Euler angles leads to the singularity. As shown in the p ceding subsection, in terms of the harmonic polynomials (x) the angu variables are avoided in the derivation of the radial equations. What is generalization of the harmonic polynomials to a threebody system? For a quantum threebody system there are two Jacobi coordinate v tors Rl and R 2 ,
Y,;,
Y,;,
(4.2
§4·9
An Isolated Quantum nbody System
173
The Schrodinger equation (4.221) becomes
where 'VkJ and 'Vk2 are the Laplace operators with respect to the Jacobi coordinate vectors Rl and R 2 , respectively. There are nine degrees of freedom for a threebody system, where three degrees of freedom describe the motion of centerofmass, three degrees of freedom describe the global rotation of the system, and the remaining three degrees of freedom describe the internal motion. The internal variables are denoted by (4.227) which are invariant in the global rotation and in the system. The potential V is the function of the three On one hand, the number of linearly independent mials of degree n with respect to the components of M(n)
space inversion of the internal variables. homogeneous polynoR, and R2 is M (n):
1
= i(n + l)(n + 2)(n + 3)(n + 4)(n + 5). 5.
(4.228)
The independent. basis eigenfunctions of angular momentum do not contain a factor of the interna.l variables, because the factor should be incorporated into the radial functions. The number of the homogeneous polynomiaJs of degree n that do not contain a function of internal variables as a factor is K(n)
= M(n) 
3M(n  2)
+ 3M(n  4) 
M(n  6)
= 4n 2 + 2,
On the other hand , a common eigenfunction of L2 and L z with the eigenvalues £(£ + 1) and /L, which is a homogeneous polynomial of degree n, can be obtained from yn~(R,)Y,~,q(R2) by the ClebschGordan coefficients,
m
where £ = n, n  1, ... , In  2ql. Ye:(n q ) (R" R 2 ) is a homogeneous polynomial of degrees q and n  q with respect to the components of R, and R 2 , respectively. When JL = £ = nand (n  1), one has yq(nq)(R te
. (RJc
R) J,
2
= (_l)e {[(2 q + 1)!(2£ 
2q + 1)!]1/2}
q.'(£_ q )'2 . e+ 2 7r
+ iR 1y )Q(R2x + iR 2y )e",
0 :; q :; £ = n,
Chap. 4 Threedimensional Rotation Group
174
(W~ { (2q + 1)!(2e  2q + 3)!} 1/2 2q( e+ 1) (£.  q + 1) q)!2i+21T} 1 (Rlx + iR 1y )q1 (R2x + iR 2y /q
y,,(n q)(R R) = if
I,
2
{(q  I)!(£. 
l~q~e=n1.
(4.23 It is evident that these expressions do not contain a function of the inter variables as a factor, neither do their partners with smaller f..L owing to spherical symmetry. The number of those eigenfunctions is (2n
+ 1)(n + 1) + (2n
 1)(n  1) = 4n 2
+2=
J((n),
n2::1.
Namely, any of the remaining eigenfunctions Yi~nq) (R 1 , R 2 ) with £ n  1 is a combination of polynomials, each of which contains a factor internal variables. For example,
Y oo11 (R1,R2 ) =  J3 41T6,
YOo22 (R I,R2)
J5 {3~2 2 } 61]2 ,
= 81T
Y2~2(RI,R2) = 5~ {1J2(R lx +iR1y )2 +6(R2x +iR2y )2 36(R1x + iR1y )(R2x + iR2y )} .
The mathematical reason for this can be seen in §9.4.4. Now, any eigenfunction of L2 and Lz with the eigenvalues e a f..L = £, which will be called the eigenfunction of angular moment e for convenience, is a combination of those homogeneous polynomi (R 1, R) h · · coe ffi cte . Y if,,(fq) (R 1, R) 2 an d y,,(£q+1) if 2 were t he com b matIve
are functions of the internal variables. Yi?q+.\) (R 1 , R 2 ) can be replac with a simpler form Q;.\(R1' R 2) by removing a constant factor which c be incorporated into the radial functions [Hsiang and Hsiang (1998)] Q ~.\
(R 1 , R 2 )
=
Xq.\y£qZ'\ (q _ A)! (£ _ q)! '
A~ q
~
e,
A = 0,1,
(4.2
Note that
R)Z Q qfl( R 1 , R2 )  Q(fl)o(R ql 1, 2 .
(4.2
Q~).(RI' R 2 ), called the generalized harmonic polynomial, is a homogeneo polynomial of degrees q and (£  q + A) with respect to the components R j and R 2 , respectively. It is the common eigenfunction of L2, L z , L Lk 2 , \7k l , \7k 2 , \7 RI . \7 R2' and the space inversion with the eigenval
§4.9
An Isolated Quantum nbody System
175
P(t+1), t, q(q+1), (tq+A)(eq+A+1), 0,0 , 0, and (_l)f+", respectively,
L1, (L1J
is the square of the partial angular momentum. Any wave where function with the given angular momentum f. and the parity (_l)f+A can be expanded with respect to Q~).(Rl' R 2 )
e
IJ!t(R1,R2 )
=L
¢~).(~1,6,T)2)Q~A(RJ,R2) '
A
= 0,1.
(4.233)
q=).
That is, for a threebody system, the generalized harmonic polynomials Q;A(RJ' R 2 ) constitute a complete set of basis eigenfunctions with angular momentum e and parity (_l)f+A. Only (e+ 1 A) partial angular momentum states are involved in constructing a function with angular momentum e and parity (_l) f+ ", and the contributions from the infinite number of remaining partial angular momentum states are incorporated into those of the radial functions. When substituting Eq. (4.233) into the Schrodinger equation (4.226), the action of the Laplace operator is divided into three parts. The first part is its action on the radial functions ¢~A (~l ' 6 )T)2) which can be calcula ted by the replacement of variables directly. The second part is its action on the generalized harmonic polynomials Q~A (Rl ' R 2 ) which is vanishing. The third part. is its mixed action 2 {(O~ ,¢;A) 2R,
+2 {(0~2¢~).) Rl
+
(O~A)~A) R 2 }
·
\lR,Q~A
+ (01)2¢~).) 2Rd·
\lR2Q~T,
where o( denotes olo~ . In terms of Eqs. (4.231) one obtains
R 1 . \l R , QfA q

Rl . \lR2Q~).
= (q 

qQf)' Q '
A + 1) Q~~l'
Rz . \l R, Q~A R2 ·\lR2Q~A
= (f.  q + 1) Q;~l' = (eq+A)Q~)..
Thus, the radial equations for the radial functions ¢~). (6,6, T)2) are [Hsiang and Hsiang (1998); Gu et al. (2001b)j: \l2¢~>'
+ 4qoE, ¢~A + 4(e  q + A)07),¢~>' + 2(q  A)8E2¢~~1 + 2(t  q)0~2¢;~1 = 2 (E  V) ¢;",
\l2¢~>'(~1,6,T)2) = {46 0 fl +4T)20~2 +6(oEI +(1)2) +(~l +T)2)OZ2 +46 (o() +(7)2)0~2}1jJ;).(~1,6,T)2)' A ~ q ~ e, A = 0, 1.
(4.234)
Chap. 4 Threedimensional Rotation Group
17(;
4.9.4
Quantum nbody System
For a quantum nbody system, there are (n 1) Jacobi coordinate vecto Arbitrarily choose two Jacobi coordinate vectors, say Rl and R 2 . In bodyfixed frame, RI is parallel to its zaxis, and R2 is located in its plane with a nonnegative xcomponent. Among 3n variables of the body system, three variables describe the motion of centerofmass, th variables describe the global rotation of the system, and the remain (3n  6) variables describe the internal motion. The internal variab cannot be chosen as R j . R k , because the number of R j . Rk is n( n  1 which is larger than (3n  6) when 11. > 4. Further, this set of inter variables is not complete because two configurations, which are related a reflection to the plane spanned by Rl and R 2, are described by the sa internal variables. The complete set of internal variables are
= Rj . R I ,
f,j
~ j ~
1
'f/j = R j
(71.  1),
'f/ l
·
R2,
=6,
(j
(1
= R j . (RI = (2 = 0,
x R2)
,
( 4.2
which are invariant in the global rotation of the system. The number the internal variables is (371.  6), where f,j and T/j have even parity, bu have odd parity. Introduce a set of functions of internal variables ,
nj
= (Rl x R j
) .
(RI x R 2 ) = f,l'f/j  6f,j,
= (R2 X R j ) . (RI x R 2 ) = 6'f/j  'f/2f,j, n1 = W2 = 0, n2 = WJ = (RI x R 2 ) 2 .
Wj
In the bodyfixed frame, due to Eq.
(4.2
(4.235), Rl is (0,0, f,;/2), R2
[ (n2/ f,tl 1/2 , 0, 6f,~ 1/2] , and the components Rjb of R j are
R jx I


nJ (C n <,1
2
)1/2
,
I R jy
/2 = (j~,,1 t2 ,
where 1 ~ j ~ 11.  l. The volume element of the configuration space be calculated from the Jacobi determinant by replacement of variables: nl
IT
j=l
dRjcdRjydRjz
= ~ngnsinj3do:dj3d'Yd6df,2d'f/2
nl
II
df,jd'f/jdC·
j=3
(4.2 The domains of definition of the Eul er angles are given in Eq. (4.6.3). T domains of definition of f,1 and 'f/2 are (0, (0) and the domains of definit of th e remaining variables are (00, (0). Because Eq. (4.222) and
An Isolated Quantum nbody System
§4.9
I Tt
(4.23!l)
the potential V is a function of only the internal variables. Denoting by R(a, (3, "I) the rotation transforming the centerofmass frame to the bodyfixed frame, one has R j = R(a, (3, "I)Rj. R(a, (3, "I) is given in Eq. (4.62), where a, ,8, and "I are the Euler angles and CIY. = cosa, SlY. = sin a, etc. Through a straightforward calculation, one obtains
= R 1:, + l'R ly = <1/2 e iIY. S(3,
v
<,j
II.
Y = R 2x + iR 2y = (D2/~dl/2 e iQ (C(3CY + isy) + 6~;1/2eiO:8.8, R 1z = ~~/2C(3,
R2z =  (D2/~1)1/2 s(3cy
Z = (Rlx + iR 1y ) R 2z R jx + iRjy
= D;l

+ 6~;1/2C(3,
R 1z (R2x + iR2y ) = _D~/2eiIY. (cy + iC(38y),
{WjX + Dj Y  i(jZ} ,
(R jx + iRjy ) Rkz  R jz (Rkx + iRky )
~kTjj) Z}. (4.240) Namely, the components of the Jacobi coordinat.e vectors R j can be expressed as a linear combination of the components of Rl and R 2 , where the coefficients depend only on the internal variables. It is reasonable because Rl and R2 determine the bodyfixed frame completely. Thus, each harmonic polynomial Y/ (Rj) can be expressed as a combination of Q;A(R 1 , R 2 ) wit.h the coefficients depending on the internal variables. This means that the generalized harmonic polynomials Q~A(Rl' R 2 ) given in Eq. (4.231) do constitute a complete set of independent basis eigenfunctions with the given angular momentum for a quantum nbody system, just like they do for a quantum threebody system. A more direct proof for this can be found in [Gu et al. (2001b)1. Therefore, any function iJt f' (R 1 , ... , Rnd with the angular momentum and the parity (_l)e+A in a quantum nbody system can be expanded with respect to the generalized harmonic polynomials Q~T (Rl , R2), where the coefficient.s
= D;l {i (rU(k

7)k(j)
X  i (~j(k

~k(j) Y
+ (~jTjk

e
e
1
iJtfA(Rt, ... ,Rnd
e
= 2:2:
=
(4.241)
Chap. 4 Threedimensional Rotation Group
178
The last equality means that the parity of ¢~;(~,7],() is ( _ l)>T. The expansion (4.241) has two important characteristics, which mak easier to derive the radial equations. One is that the generalized harmo polynomial Q~T(Rl' R 2) is a homogeneous polynomial in the compone of two Jacobi coordinate vectors RJ and R 2 , where the Euler angles do appear explicitly. The other is the well chosen internal variables (4.2 where the internal variables (j have odd parity. It is due to the existe of (j that Q~O(Rl,R2) and Q~1(Rl,R2) appear together in the expans (4.241) of the wave function. When substituting Eq. (4.241) into the Schrodinger equation (4.221
(4.2
The action of the Laplace operator to the function IJij>(R 1 , ... , Rnd separated into three parts. The first is its action on the radial functi ¢~;(~, 7], () which can be calculated by the replacement of variables direc
'V2¢~;(~, 7] , ()
=
{46 0l
1
+ 47]20~2 + (~l + 7]2) Ot2 nJ
+ 46 (oEI + 07)2) 06 + 6 (OEI + 07)2) + L [~jolj + 7]20~j j=3
+ D20~j
+ 260EjOT)j + 4 (~jOEj + (jOeJ 0EI + 4 (7]j07)j + (jOe,) 07)2 + 2 (7]jOEj + ~j07)J OE2]
(4.2
n  j
L
[(Dj7]k  Wj~k + (j(k) (OEjOEk + 07)j 07)J j,k=3  2 (Wj(k  Wk(j) 0Ej oek + 2 (Dj(k  Dk(j) O'lj 0Ck
+ D;l
+ (DjD k +WjWk
+ 6 (j(k +7]2(j(k)OejO(kJ}¢~;(~,7],().
The second is its action on the generalized harmonic polynom Q~T (R J, R 2) which is vanishing because Q~T (RJ, R2) satisfies the Lapl equation. The third is the mixed application
2 { (8".::) 2R, I (a".;;) R, , + (O(J¢~;) (R2
,
X
~ [(a",.:;) R
~ [(a".;;) R,
R j )]) . 'VRIQ;T j
+ (a,,1;;) (R,
+ 2 {(OE2¢~;) RJ + (07)2¢~;) 2R x
Rd] } . 'Vn,Q:'
§4,9
[n terms
An Isolated
body System
(4.231) and (4.240) one obtains
R1 'VH1Q~T = R 2 · VR1Q~T = (P
eo '62 _ n1 vR 1 Q q
, v R2Q~O Q~1
R j 'V
=
{ Wjq'lj Q'O I, Hj n (0{.q l)Q~~1
0;1 {Wj(q
= 0;1
+ I'), Qeo '(' Qn} q1~..,j q
+ OJ (€
{if)2(jq2Q~O
,
 q) Q~o  i(jQ~~l} ,  1)(£  q + l)Q~~1
+ l)qQ~~l + i6(jq(2£  2q +  i6(/i! 1)2Q~~1  WJqQ~~1 + OJ(£  q + l)Q~I} , (R2 X R j )· vHIQ~O = D;l {r/?,()qQ~o ~2(J (€ q + 1) Q~~l  ~w)Q~l}, x R 1 ) . v R2Q~O = D;l ~2(j(q + l)Q~~l + 6(j (€  q) Q~O Rj
.
0;1 {if)2(j(q
\]R2
+ iO)Q~~d, x R j ) '\] RI Q~1 = 0;1
1f)2Wjq2Q~0
+ iE,2Wj(2q
1) (f  q + 1) Q~~
 i~IWj(€  q +
( + 1)Q~~2 + f)2(jqQ~1 q + 1) Q~~d Q&1 = D;l {if) 2 0j(q + l)qQ~~l  i6 D jq (2£ + 1) Q~ + 1)2Q&~1  6(jqQ~~1 + ~l(j (£  q + 1) Q~I
(Rj x R J )
+ i6nJ(i! Now,
radial equations are
nl
+ L 20;1{
 OJ(€  q)OlJj
+ f)2(J
)=3
 q [Wj O')j
+ 6()
 iT}2q(q 1) [(I
 6(J(2£  2q + 1)oru
+
1)0
O.jO(,]
+ (£ 
¢f;J)J  iq
+ T}2WjqaC + 6 Dj(?f 
q)[6(j(2q+1)o~j
(€  q)O(JI ¢t;+l)l  i~l
[E  Vj ¢~8,
q) [OjO~J
e
2q
+ l)ac,]
¢~{
+6wj(2q+l)o(j
Chap. 4 Threedimensional Rotation Group
tHO
D¢;~
+ 4 {qO~1 + (£  q + 1)01)2} ¢g + 2(q 
1)0~2¢1;_1)1
nl
+2(£q)0~A1~+I)1 + L 2D21{[_WjqO~j +Dj(£q+1)o'lj j =3
+rJ2(jqO(j  (q  1)
+ 6(j(€  q + l)O(j] ¢~~ [Wj01)j + ~2(jO(,] ¢1~1)1 + (£  q)
 i [(j01))  Djo(,]
¢~~l)O
i

[Djo~j  6(jo(j] ¢1;+1
[(jO~) + WjO(,] ¢~8}
= 2 [E  V] ¢~~,
(4. where D¢;~ was given in Eq. (4.243). When n = 3, Eq. (4.244) reduce Eq. (4.234), where the radial functions ¢~~(~, rJ, () with .A # T have t vanishing because all internal variables have even parity.
4.10
Exercises
1. Prove the preliminary formula by induction:
1
1
0+ l! [a, OJ + 2! [a, [a, OJ] + ... n
L CXJ
n =O
l~
I [a, [a, ... [a, OJ ... J ], n.
where a and 0 are two matrices with the same dimension. Then, s Eq. (4.38) and prove that SU(2) is homomorphic onto SO(3).
2. Expand R(n, w) = exp (iwn· T) as a sum of matrices with the fi terms. Hint: (n· T)3 = n . T.
3. Check the following formulas in the group 0 [see the notation give §2.5.1] in terms of the homomorphism of SU(2) onto SO(3): TzRl
= 53,
TzT.'C
= R 1,
RIR2
= Rj.
4. Prove the following formulas for the group I in terms of the homom phism of SU(2) onto SO(3) (see the notation given in §2.5.4)
which were used in Prob. 21 of Chap. 3.
Exercises
I
I
5. Calculate the Euler angles for the following transformation mal.r i" 't /,' , 5, and T, and write their representation matrices in Dj of SO(3) :
(a) R(o,;3,,)
(b) 5(0,;3,,)
(c) T(o,;3,,)
=~
Cvf:l 2 V3  2
V3  2 V32
V2
V2
1(v0 + 2/3
=8
2
1 0
2V3
3)2  2
V6 + 2V3
)2  6 2)2
1(/3
=
/3) V2,
2\1'6
1 V3 0
2v0) 2)2 4)2
,
~,)
6. Calculate the Euler angles for the following rotations R, 5, and T, and write their representation matrices in Dj of SO(3). (a) R is a rotation around the direction = e) sin + e3 cos through an acute angle e; (b) 5 is a rotation around the direction n = (e 1 + e2 + e3) / V3 through 27f /3; (c) T is a rotation around the direction = (e) + e2) /)2 through 7f.
n
e
e
n
7. Calculate the Euler angles for the rotations To, T 2 , R), R 2 , R 6 , 5), 52, 56, 5 11 , and 5J2 in the icosahedron group I, where the notations for the elements are given in §2.5.4.
8. Express the representation matrix Dj (n, w) of a rotation around the direction n(e,tp) through w in terms of Dj(e3,0) and dj (;3).
9. Calculate all the matrix entries d!'J1.(w) by Eq. (4.74), where j = 1/2, 1, 3/2,2,5/2, and 3.
10. Reduce the subdured representation from the irreducible representation D3 of SO(3) with respect to the subgroup D3 and find the similarity transformation matrix.
11. Reduce the subduced representations from the irreducible representations D20 and D)S of SO(3) with respect to the subgroup I (the proper symmetry group of the icosahedron), respectively.
Chap. 4 Threedimensional Rotation Group
182
12. Reduce the subduced representations from the irreducible represen tions Dl, D2, and D3 of 80(3) with respect to the subgroup I, resp tively, and calculate the similarity transformation matrices. From results, calculate the polar angles of the axes in the icosahedron. 13. Prove that the elements u(n,w) with the same w form a class of 8U(2) group.
14. Calculate the representation matrices of the generators in an i ducible representation of 8U(2) by the second Lie theorem.
15. For anyoneorder Lie group with the composition function f(r; please try to find a new parameter r' such that the new composi function is the additive function, f'(r'; s') = r' + Sf. The Lore transformation A(v) for the boost along the zaxis with the rela velocity v is taken in the following form. The set of them form oneorder Lie group:
1 0
Av = ( 01
( )
0 0
0) 0
0 0 "f i"{v / c o 0 i"{v/c "f
Find the new parameter with the additive composition function .
16 . Directly calculate the ClebschGordan coefficients for the direct pr uct representation of two irreducible representations of 8U (2) in te of the raising and lowering operators J±: (a) DI/2 x D 1 / 2 , D 1 / 2 x Dl, (c) DI X Dl, (d) DI x D 3 / 2
17. Prove two sets of formulas for the ClebschGordan coefficients in te of the raising and lowering operators J±: (a) In the reduction of D 1 / 2 x Dj,
11(j + 1/2), M) =
. M 1/2)1/2 ( J + 2j +\ 11/2, 1/2)lj, M  1/2)
+ 11(j  1/2), M)
=
1/2)1/2 . M ( J ++ Il/2,1/2)IJ,M+1/2), 2j 1
. M 1/2) 1/2 (J . + 11/2, 1/2)lj, M  1/2)
2J + 1
. M 1/2) 1/2 _ ( J+ . + 11/ 2,1/2)lj,M+1/2). 2J + 1
Exercises
111
(b) In the reduction of Dl x Dj,
1
1( '+1) M)= {(j+M)(j+M+1)}1/2 111 )I' M1) ), 2(2j+I)(j+I) ,),
+ { (j  M + I)(j + M + I)} 1/211 0)1' M) (2j+1)(j+1)
+
II ),' M) =
{
,
),
(j_M)(j_M+I)}1/2 , 11, I)I),M + 1), 2(2j + I)(j + 1)
{(j + M)(j  M + 1) }1/211 1)1' M  1) 2j(j+1) , ), 
M
[j(j + 1)]
1/2 II ,0)lj,M)
_{(jM)(j+M+1)}1/2 11 1)1' M+1) 2j(j+1) ,),'
II(j l),M)
= {(j 
~~~~ ~ ~ + 1) r/211, I)lj,M 
1)
_ {(jM)(j+M)}1/2 11 0)1' M) j(2j+1) ,), +{(j+M)(j+M+1)}1/2 11 1)1' M+1), 2j(2j + 1) ,),
18, Calculate the eigenfunctions of the total spinor angular momentum in
a threeelectron system,
19, The spherical harmonic function Y,;,(ii) belongs to the mth row of the
representation De of SO(3) so that it is the eigenfunction of the orbital angular momentum operator L3 with the eigenvalue m, Calculate the eigenfunction with the eigenvalue m of the orbital angular momentum operator L ' a along the direction a = (el  e2) /12 in terms of combining Y,;(ii) linearly,
20, Let the function 1/J;"(x) belong to the mth row of the irreducible repre
sentation De of SO(3), Calculate the eigenfunction with the eigenvalue m of the orbital angular momentum operator L ' b along the direction b = (V3e2 + e3) /2 in terms of combining 1/;;" (x)* linearly, Hint: Use the similarity transformation between the representation Dj o SO(3) and its complex conjugate representation,
184
Chap. 4 Threedimensional Rotation Group
21. QIt is the rotational transformation operator in the spinor space. 11 the rotation Qn, the basis spinor e(sl(p) belongs to the pth row of till irreducible representation D S , so that it is the common eigenfunction ( l
the spinor angular momentum 52 and 53 with the eigenvalues s(s + J and p, respectively. Based on this property, calculate the eigenfunctiol of the spinor angular momentum S· r along the radial direction, when r is the unit vector in the radial direction. 22. There are three sets of the mutual commutable angular momentulI operators. One set consists of L2, L 3 , 52, and 53. The other sc consists of J2, J3, L2, and 52, and the third set consists of J2, J3, 52 and S . r. Calculate the common eigenfunctions of the three sets o operators, respectively. 23. Calculate {de (e)
(11)2 df(e)l}
,where de(e) is the representatiol mm
11
matrix of R(e2, e) in De of SO(3) and is the third generator in th representation. H int: Use the property of the adjoint representation.
24. Establish the differential equation satisfied by the matrix entrie D~,t(o,fJ,')') of the representation Dj of SO(3). 25. Discuss all inequivalent and irreducible unitary representations of th SO(3) group and the SO(2,1) group.
Chapter 5
SYMMETRY OF CRYSTALS
The study on the symmetry of crystals is a typical example of the physical application of group theory. Through a systematic study by group theory on ly based on the translation symmetry of crystals, the crystals are classified completely that there are 11 proper crystallographic point groups, 32 crystallographic point groups, 7 crystal systems, 14 Bravais lattices, 73 symmorphic space groups, and 230 space groups. In this chapter we will study the symmetry group of crystals, theif representations, and the classification of crystals.
5.1
Symmetric Group of Crystals
The fundamental character of a crystal is the spatial periodic array of the atoms composing the crystal, called the crystal lattice. By the periodic boundary condition, the crystal is invariant in the following translation (see [Ren (2006)] for the crystals of finite size): r t T(f)r = r +f,
(5 .1)
where f is called the vector of crystal lattice. Three fundamental periods of a crystal lattice, which are not coplanar, are taken to be the basis vectors of crystal lattice, or briefly called the lattice bases. The lattice bases are said to be primitive if each vector of crystal lattice is an integral combination of the lattice bases. For simplicity, we only use the primitive lattice bases if without special not.ification. For three chosen lattice bases ai , a vector of crystal lattice is characterized by three integers €;: aj
3
f
= alPI + a2€2 + a3€3 = L ·,=1
ai€i,
€; are integers.
(5.2)
186
Chap. 5 Symmetry of Crystals
The multiplication of two translations is defined to be a translation wh two translation vectors are added. The set of all translations T(.e) whi preserve the crystal invariant forms an Abelian group, called the translati group T of the crystal. Usually, in addition to the translation symmetry, a crystal also reserv invariant under some other symmetric operations composed of the spat inversion, the rotation, and the translation. A general symmetric operati is denoted by g(R, a), r ~ g(R,a)r
= Rr + a,
(5
where R E 0(3) is a proper or improper rotation, and a is a translati vector, not necessary to be a vector of crystal lattice.e. If a = 0, g( R, 0) R is a proper or improper rotation which preserves the origin invariant. R = E, a has to be a vector of crystal lattice .e and g(E ,.e) = T(.e). T multiplication of two symmetric operations is defined as their success applications, g(R,a)g(R',{3)r = g(R,a) {R'r g(R,a)g(R',{3)
+ {3}
= RR'r
= g(RR',a + R(3).
+ a + R{3,
(5
The inverse of g(R, a) is
(5
The set of all symmetric operations g(R, a) for a crystal with the m tiplication rule (5.4) forms a group 5, called the space group of the cryst In the multiplication (5.4) of two symmetric operations, the rotational p obeys the multiplication rule of two rotations, but the translational part affected by the rotational part. Therefore, the set of the rotational parts in g(R, a) forms a group G, called the crystallographic point group. Removing the vector of crystal lattice .e from a, the general symmet operation can be expressed as g(R, a) = T(£)g(R, t), 3
t
=L
(5 ajtj,
j=l
For a given crystal with the space group 5, it is easy to show by reducti to absurdity that t in the symmetric operation g(R, t) depends upon
§5.2 Crystallographic Point Groups
IK 'l
uniquely. In fact, if g(R, i) and g(R, t') are both the symmetric operatiolls of the crystal, g(R, t)lg(R, i') = T( R 1t
+ R1t')
= T(f.),
t'  t = Rf. = f' .
Due t.o the restriction (5.6) on t , t' = t. Generally, R is not an element of S, and G is not a subgroup of S. The space group S is called the symmorphic space group if G is the subgroup of S. Namely, in a symmorphic space group, t in each g(R, t) is vanishing, and any element in S can be expressed as g(R, f.) = T(f.)R. The conjugate element of a translation T(f.) is still a translation, g(R, a)T(f.)g(R, a)l = 9 (E, a
+ R(f. 
R  1a)) = T(Rf.),
Rf. = f'.
(5.7)
Namely, the translation group 7 is the invariant subgroup of the space group S. Due to Eq . (5 .6), the coset of 7 is completely determined by the rotation R. Thus, the quotient group of 7 with respect to S is the crystallographic point group,
G = 7/S.
(5.8)
It will be seen that Eq . (5 .7) is a fundamental constraint for the possible crystallographic point groups, the crystal systems, and the Bravais lattices.
5.2 5.2.1
Crystallographic Point Groups
Elements in a Crystallographic Point Group
In the crystal theory, it is convenient to choose the lattice bases aj to be the basis vectors. The merit for this choice is that, due to Eq. (5.7), any matrix entry Dij (R) of R in the bases is an integer: 3
Raj
=L
aiDij (R)
= f.
(5.9)
i=1
The shortcoming is that aj are generally not orthonormal and D(R) is real but not orthogonal. Let eo be the orthonormal basis vectors in the real threedimensional space. The matrix of the rotation R in the basis vectors eO. is denoted by D(R),
Chap. 5 Symmetry of Crystals
188
3
= 2:=
Rea
ebDba(R).
(5.1
b=1
D(R) is a real orthogonal matrix, relat.ed wit.h D(R) by a real similar transformation X, 3
ai
= 2:=
D(R)
edXdi,
= XI D(R)X.
(5.1
d=l Define another set of basis vectors b j satisfying 3
bj =
2:=
(5.1
(XI) jae",
a= 1
b j is called the basis vector of the reciprocal crystal lattice, or briefly call the reciprocal lattice basis. The matrix form of R in the basis vectors bi 3
Rbi
= 2:= (X l Ld Red = 2:= (X 1 lid eeDcd(R) de
d=l
=;;= {~
(XI)id [D(R)lLe XCj } bj
t.
=
[D(R)lL j b j .
(5.1 A rotation R in the crystal theory is usually expressed in the form a doublevector. A doublevector can be simply understood as two merg vectors, called leftvector and rightvector, respectively. Two vectors c make the usual vector calculation independently. The doublevector fo of a rot.ation R is
ij
ij _
::::
Dij(R)  bi · R·
_ aJ 
(5.1 ::::1
aj . R
. bi ·
Equation (5.14) shows the method how to write the doublevector of The coefficient of bj in identical clement is
R is Raj = Lij Dij(R)ai.
f = 2:=
ajbj
= 2:=
bjaj,
The doublevect.or of t
(5.1
j
and the doublevector of the spatial inversion (J is that multiplied with negative sign. The doublevector of a rotation R(n, w) around the direct.i n through an angle w is (see Prob. 3 of Chap. 5 in [Ma and Gu (2004)]
§5. 2 Crystallographic Point Groups
fl(n,w)
nn +
=
(f  nn) cosw + (f
18
X
n) sinw.
(f  nn)
nn
is a projective operator along the direction n, and the plane perpendicular to n. From Eq. (5.9), Dij(R) are all integers, so is its trace:
= Tr D(R) = ±(1 + 2 cosw)
Tr D(R)
= integer,
(5.16
is that t
(5.17
where the positive sign corresponds to the proper rotation, and the negativ sign to the improper one. Thus, cosw is a half of integer: 0, ±1/2, and ±1 w
= 27rm/N,
N = I, 2, 3, 4, or 6,
OSm
(5.18
The element in a crysta.llographic point group has to be an N fold prope or improper rotation, where N is I, 2, 3,4, or 6. An integral combination K of the reciprocal lattice bases bj is calle the vector of the reciprocal crystal lattice. The inner product of K and is an integer and invariant in the rotation,
K . f = (RK) . (Rl) = integer,
(5.19
namely, RK is also a vector of the reciprocal crystal lattice
RK=K'.
(5.20
Both the crystal lattice and the reciprocal crystal lattice have the sam cryst.allographic point group.
5.2.2
Proper Crystallographic Point Groups
First, if a proper crystallographic point group contains only one prope rotational axis, it is a cyclic group. According to the Schoen flies notations they are eN, N = 1, 2, 3, 4, and 6. They are denoted by N in th international notations. The generator of eN is a rot.ation C N aroun the rotational axis n through an angle 27r / N. The direction n of th rotational axis in eN is usually taken to be the zaxis. The group eN an Abelian group with order N. There are N inequivalent onedimensiona representations
Din (C N
)
= exp (i27rm/ N) ,
OS m < N.
(5.21
Second, if a proper crystallographic point group G contains two or mor than two proper rotational axes, we are going to study the restrict.ion o
l
Chap. 5 Symmetry of Crystals
190
th e number of axes. Denote by nN the number of the N fold proper contained in G, where N = 2, 3, 4, and 6. Except for the identical ele E, there is no common element contained in the cyclic groups with diff rotational axes. Thus, nN are related to the order 9 of G
(
For each Nfold proper axis, the doublevector of the sum of elem in the cyclic group CN is NI
{CN}
=
L
(CN)m,
(
m=O
because its application to n is Nn and its application to a vector per dicular to n is zero. In a set of orthonormal bases including a basis v n, Tr D ( {CN }) = N. Removing the trace of the identical element trace of the sum of the remaining elements in C N is N  3. Thus , the of the sum of all elements in G is
3 + (2  3)n2
+ (3  3)n3 + (4  3)n4 + (6  3)n6
= 3  n2
+ n4 + 3
On the other hand, from the rearrangem ent theorem (Theorem 2.1), f arbitrary vector r one has
5{~ Rr} = {~ Rr},
5 E G.
Since G contains at least two different rotation axes, the vector in the c bracket has to be null. Thus, the sum of all elements in G is vanishin
(
It is another restriction on the number of axes in G .
Since SO(3) ~ SU (2), the calculation of the product of two elem in SO(3) can be replaced with that in SU (2), which is much simpler. method can be used to calculate the product of two rotations along diff axes in G where th e sign in u(n,w) does not matter with the calcula The fundamental calculation formulas are
= 1 cos(w/2)  i (0". n) sin(w/2), cos(w/2) = ~Tr {u(n,w)},
u(n,w)
(0" . nd (0" . n2)
= 1 (nl
. n2)
+ iO"
. (n[ x n2),
(
§5. 2 Crystallographic Point Groups
191
where cos(w/2) can take only the following values
cos(w/2)
= 0,
1 ±2'
±v'3
±l.
2 '
(5.26)
Third, if G contains only 2fold proper axes, one obtains n2 = 3 from Eq. (5.24). The matrix of a 2fold proper rotation in SU(2) is u(n,rr)= i( a·n).
T he product of two 2fold proper rotations along different axes is (5.27) Thus, n] . n2 = 0, and n3 = n] x n2. The three 2fold axes are orthogonal to each other. The group G is D 2 , which is an Abelian group and isomorphic onto the inversion group V4. Its group table is given in Table 2.4 and its char acter table is given in Table 3.6. Fourth, if G contains, in addition to the 2fold axes, only one N fold proper axis where N is larger than 2, then the Nfold axis is called the principal axis and taken to be th e zaxis. The remaining elements in G are 2fold proper rotations with the axes perpendicular to the principal axis. Due to the restriction (5.24) , n2 = N. The product of two 2fold rotations has to be an N fold rotation, W = 2mrr / N, and due to Eq. (5.27) nl . n2 = cos(mrr/N), namely, N 2fold axes welldistribute in the x y plane, where the angle of two neighbored axes is rr / N. This group G is j us t D N, where N = 3, 4, or 6. The D N group contains 2N elements 'whose multiplication rule is given in Eq. (2.13). Its character table is given in Tables 3.7, 3.12, and 3.13. DN can be expressed as a product of two subgroups, CNCS, where is the cyclic group of 2fold proper rotations along th e xax is. D N is denoted by N2' in the international notations. Fifth, G contains only 2fold and 3fold proper axes. The matrix of 3fold rotation in SU(2) is
C;
u(n,±2rr/3)
= 1/2=Fi(a·n)V3/2.
For the product of two 3fold rotations with different axes, the cosine of half of its rot ational angle is 1 "2Tr{u(nl,±2rr/3)u(n2 , 2rr/3)}
Due to Eq. (5.26),
n] . n2
n] . n2 =
= 1/4=F3(n]
·n2)/4.
(5.28)
= ±1/3. vVithout loss of generality, one has
cosf) = 1/3 ,
(5.29)
192
Chap. 5 Symmetr'Y of Crystals
Thus, from the products of 3fold rotations one obtains four 3fold axes welldistributed in th e threedimensional space, n3 = 4. The angl e of two neighbored 3fold axes is e. Due to Eq. (5.24), n2 = 3. Each 2fold axis has to be along the angular bisector of two neighbored 3fold axes. Since
= icr . (V3nl  V3n2 + 3nl + Jnl x n2) / 4 = 1/V3,
'U(nl'  21f /3)'U(n2' 21f /3) nl . (V3nJ  V3n2
x n2)
/4,
the cosine of the angle between the neighbored 3fold axis and 2fold axis is 1/V3 Three 2fold axes are equivalent to each other, and four 3fold axes all are polar. This group G is the proper symmetric group T of a tetrahedron. Its group table and character table are given in Tables 2.9 and 3.8. T can be expressed as a product of threp sllbgroups, C~C2C~. T is denoted by 3'22' in the international notations. At las t , if G contains more than one 6fold axes, the angle between two 6fold axes has to be 109°28' , because a 6fold axis is also a 3fold axis. On the oth er hand , since a 6fold axis is also a 2fold axis, the product of two 2fold rotations with this angle cannot satisfy the condition (5.26) . Therefore, the only remaining case is that G contains more than one 4fold proper axes . The matrices of 4fold rotations in SU (2) are
u(n, ±1f/2)
= 1/V2 ~ i
(cr·
n)/V2,
u(n,1f)
= i (cr· 'n)
.
For a product of two 4fold rotations along different axes, the cosine of half of its rotational angle is Tr {u('nl,C::7r/2)'U(n2,7r/2)} /2 Tr {1t(nl' 7r)'U(nz, 7r/2)} /2
=
= 1/2 ~
(nl' n2)
(nl' n2)
/2,
/V2.
The only solution satisfying the condition (5.26) is III . i1. 2 = 0, namely, there are three 4fold axes perpendicular to each other. They are nonpolar and equivalent to each other. In the products of two 4fold rotations , there are 3fold rotations and 2fold rotations, u(nl,7r/2)u(n2,7r/2)
= 1/2 
icr· (nl
+ n2 + nl
x n2)
/2 ,
+ n2 + nl x n2) /V3 = 1/V3, n(nl' 7r)n(n2' 7r /2) = icr . (n] + nl x n2) /V2. i1. l . (nl
The rosin e of the angle between the neighbored 3fold axis and 4fold axis is 1/ vI Four 3fold axes welldistribute around a 4fold axis They are nonpolar and equivalent to each other. Since n3 = 4 and nq = 3, from Eq . (5.24), one has n 2 = 6. Each 2fold axis is along the angular bisector of two
193
§5. 2 Crystallographic Point G'r oups
neighbored 4fold axes and perpendicular to the third 4fold axis. Six 2fold axes arc equivalent to each other. This group G is the proper symm(~ tric group 0 of a cube. Its group table is given in Tables 2.9 and 2.11. Its character table is given in Table 3.9. 0 can be expressed as a product of three subgroups C; C 4 C~. 0 is denoted by 3'42" in the international notations. Table 5.1
The proper crystallographic point group n, N
e Cl C2 C3 C4 C6 Ih
D3 04 D6 T
0
9 1 2 3 4 6 4 6 8 12 12 24
9c
2
3
4
6
1
2 3 4 6 4
3 5 6 4 5
No. of re p. Id 2d 3d 1
1 1 1 1
3 3 4
1 1
6
3 6
I
4 4
3
2 3 4 6 4 2 4
1 1
4
2
3 2
1
1 2
Product of s u bgroups
Gencratol'S
Cl C2 C3 C4 C6 C2 C2
C1 C2 C3 C4 C6 C2, C~
C3C~ C4 C~
C4, C;
C6 C2
C6,C~
C2 C3 C2 C3 C4 ,D2 C6,03
C~C2C2
C~,C2 C~ , C4
T
C~C4C~
c 3 ,q
H
CI
There are 11 proper crystallographic point groups G listed in Table 5.1, where g is the order of G, gc is the number of classes in G, nN is the number of Nfold proper ax es contained in G, H is the invariant subgroup wit h index 2 in G, "No . of rep." denotes the number of the inequivalent irreducible representa tions of the given dimension, and G can be expressed as the product of subgroups, which is related to the symbol of G in the int.ernational notations. In the Lheory of crystals, an Nfold rotation along the zaxis is denot.ed by CN , the remaining N fold ro t ation with N > 2 by C;v, a 2fold rotation along the .Iaxis is denoted by C~ , and the remaining 2fo ld rotation by C;~.
5.2.3
Improper Crystallographic Point Group
In §2.6, we discussed the general method for finding the improper point groups from a proper point group, and introduce the meaning of the subsc ripts in the Schoenflies notations. There are two types of improper point groups. An improp er crystallographic point group G' of Itype is a direct product of a proper crystallographic point group G and the inversion group V 2 , composed by the identical el ement E and the spatial inversion ()'. There
194
Chap. 5 Symmel.ry of Crystals
are 11 improper crystallographic point groups of I type,
Ci , C 4h == C4 ® Ci , D3d
== D3 ® Ci ,
Th==T ® C i ,
C 2h == Cz ® Ci ,
C3 , == C 3 ® Ci ,
C 6h == C6 ® C" D4h == D4 ® C i ,
D2h == Dz ® Ci , D6h
(5.30)
== D6 ® C i ,
Oh == O ®C i ·
If a proper crystallographic point group G contains an invariant subgroup H with index 2, one can construct an improper crystallographic point group G' of Ptype by multiplying the elements in the coset of H in G with IJ and remaining H invariant. G' is isomorphic onto G. From Table 5.1, there are 10 improper crystallographic point groups of Ptype (5.31) The property of 32 crystallographic point groups are listed in Table 5.2.
Table 5.2
Crystallographic point groups
Proper crystallographic point groups Product of subgroups C H
C1 C2 C3 C4 C6 D2 D3 D4
C4C~
D6
C6C~
T 0
C1 C2 C3 C4 C6 C2C~
C3 C,
C3C2C~ C3C 4 C~
Improper crystallographic point groups Itype Ptype Product Product G' G'
C1
Cs
Cs
C2 C3 C2 C3 C4 D2 C6 D3
S4 C3h C2v C3v C4v D2d C6v D3h
S4 C3h
C3hC'2
T
Td
C3S4C~
C2C~ C3C~ C4C~ S4C~ C6C~
Ci C2h C3i C4h C6h D2h D3d D4h
Ci CiC2 C3i CiC4 CiC6 CiC2C~ C3iC~ CiC4C~
D6h
C;C6C~
Th Oh
C3,C2C~ C;C4C~
An improper crystallographic point group G' of Ptype is isomorphic onto the corresponding proper crystallographic point group G. They have the same character table and the irreducible representations. An improper crystallographic point group G' of Itype is a direct product of the corresponding proper crystallographic point group G and the inversion group V2· The irreducible representation of G' is the direct product of two irreducible representations of G and V 2, respectively.
§5.3 Crystal Systems and Bravais Lattice
5.3
195
Crystal Systems and Bravais Lattice
In the previous sections we studied the translation symmetry of a crystal and obtained the constraint (5.7), which restricts the possible rotation R in the symmetric operation g(R, a). The conclusion is made that there are only 32 crystallographic point groups. The constraint (5.7) also restricts the choice of the basis vectors ai of crystal lattice for each given crystallographic point group. In this section the lattice bases ai are not assumed to be primiti ve. The crystals are classified into seven crystal systems based on the restriction to the choice of ai. For each crystal system we will discuss the possibility of the vectors of crystal lattice f which are the fractional com binations of the lattice bases ai. A crystal system contains different Bravais Lattices and the symmorphic space groups based on the vectors of crystal lattice f. There are 14 Bravais Lattices and 73 symmorphic space groups, where the crystallographic point group is a subgroup of the space group. The general space groups will be discussed in the next section.
5.3 .1
Restrictions on Vectors of Crystal Lattice
T he conditions (5.7) and (5.20) give the restriction on the lattice basis ai and the reciprocal lattice basis bj when the crystallographic point group G is given. Obviously, the restriction for some crystallographic point groups may be the same. For instance, the spatial inversion (J makes no restriction on the lattice basis so that the restriction for the improper crystallographic point groups, both Ptype and Itype, are the same as that for the corresponding proper crystallographic point group. It will be shown that except for N = 2, the restriction for C N is the same as that for D N, and the restriction for T is the same as that for O. Based on the restrictions, the crystals are classified into seven crystal systems listed as follows, where the crystallographic point groups are also listed. (a) (b) (c) (d) (e) (f) (g)
Triclinic crystal system, C] and Ci . Monoclinic crystal system, C 2 , C s , and C 2h . Orthorhombic crystal system, D 2 , C 2v , and D 2h . Trigonal crystal system, C 3 , C3i , D 3 , C3v , and D 3d . Hexagonal crystal system, C6 , C3h , C6h , D 6 , C6v , D 3h , and D 6h . Tetragonal crystal system, C 4 , 54, C4h , D4 , C4v , D 2d , and D4h . Cubic crystal system, T, T h , 0, T d , and 0 10 .
In the study, one finds that for some cases in the trigonal crystal system
Chap. 5 Symmetry of Crystals
196
the restriction on ai is the same as that in the hexagonal crystal system , an those cases are merged into the hexagonal crystal system. The remainin cases in the trigonal crystal sys tem are called the rhomboh edral crysta system. For each crystal system, the lattice bases a i are chosen in a given wa according to the restriction, and then , the possibility of the vectors of crys tal latti ce f which are the fractional combinations of ai is studied,
0S;J;<1.
(5.32
Denote by T(f) == T(h ,h,h) the translation operator with f. The ral cul at ion shmvs t.hat fi can only be or 1/2. The set of the translation where the translation vectors are the illLegral rombinations of a i forms a invari ant subgroup Te of the translation group T. The cosets of Te a r generated by T(f) and its powers. The crystal system is divided into a few Bravais lattices, based on the cosets as follows: (a) Primitive translation gro up, denoted by P:
°
T == Te.
(5.33
The primitive translation group for the rhombohedral crystal system denoted by R. (b) Bodycentered translation group, denoted by I:
T== Te {E, T (~, ~,D }. @
(5 .34
(c) Basecentered translation group, denoted by A, B, and C:
A.:
B:
C:
T == Te 0 { E , T
(o, ~,
D},
T: Te @{E' T q,~m, T  Te E, T C ,2' 0) }.
(5.35
@ {
(d) Facecentered translation group , den oted by F:
T == Te @{E, T
(o,~,
D 0,0, n (~ , ~,o) }. ,T
,T
(5.36
Obviously, if two of A, B, and C are the symmetric subgroups of the crysta simultaneo usly, its tran slat ion group is t.he face cen tered translation grou F. Combining the translation group with the crystal system, one obtain 14 Bravais lattices. A cryst allographic point group headed by the symbol o
§5.3 Crystal Systems and Bravais Lattice
197
the Bravais lattice denotes the symmorphic space group. In the follow ing, we are going to count the 73 symmorphic space groups. There are a few notationsystems for the space groups S. The Schoenflies notations are convenient for the crystallographic point groups, but not vcry convenient for the space groups. Through some improvement there are international notations for the point groups, the Mauguin  Hermann notations and the international notations for the space groups. Their comparison is listed in Table 5.3. We recommend the international notations for the space groups, where a proper cyclic subgroup eN is denoted by a number N, an improper one by a number with a bar N, and the inversion group V 2 (C i ) is denoted by ±. N (or N) stands for an Nfold cyclic subgroup along a3 . N' (or N') with N > 2 stands for the remaining Nfold cyclic subgroup. 2' (or 2') stands for a 2fold cyclic subgroup along al. 2/1 (or 2/1) stands for the remaining 2fold cyclic subgroup. We will explain the phenomena later that one crystallographic point group (e.g., D 3 ) corresponds to two notations in Table 5.3. Table 5.3
Comparison between notations for space groups.
Sch. = the Schoenflies notations, INPG = the international notations for point groups, MPN = the Mauguin  Hermann notations, INSG = the international notations for space groups. Seh.
INPG 1
MHN
INSG
Seh.
C]
I
1
C 3v
Ci
T
T
T
03d
C2 Cs Cn C3 C3i C4
2 m
2
2
03d
m 2/ m 3 3
2
04
2/m 3 3 4
±2 3 3
C 4v 02d
4
4
04h
S4
4
4
4
06
C 4h
4/ m 6 6 6/m 2'22 2mm mmm 32 32 3m
4/m
±4 6 6 ±6 22' 22' ± 22'
C6 v
C6 C 3h C 6h
O2 C2 v 02 h
D3 D3 C 3v
6 6 6/m 222 2mm 1..1..1.. Tn 171
m
321 312 3m1
02d
03h 03h 06h
T Th 0
32'
Td
32"
Oh
32'
INPG 3m 3m 3m 422 4mm 42m '12m 4/ mmm 622 6mm 6m2 6m2 6/ mmm 23 m3 432 43m m3m
MHN 31m 31.. 1 312m 422 4mm 42m 4m2 .i.1..1.. 1n m m 622 6mm 62m 6m2 ~1.. 1.. m Tn m
23 1.. 3 m 432 '13m .i.3 1.. nt
til.
INSG 32 32' 32" 42' 42'
42' 42" ±4 2' 62' 62' 62' 62" ± 62' 3'22' 3'22' 3'42" 3'42" 3'42"
198
Chap. 5 Symmetry of Crystals
Except for the trivial point group C l , there are vectors of both crys tal lattice and reciprocal crystal lattice along each rotational axis, and a least two noncollinear vectors of lattice bases and those of reciprocal crys tal lattice at the plane perpendicular to a rotational axis. In fact, eac crystallographic point group G, except for C l , contains C 2 , C s , or C 3 as subgroup. HG contains C2 , i+C2 i is parallel to the 2fold axis, and iC2 is perpendicular to it. Find another vector of crystal lattice i' outside o the plane spanned by i ± C2 i. i'  C2 i' is another vector of crystal lattic at the plane perpendicular to the 2fold axis. The situation is similar whe G contains C s . If G contains C3 , i + C3 i + Ci i is along the 3fold axi and i  C3 i and i  Cii are two noncollinear vectors of crystal lattice a the plane perpendicular to the 3fold axis. Since the crystallographic poin group for the reciprocal crystal lattice is the same as that for the crysta lattice, the situation is also the same. In the following the length of ai is denoted by ai and the angle betwee al and a2 is denoted by 0:3 and its cyclic.
5.3.2
Triclinic Crystal System
The crystallographic point groups in the triclinic crystal system are C j (1 and C i (I), where the number in the bracket is the symbol of internationa notations. The matrices of the identical element E and the spatial inversio (J" are constant ones, which make no restriction on the choice of the lattic bases. Choose three vectors of crystal lattice, which are not coplanar, t be the primitive lattice bases. Thus, there are one Bravais lattice P an two symmorphic space groups PI and pI in the triclinic crysti11 system.
5.3.3
Monoclinic Crystal System
The crystallographic point groups in the monoclinic crystal system are C (2), Cs (2), and C2h (±2). The shortest vector of crystal lattice along th 2fold axis is chosen to be a3. In the plane perpendicular to a3, choos two smallest vectors of crystal lattice to be al and a2 such that at x a is along the positive direction of a3, where "two smallest vectors of crysta lattice in a plane" means that each vector of crystal lattice in the plan is an integral combination of those two smallest vectors of crystal lattice The length al and a2 may not be the shortest among the vectors of crysta lattice in the plane. Thus, in the monoclinic crystal system the lengths o
§S.3 CT1.jstal Systems and Bravais Lattice
199
the lattice bases have no restriction, and the restrictions on the angles are O'j
=
0'2
=
(5.37)
K/2.
From Eq. (5.14) the doublevector and the matrix of the generator C 2 in the set of lattice bases are calculated as 1
0 0)
D(C2 ) = ( 0 1 0 o 0 1 Discuss the possibility of the vector of crystal lattice tional components
o ::; If
(5.38)
f with the frac
fJ < 1.
(5.39)
f is a vector of crystal lattice, the following are ones, too f + Z· f =
a3 (2!J) ,
Hence, three fj all can be 0 or 1/2. However, when h = 0, h = 12 = 0 because al and a2 are two smallest vectors of crystal lattice in the plane perpendicular to a3. When h = 12 = 0, h = 0 because a3 is the shortest vector of crystal lattice along the 2fold axis. Therefore, there are four possibili t ies for f , f = 0 (Ptype), f = (a2 + a3) /2 (Atype), f = (al + a3) /2 (Btype), and f = (aj + a2 + a3) /2 (Itype). Since Ctype translation group does not exist in the crystal system, those of Atype and Btype cannot exist for one crystal, neither can Ftype one. The Btype tram;lation group becomes the Atype one if a2 is chosen to be new aj and  a] to be new a2. Similarly, the Itype translation group becomes the Atype one if a] + a2 is chosen to be new al and a2 remains invariant. Thus, there are two Bravais lattices P and A, and six symmorphic space groups P2, P2, P ± 2, A2, A2, and A ± 2 in the monoclinic crystal system. 5.3 .4
Orthorhombic Crystal System
T he crystallographic point groups in the orthorhombic crystal system are D2 (22') , C2v (22'), and D2h (±22'). The point groups all contain three per pendicular 2fold axes (proper or improper). The shortest vector of crystal lattice along the one 2fold proper axis is chosen to be a3, an d those along the remaining two 2fold axes to be at and a2 such that al x a2 is
200
Chap. 5 S ymmetry oj Cry stai s
along the positive direction of a3. Thus , in the orthorhombi c crystal system the lengths of the lattice bases have no restriction , and the a ngles are (5.40 ) In the set of lattice bases the doublevector and t he matrix of the 2fold rotation along a3 are given in Eq. (5.38), and those along aj are calculated from Eq. (5.14)
D(C~) = (~~l ~ o
If
f
0 1
)
given in Eq. (5.39) is a vector of cryst al lattice, ~
~
(5.41)
f + Ii . f
also
~
is the one where Ii is 2, 2' or 2· 2'. Hence, three fj can be 0 and 1/ 2, respectively. Thus, for the orthorhombic crystal system, in addition to the Ptype, Atype, and Btype translation groups, there are Ctype and Ftype translation groups. For the point groups D2 and D 2h , A, B, and Ctype translation groups are the same. For the point group C2v , the 2fold axis along a.} is proper, but two 2fold axes along aj and a2 are improper so that A and Btype translation groups are the same, but Ctype is different. Those different space groups are usually classified into the same Bravais lattices. Thus , there are four Bravais lattices P, C (A), 1, and F, and 13 symmorphic space groups P22', P22', P ± 22', C22', C22', A22', C ± 22') 122') 122' ) 1 ± 22' , F22', F22') a nd F ± 22' in the orthorhombic crystal system.
5.3.5
Trigonal and Hexagonal Crystal System
The crystallographi c point groups in the hexagonal crystal system are C6 (6), C3h (6), C6h (± 6), D6 (62'), C6v (62\ D3h (62'), and D6h (±62'). The shortest vector of crystal la ttice along the 6fold axis is chosen to be a3 ' For the cases with th e point groups C 6 , C3h , and C6h , the shortest vector of crystal latti ce in the plane perpendicular to a3 is chosen to be aj. For the cases wi t h the point groups D 6 , C6v , D 3h ) and D 6h , the shortest vector of crystal lattice along the 2fold axes in the plane perpendicular to a 3 is chosen to be a j . Then, a 2 = C~al' Thus, 0:1
=
0:2
= 7r / 2,
0:3
= 27f / 3.
(5 .4 2)
§5.3 CTyslal Syslems and Bravais Lattice
201
For the cases with the point group D 3h , two space groups whether al is along a proper or an improper 2fold axis are different. From Eq. (5.14), the doublevectors and the matrices of the generators C6 , C~, and C~' in the set of latti ce bases are
(5 .43 )
If
f
given in Eq. (5.39) is a vector of crystal lattice,
f + (~3 . f = 2 h a3, f +
(~2 . f
f 
6·
f
+
(~4 . f
= hal + (/2
= 3 h a3 ,
(5.44 )
 h) a 2,
are all vectors of crystal latt ice. From the first two conditions in Eq. (5.44), For the cases with the point groups C 6 , C 3h , and C6 h , h = 12 = 0 because the len gth of the vector in the third formula of Eq. (5.44) is 0 or less than al. For the cases with the point groups D6 , C 6v , D 3h , and D 6h ,
h = O.
f + 2' . f = (2h  h) a], f+
U·2~3.§)
(5.45) ·f=(2hh)a2,
are also the vectors of crystal lattice. Thus, 2i]  12 and 2/2  it are 0 or 1, respectively. They cannot be 1 at the same time because both hand h are less than 1. There are three solutions to Eq. (5.45):
it = 12 = 0,
il = 212 ==
2/3 ,
2h
= 12 = 2/3.
(5.46)
However, both vectors f == (2 al + a2) /3 and l' = (al + 2a2) /3 are a long a 2fold axis with the length less than aj , so that they are ruled out. Thus, t here is one Bravais lattice P and eight symmorphic space groups P6, P6, P ± 6, P62' , P62' , P62', P62/1, and P ± 62' in the hexagonal crysta l system. The crystallographic point groups in th e trigonal crystal system are C 3 (3), C 3 , (3), D3 (32'), C 3v (32'), and D3d (32'). The shortest vector of
202
Chap. 5 Symmetry of Crystals
crystal lattice along the 3fold axis is chosen to be a3. For the cases wi the point groups C 3 and C 3i , the shortest vector of crystal lattice in t plane perpendicular to a3 is chosen to be al. For the cases with the poi groups D3 , C 3v , and D 3d , the shortest vector of crystal lattice along t 2fold axes is chosen to be al. Then, a2 = C3a1. Thus, the conditio (5.42) is satisfied. The doublevectors and the matrices of the generato C3 = Cl and C~ in the set of lattice bases are given in Eq. (5.43).
If
I
given in Eq. (5.39) is a vector of crystal lattice, 3ha3 is also a vector of crystal lattice. Thus ,
h = 0,
 
1+:3· I +:3.:3. I
(5.4
1/3, or 2/3.
For the cases wi th the point groups D3 , C 3v , and D 3d , one also obtai Eqs. (5.45) and (5.46). When h = 12 = 0, h has to be 0. For the cases with the point groups C3 and C3i , the followin g vector also a vector of crystal lattice
1 :3. I = (h + h) aj + (212  fd
a2·
After removing the integral part, if (h + h) and (212  fd both are n equal to 0, the length of the vector is less than aI, which is in conflict wi the assumption. If h + 12 = 0, one has the first solution in Eq. (5.46), an if h + 12 = 1 and (212  fd = or 1, one has the remaining solutions Eq. (5.46) . From Eqs. (5.46) and (5.47), there are four possibilities fur I in t trigonal crystal system, where f' is calculated from 2/,
°
1 =0, (b) 1= (2al + a2) /3, and II = (a] + 2a2) /3, (c) 1= (2al + a2 + a3) /3, and f' = (aj + 2a2 + 2a3) /3, (d) 1= (aJ + 2a2 + a3) /3, and f' = (2aj + a2 + 2a3) /3. (a)
( 5.4
The length of t.he vector I in (b) is less than aj. It is not a vector of cryst lattice for the cases with the point groups C 3 and C3i . For the cases wi the point groups D 3 , C 3v , and D 3d , a set of new lattice bases can be defin as follows such that. new lattice bases are primitive,
§5. 3 Crystal Systems and Bravais Lattice
(2a]
+ a2) /3,
a~ = (al
+ a2) /3
a~ =
a'1
a~
203
=
:3 . a~,
a] = a~  a~,
Fig. 5.1
New lattice bases
(5.49)
a2
= a~ + 2aS,
f'
= a~
+ a~.
However, in these cases new lattice bases a~ and a~ are not along a 2fold axis, but along the angular bisector between two neighbored 2fold axes (a direction wi~h an angle n/6 to a 2fold axis) . ..Now, the 2fold rotation is denoted by 2" given in Eq. (5.43) instead of 2'. Two space groups are different, whether al is along a 2fold axis or not. This Bravais lattice is merged into the Ptype Bravais lattices in the hexagonal crystal system. In this case there are eight symmorphic space groups: P3, P3, P32', P32", P32', P32", P32', and P32". The length of a] + a2 is the same as a). a] + a2 is along a 2fold axis for the cases with the point groups D 3 , C3v , and D 3d . Thus, for the cases of all five crystallographic point groups, a] + a2 can also be chosen as a lattice basis instead of a] such that the case (d) becomes the case (c). For the case (c), one defines new lattice bases, which are primitive
a~ a~ a~
= (2a] + a2 + a3) /3 =:3. a~, = (at + a2 + a3) /3 = :3. a't, = (at  2a2 + a3) /3 = :3. a~,
(5.50) a3
= a~ + a~ + a~.
From Eq. (5.14), the doublevectors and the matrices of the generators C 3 and C~ in the new lattice bases are
D(C 3 )
=
(01 00 1)0 , 010
D(q) =
(~l ~1 ~ o
0 1
)
(5.51)
It is called the Rtype (rhombohedral) Bravais lattice where (5.52) T here are five symmorphic space groups belonging to the Rtype Bravais lattice: R3, R3, R32', R32', and R32'.
l
204
Chap. 5 SymmeLTY of CTystai s
5.3.6
Tetragonal Crystal System
The crystallographic point groups in the tetragonal crystal system are (4), S4 (4), C~h (±4), D4 (42'), C 4v (42'), D 2d (42'), and D4h (±42'). T shortest vector of crystal lattice along the 4fold axis is chosen to be a For the cases with the point groups C4 , S4, and C4h , the shortest vec of crystal lattice in the plane perpendicular to a3 is chosen to be al. F the cases wi th the point groups D 4 , C 4v , D 2d , and D4.h, the shortest vec of crystal lattice along the 2fold axes in the plane perpendicular to a3 chosen to be al· Then, a2 = C4al. Thus,
(5.5
For the cases with the point group D 2d , two space groups whether al along a proper or an improper 2fold axis are different. From Eq. (5.14), the doublevectors and the matrices of the generato C 4 , C~, and C~' in the set of lattice bases are D(C4 )
=
1 ~) , (~0 o0 1
D(C~) = (6o ~10 1~ )
,
(5.5
D(C~') = (~o 60 1~ ) . If f given in Eq. (5.39) is a vector of crystal lattice, is also a vector of crystal lattice, so that
h = 0,
f + 2· f = 2h
(5.5
or 1/2.
For the cases with the point groups D 4 , C4v , D2d , and D4h ,
f + 2' . f f +
= 2h a),
{~. i' . (~3}
f + 2" . f
= (11
.f
= 2h a 2,
+ h) (a t + a2)
,
are also the vectors of crystal latti ce. Thus, h = 0 or 1/2, 12 = 0 or 1 but It + 12 =1= 1/2 because the vector of crystal lattice (a) + a2) /2 is alo a 2fold axis and with the length less than aI, namely, h = 12 = 0 or 1
~53
205
Crystal Syst em s and Bravais Lattice
Note that h = 0 when it = 12 = O. Conversely, fr = 12 = 0 when Therefore, there are only Ptype and Itype Bravais lattices. For the cases with the point groups C4 , S4, and C4/LJ
f  4· f =
aj (it
+ h) + a2 (12 
fl)
== 1',
h = O.
(5.56)
f'  4· l' = al (212) + a2 (2fJ) , are also the vectors of crystal lattice. If h = 12, they have to be 0 or 1/2. Since h = 0 when h = 12 = 0, and h = 12 = 0 when h = 0, there are only Ptype and Itype Bravais lattices. \Ve are going to show that it i 12 should be ruled out, otherwise, due to Eq. (5.56), there exists a vector of crystal lattice in the plane perpendicular to a3 such that its l<ength is less t han al. In fact, if one of ft and 12 is vanishing, the other has to be 1/2 such that the length of l' is less than al· If ft + 12 :::: 1,
Namely, the length of l' a nd 212 < l. Then, if fr
 al is less + 12 < 1,
than
al.
For the same reason,
2h < 1
1f'12 = af {(it + 12)2 + (12  it)2} = ai {2f[ + 2fi} < ai· Thus, there are two Bravais lattices P and I, and 16 symmorphic space grou ps P4 , P4, P ± 4, P42', P42', P42', P42//, P ± 42', I4, 14, I ± 4, I 42', 142', 142', 142//, and I ± 42' in the tetragonal crystal system.
5.3.7
Cubic Crystal System
T he crystallographic point groups in the cubi c crystal system are T (3'22'), T it (3'22'), 0 (3'42//), T d (3'42//), and 0" (342//). The shortest vectors of crystal lattice aloni"\ the three 2fold 3.v'{es (for T and Til) or along the three 4folcl axes (for 0, T d, and 0 h) are chosen to be the lattice bases, where al x a 2 is along the positive direction of a3. Thus, (5.57) In the se t of lattice bases, the doublevectors and the matrices of C 4 along a;J , q along ai, and q' along (a] + a2) arC' given in Eq. (5.54), and those of OJ along the direction (at + a2 + a3) are given in Eq. (5.51). If f given in Eq. (5.39) is a vector of crystal lattice, f + C 2 f is also a vector of crysta l lat.tice', and then, h = 0 or 1/2. Since three 4fold
Chap. 5 Symmetry of Crystals
206
axes are equivalent, It and 12 are also equal to 0 or 1/2, and the Brava lattices A, B, and C are merged to the Ftype Bravais lattice. Thus, the are three Bravais latti ces P, I, and F , and 15 symmorphic space grou P3'22' P3'22' P3'42/1 P3'4 2/1 P3'42/1 13'22' 13'22' 13'42/1 13'4 2 13'42/1, F3'22' , F3'22', F3'42/1 , F3'4 2/1, and F3'42/1 in the cubic crys system. Seven crystal systems, 14 Bravais lattices and 73 symmorphic spa groups as well as the relations between the lattice bases and the rotation axes are listed in Table 5.4. The matrices of the generator are listed in t note. )
)
Table 5.4 Crystal system, Bravais Triclinic, P PI, PI Monoclinic, P, A, no. = 6 P2, P2, P ± 2, A2, A2, A ± 2 Orthorhombic, P, C(A), [ , F no. = 13 see Note (e) Tetragonal, P,1 no. = 16 P4, P4, P ± 4, P42', P42', P42', P42/1, P ± tl2',14, [4, f ± 4, 142', [4 2', [42', 142" , [ ± 42' Cubic,
)
[
,F
no. = 15 see Note (f) Rhombohedral, R no. = 5 R3, R:3, R32', R 32' , R:32'
,
)
)
)
The property of symmorphic space groups Point groups Sch INSG C, 1 Ci I
Prod uc t of subgroups Ct Ci
Lattice bases
Direction of rotational axes
a3
C2 Cs
Cn D2 C lv D2h
2 2 ±2 22' 22' ±22'
C2
(Xl
Cs
= (X2 = 7r /2
2 2 ±2
CiC2 C2C~
(Xl
= (X2
=(X3
C2C~ CiC2C ~
= 7r /2
a3
C4 S4 C 4h
D4 C4v
4.
4 ±4 42' 42'
Old
42'
D 2d
42" ±42'
04h T
P,
)
Th
0 Td Oh
C4 S4 C;C 4
at (Xl
= a2 = (X2
= 0:3
= 7r /2
C4C~ C4C~ S4C~ G;C4C~
C~C2C~
3'22' 322' 3'42" 3'42" 3'42"
C~C4C~
3
C3
a,
= a2
(Xl
= 0:2
C;; C2C 2 C~C4C~ C3S4C~
= =
a3 0:3
= 7r/2
C3 v 03d
:3
C3i
32' 32' 32'
C3Cz
C3C~
C3iC2
aj
= a2
O:j
= 0:2
2 ±2 2
at
al
+a2
4 ±4 4 4 4 ±4.
2 2 2 2 ±2
a3
a'
2 ±2
3 3 3 3 3
4.
4 ±4 a'
C3 C3i 03
at
2 ±2 2 4.
4
S4C~
a3
2 2 2 2 ±2 al
+al 2 2 ±2
a,  a 2
3
= a3
:3
= (X3
3 3
3"
2 2 ±2
207
§S.3 Crystal Systems and Bravais Lattice
Crystal system, Bravais Hexagonal,
Point groups Sch. INSG
P
C3
3
C3i
3 32' 32" 32' 32" 32' 32/1 5 6 ±6 52' 52' 62' 62/1 ±62'
D3
03 C3u
no.
C3v D3d
= 15
DJd C6
P3, P3, P32', P32/1, P3'2', P32", P32', P32/1 P5, P6, P ± 6, P62', P52', P62', P 62 /1 , P ± 52'
Note .
C 3h C Gh
D5 C6v
DJh D3h D6h
(continu ed) Direct ion of rotational axes
Product of subgroups
Lattice bases
C3
= a2 0<1 = 0<2 = 1f /2 0<3 = 21f / 3
a3
a1
3 3 3 3 3
2
a]
C 3i
C3 C; C3C~ C3C~ C3C~
3
CJiC; C3;C~
3 3
C6
6
C Jh C i C6 CGC;
6 ±5 6 5
C6C~
6
C 3h C ; C3hC~
6 ±6
C i C6 C ;
at 
a2
2
'2 2 ±2 ±2
2 2 2 2 ±2
2 2 2 2 ±2
(a) Generators in the hexagonal crystal system:
0 1
Gil 2  ( 1
,
0 0
o
0 ) 0 1
(b) G e ne rators in t.he rhombohedral crystal syste m :
C~ = ( ~l ~l ~ o
0
)
 J
(c) G ene rat.o rs of o ther cycl ic s ub groups:
C2 =
C4 =
( 1o 00) (0o 1 0) 0
10 0 1
0 1 0 )
, C~' = ( 1 0 0
o
,
0  1
1 0 0 0 1
(d ) a ' = a, + a2 + a3 . (e) Th e re a re 13 sYIl,moq)h ic , pa ce grou ps P22', P2'5.', P ± 22' , C22', C2'5.', A 2'5.', C 122' , 12'5.', 1 ± 22', F22', F2'5.', a nd F ± 22' in the o rth o rhombic crys t.al syst e m.
± 22',
(f) Th e re a rc 15 syrnrnorphic space groups P3'22' , P3'22', P3'42", P3'4 '5.", P3'42",
13'2 2', 1'3'22', 13' 42", /3'4 '5.", [3' 42" , F3'22', F3'22' , F3' 42", F3 ' 4 '5.", and F3' 42" in the cu b ic cr ys t al sys tem.
Chap. 5 Symmetry of Cr·ystals
208
5.4
S pace Group
In this section we will study the space group S where there is at least one symmetric operation g(R, t) with nonvanishing t. We will first study the constraint on t from R and the dependence of t on the choice of the origin . Then, we will briefly introduce the method of finding the inequivalent space groups . At last, we will analyze the symmetric property of a crystal from t.h e sy mbol of its space group. 5 .4.1
Symmetric Elements
For a given crystal with a crysta.llographic point group G, the symmetric operat.ion in a coordinat.e frame is expressed generally
g(R, ex)
= T(£)g(R, t) ,
ex
= £ + t,
3
t
=L
(5.58) ajtj,
j=l
where t depends on R uniquely. We first study the constraint on t from R. In this section we denote the proper rotation R by C N and the improper rotation R by SN for convenience, where N = I, 2,3,4, or 6. The doublevector of the sum of the cyclic group generated by R is denoted by
{R}.
For an arbi trary vector r, one has (5.59)
It means that th e vector
{11} .
r does not change in the rotation R. On
the other hand, C N (N f:. 1) preserves only the vector along the rotational ilxis n invariant. S2 is a reflection with respect to il plane pe rpendicula r to the improper axis and preserves only the vector on the plane invariant. There is no vector invariant in SN (N i= 2). Therefore,
{6 = I, {S2 } = 2 (I  nn) , 1}
{6
N }
= NfLn,
{SN} = 0,
Sin ce (R)N' = E, where N' = 2N for R = SN with N iV' = N for the remaining cases, the constraint
N
f:.
I, ( 5.60)
N
i=
2.
= 1 or
3, and where
§5. 4 Space Gmup
{g(R,t)}N' = T ({
R}. t)
209
= T(£),
(5.61)
gives a strong restriction on t: t
= 0,
when R
= C 1 = E,
{n (n· t)} = Nt ll = ma ll, 2 {t  n (n· t)} = 2t..L = ma ..L ,
when R= C N , N
t no restriction,
when R= 5 N , N
N
when R
i I,
= 52, i
(5.62)
2,
where m is an integer, til = n (n . t) is the parallel component of t, and t..L = t  t il the perpendicular component of t. When R = C N , N i I, a is t he shortest vector of the crystal lattice along the rotational axis of C N· When R = 52, a..L is the shortest vector of the crystal lattice along t he direction of t..L. We are going to show that the part of t restricted by t he constraint is independent of the choice of the origin, and the remaining pa rt can be removed by a suitable choice of the origin. In the laboratory frame K,
n
g(R, t)T
= RT + t,
and a symmetric operation g(R, t) moves the origin 0 with a translation vector t. Denote by TO the position vector of 0' in K. The operation g( R , t) moves 0' to the position RTo+t, namely, the translation of 0' in g(H. t) is (R  E)TO+t. Move in parallel the coordinate frame K to the coord inate frame K' with the origin 0', where the symmetric operation g(R , t) becomes g',
g'
= 9 [R, (R 
E)TO
+ t j = T( To)g(R, t)T(TO).
(5.63)
Hence , the operator forms of the same symmetric operation depend upon the position of the origin of the coordinate frame. The suitable choice of the pos ition of the origin may simplify the operator forms of the symmetric operations. Obviously, only one origin can be chosen for a given crystal. For R = C N, N i I, the shift TO of the origin changes t with (C N  E)TOl which is perpendicular to the rotational axis of C N. In the perpendicular pl a ne the eigenvalue of C N is not equal to one such that the equation
(eN 
E)TO = t..L has a unique solution TO. Thus, t. = g( C N , t) can be removed by the choice of the origin.
(r  nn) . t in
210
Chap. 5 Symmetry of Cr'ystaLs
For R = S2, which is a reflection with respect to the plane perpendicu to the improper rotational axis of S2, the shift ro of the origin change with (S2  E)ro. The equation (S2  E)ro = til has a solution, ro = tll/ Thus, t il = nn· t in g(S2' t) can be removed by the choice of the origin. For R = SN, N ::J 2, the shift TO of the origin changes t with (SN E)ro. Since the eigenvalue of SN is not equal to one such that the eq uati (SN  E ) TO = t has a unique solution ro , the vector t in g(SN, t) can removed by the choice of the origin. In summary, the part of t which is not restricted by the constraint c be removed by a suitable choice of the origin, but the part of t restrict by the constraint is independent of the choice of the origin. Remind th the position of the origin can be furthe r chosen in the condition that t simplified t does not change. We will discuss this with examples later. A point is called the symmetric center of a symmetric operation g(R, if the point does not change in g(R, t). A symmetric operation g(R, t) called closed if it has a symmetric center, otherwise, it is called an op operation. The position vector ro of a symmetric center satisfies
[g(R, t)  E]
TO
= (R 
E)
TO
+ t = O.
(5.6
In comparison with Eq. (5.63) , the operator form of the symmetric op ation g( R, t) becomes R if the symmetric center is chosen to be the orig of a new coordinate frame. Since R N ' = E and the operator form of E independent of the choice of the origin, one has
[g(R, t)]
N'
=
{E,T(£)
::J
E,
closed operation, open operation.
(5.6
This is another definition for a closed or an open operation. A symmet operation g(R, t) is called closed if its power is equal to E, otherwise, is called a n open operation. Due to Eq. (5.61), the vector t in a clos operation g(R, t) satisfi es
( 5.6
Hence, for a closed operation, Tn in Eq. (5.62) has to be 0 such that t vector t in a closed operation g(R, t) can be removed by a suitable cho of the origin. The position vector TO of the symmetric center of a closed symmet operation can be calculated from Eq. (5.64). For a closed symmetric op ation q(C N , t ), N ::J 1, every point on a straight line, which is parallel
211
§5.4 Space Group
the rotational axis of C N and across the symmetric center of g(C N , t), is a symmetric center. For a closed symmetric operation g(S2, t), every point on a plane, which is perpendicular to the improper rotational axis of S2 and across the symmetric center of g(S2, t), is a symmetric center. There are only two kinds of open operations where m in Eq. (5.62) is not vanishing. In a suitable choice of the origin, two open operations can be expressed as g(CN,tll)'
9 (S2, t.J..) ,
= mall/N f 0, t.J.. = a1../ 2 f O. til
o < m < N f
1,
(5.67)
In the first kind of open operations, til is a multiple of the shortest vector of the crystal lattice along the rotational axis n of C N divided by N. This axis is called a screw axis, which is parallel to n and across the point TO, where TO is the solution of the following equation: (5.68) t il is called the gliding vector of the screw axis. In the second kind of open operations, t1.. is on the reflection plane of S2 and equal to half of the shortest vector of the crystal lattice along the direction of t1... This reflection plane is called a gliding plane, which is perpendicular to the improper rotational axis of S2 and across the point TO, where TO satisfies TO
= t ll /2.
(5.69)
t1.. is called the gliding vector of the gliding plane. The gliding vectors of
both the screw axis and the gliding plane have to satisfy the restriction (5 .62). A straight line is called a symmetric straight line of a symmetric operation g(R, t) if the straight line does not change in g(R, t). A screw axis is a symmetric straight line although there is no symmetric center on it. A plane is called a symmetric plane of a symmetric operation g(R, t) if the plane does not cbange in g(R, t). A gliding plane is a symmetric plane alt hough there is no symmetric center on it. 5.4.2
Symbols of a Space Group
For a given crystal, its symmetric operations g(R , t) and the distribution of t he vectors of the crystal lattice determine its crystallographic point group, and then, determine its crystal system and its Bravais lattice. Neglecting
212
Chap. 5 Symmetry of Crystals
the translation vectors t in g(R, t) temporarily, one has the symbo l of symmorphic space group. The symbol of the space group S of the crys is obtained by attaching some subscripts to the symbol of the symmorp space group. The basis vectors of a crystal lattice aj and the possible vector of crys lattice 1 which is a fractional combination of aj are determined from crystal system and its Bravais latti ce. A symmetric operation of a crys is generally written as
g(R, 0')
= T(L )g(R, t),
L=£ or £+1,
(5.7
where £ is an integral combination of aj . When the crystallographic po group is C 1 , t = 0, and there are only the symmorphic space grou The remaining 31 crystallographic point groups G, from Table 5.3, can expressed as a product. of one, two, or three cyc li c subgroups, respective For a given space group, each generator R of the cyclic subgroups has corresponding translation vector t. Thus, the general form of the elem in a space group S is
T(L) {g(R, t)}",
(5.71
T(L) {g(R,t)}n {g(Rl,p)}n\,
(5.71
T(L) {g(R,t)}" {g(R 2 ,q)r 2 {g(R1,p)}n\,
(5.7
where Tt, Ttl, and Tl2 arc integers, and t, p, and q are the fractional com nations of aj with the coefficients:
o ::; Pj < 1,
o ::; qj < l.
(5.7
In the international notations for the space groups (INSG), the symbol fo space group S is obtained from the symbol of its corresponding symmorp space group by attaching the components of t, p, and q as subscripts the numbers of the cyclic subgroups. If one of t, p, and q is vanishing, correspond ing 0 subscripts can be omitted. The translation vectors t, p, and q have to satisfy three conditions. T first condition comes from the constraint (5.61) which was discussed in preceding subsection. The second condit ion comes from the group proper Th e product of two elements of S given in Eq. (5.71) has to be an elem of S and can also be written in the form (5.71). Especially, the translati vectors t, p, and q do not change in the product. The third conditi comes from the choice of the origin of the coordinate frame. The choice
213
§5.4 Space Group
the origin of the coordinate frame affects the translation vectors i, p, and q , as well as the symbol of the space group. The different symbols due to the different choices of the origin represent the same space group. They are called the equivalent symbols. For a given crystal, one has to choose one common origin of the coordinate frame. The task of group theory is to ta ke the simplest symbol among the equivalent ones by the suitable choice of the origin and to find out all inequivalent symbols for the space groups. There are 230 space groups listed in Appendix C, where the Schoenflies notations and the international notations for the space groups are given for comparison.
5.4. 3
Method for Determining ihe Space Groups
First, the origin of the coordinate frame is chosen to make the translation vcctor i in the factor g(R, i) of Eq. (5.71) as simple as possible. According to g(R, i), the space groups are classified into three types. Typ e A: R = SN, N oj:. 2. There is a symmetric center for g(SN, t). Choos ing the symmetric center to bc the origin of the coordinate frame, olle has t = O. Under the condition t = 0, the origin can be further chosen to sim plify p and q, where the shift ro of the origin satisfies
(SN  E)ro
= L.
(5.73)
The re are 15 crystallographic point groups belonging to type A: C i , C3i , S'I , C31Ll C n ) C 4h ) Ceh) D 2d ) D 3d ) D 3h , Dn ) D 4h , D 6h , T h , and 0". Type B : R = CN, N oj:. 1. There is a symmetric: straight line for g( C/v , i) . Choosing the origin of the coordinate frame at the symmetri c straight line, one has t = til = mall / IV, 0 :::; m < N, where a ll is the shortest vector of the crystal lattice along the rotational axis of C N . The origin can be further chosen with a shift TO satisfying
n
0:::; m' <
N.
(5.74)
Note t hat the lefthand side of E q . (5.74) is perpendicular to the rotational axis The second term in the righthand side of Eq. (5.74) appears in the case where L con tains a component along This additional term with m' i: 0 in Eq. (5.74) can cancel til or partly. Furthermore, the origin can still be chosen to simplify p and q under the condition that the simplified t does not change. We will discuss this simplification with example later.
n.
n.
Chap. 5 Symmetry of Crystals
214
There are 15 crystallographic point groups belonging to type B: C 2 , C C 4 , C 6 , D 2 , C 2v , D 3 , C 3v , D 4 , C4v , D 6 , C 6v , T, 0, and Td.
Type C: R = 52 and the crystallographic point group is C s . There is symmetric plane for g(52 , t). Choosing the origin of the coordinate fram at the symmetric plane, one has t = 0 or aJ.../2. For the Bravais lattice P if t i= 0, one may choose aJ... to be the lattice basis al. For the Brava lattice A, if t contains a component a2/2, this component can be remove by a suitable choice of the origin because a2/2 = f  a3/2. Therefore, addition to the symmorphic space groups, there are two space groups type C: P212 00 and A2.!.oo. 2 For the space groups in type A and type B, the restrictions on th translation vectors p and q need to be studied further. It is found th the condition Rl R 1 = RRI holds for the space groups in the form of E (5.71b) because in those space groups either R = C i or Rl is a prop or improper 2fold rotation with the rotational axis perpendicular to th rotational axis of R (see Table 5.3). Thus, p satisfies g (R1,p
+ L) = g (R, t) g (Rj,p) g (R, t).
(5.7
Under the restrictions (5.62) and (5.75), the origin of the coordinate fram is chosen according to Eq. (5.73) or Eq. (5.74) to make p as simple possible. For the space groups in the form of Eq. (5.71c), the restriction like E (5.75) becomes more complicated. Through a direct calculation one obtai
R2R = R i' i i i
= k = i' = j' = 1, j = k' = 0, = j = i' = j' = k' = 1, k = 0, = i' = j' = 1, j = 1, k = k' = 0,
Rf Rf,
for D 2h , D 4h , and Doh, for T and T h , for 0, T d , and Oh·
(5.7 Thus, p and q satisfy g (R 1, P
+Ld
g(R 1,p+L 2 )
= g (R2, q) g (Rl , p) g (R2' q) ,
= g(R,tfg(R 2 ,q)j g(Rj,p)kg(R,t)l,
(5.7
Under the restrictions (5.62) and (5.77), the origin of the coordinate fram is chosen according to Eq. (5.73) or Eq. (5.74) to make p and q as simp as possible.
§5.4 Space Group
215
Example for the Space Groups in Type A
5.4.4
with the crystallographic point group C 4h (±4), which as the product of subgroups C i C 4 . The crystal belongs crystal system, a] = a2 and 0] = 02 = 03 = 7r /2. The along a 4fold axis. The doublevector of C4 is
Discuss a crystal can be expressed to the tetragonal lattice basis a3 is
There are two Bravais lattices P and I. f = 0 for the Bravais lattices P, and f = (a] + a2 + a3)/2 for the Bravais lattices I. The origin of the coordinate frame is taken at the symmetric center of g(5],t), then, t = O. The origin is allowed to make further shift ro satisfying (5]  E) ro = 2ro = L. P lattice,
I lattice.
Because (C4  E)£/2 (C4  E) (at
=  (£2 + £d aJ/2 + (£]  £2) a2/2, + a2 + a3) /4 = aJ/2,
p changes in the shift ro or the origin
(C 4

E)ro = {
/2
+ a2) /2, (a1  a2) /2, (a] + a2) /2, or aJ/2,
(aj  a2)
or (al
P lattice, I lattice.
(5.78)
From Eq. (5.75) one has 5]g(C4 ,p)5] L' p
= 2 =
= g(C4 , p) = g(C4 ,p \
{ £/2, £/2 or (a1 + a2 + a3) /4,
L'), P lattice, I lattice.
Thus, in the Bravais lattice P, p can be 0,
al/2,
(aj + a2) /2,
a2/2,
a3/2,
(a2 + a3)
/2,
(a1 + a3) /2, (a] + a2 + a3) /2.
Due to Eq. (5.78), aJ/2 and a2/2 are equivalent, (a1 +a3)/2 and (a2+a3)/2 are equivalent, (a] + a2 + a3)/2 and a3/2 are equivalent, and (a] + a2)/2 can be removed, namely, p can be 0, aJ/2, a3/2, and (a2 + a3)/2 in the Bravais lattice P. In the Bravais lattice I, because (a] + a2 + a3)/2 is a vector of crystal lattice, from Eq. (5.78) aJ/2, a3/2, and (a2 + a3)/2 can
216
Chap. 5 Symmetry of Cry stals
be removed, but there is another p = (a) + a2 + a3)/4. In summary, t space groups with the crystallograp hic point group C4h (± 4) are
P ± 4,
5.4.5
P ± 400l ,
P ± 4~oo,
Example
2
fOT
P±40 l1 , 2 2
i±4,
i ± 4 41 41
1.
4
(5.7
the Space Groups in Type B
Discuss a crystal with the crystallographic point group D 3 , which can expressed as the prod uct of subgroups C 3 C~ or C 3 C~, depend ing on wheth the lattice basis al is a long a 2fold axis or not. Correspondingly, international symbol is 32' or 32/1. The crystal belongs to the hexagon crystal system or the rhombohed ra l crys tal system . In both cases, f =
(a) Th e Bravais lattice P in the hexagonal system. a3 is alo ng the 3fold axis, al = az, CXI = CXz = rr/2, and The doublevect.ors of t he generators are
:3 =
+ a2 )b2 + a3b3,
a2bl  (aj
2' = alb ]  (aj
+ a z )b 2

a 3b3,
2/1 = a2bl alb2 a3 b3,
CX3
= 2rr
for 32' or 32/1,
for 32' , for 321'.
The origin of the coordinate frame is taken at the symmetric straig line of g(C3 , t) such that t = 0, a3/3, or 2a3/3. The origin is allowed make further shift TO satisfying (C 3 TOJ

E ) TO
= a] (1'01 + T02) + a2(Tol
= (2£ 1 + £2)/3 ,
 2T02)
= £,
1'03 arbitrary,
where £1 and £2 are integers. For the crystallographic poin t group 32', changes in the shi ft TO of the origin (C~  E)TO
= a1T02  a 22r02  a32T03 = (a] + 2a2)( £1 + e2)/3  a32r03.
For the crystallographic point group 32/1, p changes in the sh ift origin
(q' 
E)T O
= (al + a2)(rol + 1'02) = (al + aZ)£1  a 32T03·
 a32r03
Note that (aJ + 2a2) is perpendicular to the rotational axis of (al + az) is perpendicular to the rotational axis of C~'.
( 5.8 TO
of t
( 5.8
q,
a
§5 .4 Space Group
217
On the other hand, for the crystallographic point group 32', from Eq. (5.62), PI = 0 or 1/2. Due to Eq. (5.75) P has to satisfy g(C~)
P
+ £')
= g(C3, t)g(C~,P)g(C3, t)
= g(q, t + C3p + C3C~t) wh ere t
+ C:] C~ t
= g(C~,
C 3 P),
= 0 because t is along a3. Hence,
The solution is that P3 is arbitrary, PI = (2e~ + e;)/3, and P2 = (e~ + e;)/3. Since e~ and e; are integers, Pl = 1/2 is ruled out, and then, Pl = P2 = O. P3 can be removed by Eq. (5.80). For the crystallographic point group 32", the shortest vector of crystal lattice along the rotational axis of C~' is (al  a2). From the constraint (5.61) the following vector is a vector of crystal lattice
Thus, Pl
= P2.
Due to Eq. (5.75), P has to satisfy g(C~',p+£/)
= where t
+ C 3 C~'t
g(C~/, t
= g(C3 ,t)g(q',p)g(C3 ,t)
+ C3 p + C3C~'t) = g(C~', C3P),
= 0, because t is along
a3.
Hence,
nam ely, PI = ])2 = O. ])3 can be removed by Eq. (5.81). In summary, the space groups ,vith the crystallographic point group D3 in the hexagonal crystal system are
P32', P3 00 1 2', P3 00 l2', 3
3
P32", P3 001 2", P3 00 l2". 3
3
(5.82)
(b) The Bravais lattice R in the rhombohedral crystal system. Three lattice bases aj are welldistributed around the 3fold axis, 02 = 03 and exl = ex2 = ex3. The doublevectors of the genE'rators are
01
=
The origin of the coordinate frame is taken at the symmetric str aight linn of g(C 3 , i), then, m = 0, 1, 2.
218
Chap. 5 Symmetry of Crystals
The origin makes a further shift ro according to Eq. (5.74) f
+ (a] + a2 + a3)m' /3 = (C3  E) ro = al (TOI + T03) + a2(TOl  T02) + a3(To2
 T03).
(5.8
Since the sum of the three coefficients with respect to the lattice bases on the righthand side of Eq. (5.83) is 0, that on the lefthand side has be zero, m' =  ~j i j , namely, t in g(C3 , t) can be removed by a suitab choice of the origin. The reason is that the component of aj along t 3fold axis is equal to a ll /3. Under the condition t = 0, the origin is allowed to make further shif
The solution is that T03 is arbitrary, and both integers. p changes in the shift of the origin
T03 and T02  T03 a
TO] 
(q 
E)ro = (a] + a2)(Tol + T02)  2a3T03 = (al + a2 + a3) 2To3  (al + a2)(rQl + T02  2To3),
namely, p changes an arbitrary multiple of (al + a2 On the other hand, from Eq. (5.61), P satisfies (C~
+ E)p
(5.8
+ a3).
= (a]  a2)(Pl  P2) = f.
Thus, PI = P2· From Eq. (5.75) one has
Hence,
The solution is Pl = P2 = P3 which is just removed by a suitable choice the origin [see Eq. (5.84)]. Therefore, the space groups with the cryst lographic point group D3 in the rhombohedral crystal system is only t symmorphic space group R32.
5.4.6
Analysis of the Symmetry of a Crystal
We have introduced the method for finding the space groups of cryst by group theory. Perhaps, the important problem for most readers is h to analyze the symmetry of a crystal from its international symbol of t space group. We will demonstrate this problem through an example.
§5.J,
Space Group
21
We study a crystal with the space group
From the international symbol one knows that the crystal belongs to the Bravais lattice I in the tetragonal crystal system with the crystallographi point group D 4h . a3 is along the 4fold axis, a1 is along a 2fold axis, and a2 = C4a1· 0'1 = 0'2 = 0'3 = 7r/2 and a1 = a2. The shortest vector o crystal lattice along an inequivalent 2fold axis is a1 ± a2 with the length .J2al. The lattice bases are not primitive. There is a vector of crysta lattice J which is a fractional combination of the lattice bases aj
L=£+J, The doublevectors of the generators in the space group are C4 C~
= a2b1  a l b2 + a3 b3, = a1bl  a2b2  a3b3.
The general form of an element in the space group is
where Sl is the spatial inversion with respect to the origin of the coordinat system. n1, n2, and n3 are equal to or 1, respectively. m is taken to b 0, 1, 2, or 3. The rotational axis of g(C4 , q) is parallel to a3, and qll = a3/ 4, q.l (al + a2) /4. Its screw axis is parallel to a3 and across the point TO
°
(C4

E) TO
+ q.l = 0,
rOI
+ r02 = rOI + r02 = 1/4.
The solution is rOI = 0, r02 = 1/4, namely, the screw axis of g(C4 , q) i parallel to a3 and across the point a2/4 with the gliding vector a3/4. The rotational axis of g(q,p) is parallel to aI, and p = PII = al/2 The screw axis of g(C~,p) is parallel to al and across the origin with th gliding vector al/2. Since Slg (q, al/2) = T( al)g (S~, al/2), where S~ is an improper 2fold rotation along aI, P = PII = aj /2, and the symmetri plane of g (S2' aJ/2) is perpendicular to a] and across the point aJ/4. From an arbitrary point T = a1xI + a2X2 + a3x3, one can obtain eight equivalent points in a crystal cell through the symmetric operation
220
Chap . .5 Symmetry of Crystals
+ 1/4) + a2 (.T] + 1/4) + a3 (X3 + 1/ 4) , r3 = alXI + a2 (X2 + 1/2) + a3 (.73 + 1/ 2), r 4 = a1 (.72  1/4) + a2 (Xl + 1/4) + a3 (X3 + 3/4) , r5 = at (Xl + 1/2)  a2X2  a ;IXJ, r6 = a] (X2 + 1/4) + a2 (Xl + 3/4) + a3 (.r3 + 1/4), r7 = al (X1 + 1/2) + a2 (X2 + 1/2) + a3 (X3 + 1/2), rs = al (X2 1/4) + a2 (Xl + 3/4) + a3 (X3 + 3/4). r2 =
al
(X2
Then, the remaining 24 equivalent points can be obtained through the symmetric operations T(f) and 51 .
5.5
Linear Representations of Space Groups
The translation grou p T is an Abelian invariant subgroup of the spac group S. The irreducible representations of T are onedimensional. The irreducible representations of S will be established by two steps. First, find the subclu ced representation D(£) of an irreducible representation of S with respect to T which is taken to be diagonal. Then, calculate the represen tation matrices of the elements g(R, t) in the irred ucible representation o S. In this sect.ion, the lattice bases aj are primitiv e for convenience.
5.5.1
Irreducible Representations of T
Denot.e by a J t.h e primitive basis vectors of crystal latti ce and by b; th basis vectors of reciprocal crystal lattice. The inner product of a vecto of crystal lattice £ = ~j aje j and a vector of reciprocal crystal lattic K = ~, b,1(;, where ej and 1(; are all integers, is an integer:
(5.85 If R is an element. of the crystallographic poin t group G,
(RK) . £ = K . (R J £) = K . e'
= integer,
(5.86
namely, RK = K'. The cryst.allograph ic point group G is the same fo both the crysta l lattice anel the reciprocal crystal lattice.
§5. 5 Linear Representations of Spa ce Groups
221
The real crystal is finite so that there is no rigorous translational symmetry for it (see [Ren (2006)]) . Usually, the periodic boundary condition is assumed to restore the translational symmetry of the crystal such that the translation group T is a finite Abelian group. A parallelepiped spanned by three lattice bases is called a crystal cell. Assume that a.long three lattice bases aj there are N j crystal cells, respectively, where N j are very large natural numbers. Thus, the translation group T is a direct product of three translation subgroups (5.87)
The order of TU) is N j , and the order of Tis NI N2 N 3 if there is no common factor among N 1 , N 2 , and N3 [or simplicity. When N j goes to infinity, the finite crystal becomes an ideal crystal. T (1 ), for example, is a cyclic group and has Nt onedi mensional inequivalent representations. The representation matrix of its element T( al ed is (5 .88)
T he onedimensional representation Dkl of T(1) is denoted by kl = pI/NI , where PI is an integer, 0 :S PI < N J . Thus, t.here are (NJN2N3) onedi mensional inequivalent representations characterized by a vector k in the space of the reciprocal crystal lattice 3
k= L
bjk j
(5.89)
,
j=t
T he representation matrix of an elem ent of T in the represen tation is
o :S kj
<
1.
(5.90)
k is called t he wave vector, and the space of reciprocal crystal lattice is call ed t he spa ce of wave vectors, or k space. In the region, 0 :S k j < 1, there are (Nt N 2 N 3 ) different Wilve vectors k denoting (NJ N 2 N 3 ) inequiva lent irreducible representations of T. \iVhen N j goes to infinity, k varies cont inuously in that region. If the difference k  k' is a vector of the reciprocal crystal lattice K , bot h k and k' characterize the same representation of T, and are said to be the equivalent wave vectors. In this meaning, there is the translation symmetry in the k space. The region 0 :S k j < 1 plays the role of a
222
Chap. 5 Symmetry of Crystals
reciprocal crystal cell in the k space, just like a crystal cell in the cryst lattice. However, this region is not very convenient in practice because th point in the region may move out of the region in the rotation R of th crystallographic point group. In solidstate physics the Brillouin zone introduced as a reciprocal crystal cell. The vertical bisector plane of th line from the origin to a point of reciprocal crystal lattice is called the Brag plane. The region surrounded by the nearest Bragg planes to the origin called the first Brillouin zone. The region from the first Brillouin zone to th next Bragg planes is called the second Brillouin zone, and so on. A Brillou zone, except for the first one, is not connected, but each Brillouin zone ha the same volume and contains (Nl N 2 N 3 ) wave vectors equivalent to th given in Eq. (5.89), respectively. Under a rotation R, a wave vector insid a Brillouin zone moves to another wave vector inside it, and a wave vecto on its boundary moves to another wave vector on the boundary. There a onetoone correspondence between a wave vector inside a Brillouin zon and an inequivalent irreducible representation of T. But the wave vecto on the boundary of a Brillouin zone may be equivalent, namely they ma characterize the same representation.
5.5.2
Star of Wave Vectors and Group of Wave Vectors
Denote by D(S) an mdimensional irreducible representation of the spac group S, whose subduced representation with respect to the translatio group T is reduced,
(5.9
Note that the subscript in kp is not the component index, but the row (co umn) index of the matrix. The wave vectors kp with a different subscri p is not necessary different. Since T is an invariant subgroup of S, g(E,£)g(R,a) =g (R, a)g(E,R 1 £), D(E , £)D(R, a)
= D(R, a)D(E, R 1 f).
(5.9
This conjugate transformation changes kp in the exponential function in Rkp. On the other hand, the eigenvalues do not change in a similari transformation, so Rkp has to be equivalent to one kT in Eq. (5.91):
(5.9
§5.5 Linear Representations of Space Groups
223
The wave vectors in the first Brillouin zone is called mutual conjugate if they can be related by a rotation R in the crystallographic point group G through Eq. (5.93). The set of conjugate vectors is called the star of wave vectors. The number of inequivalent wave vectors in a star of wave vectors is called the index of the star, denoted by q. Since D(E,£) is diagonal, one obtains from Eq. (5.92)
Thus, DTP(R, a) =J. 0 only if Eq. (5.93) holds. Since the representation D(S) is irreducible, any two wave vectors kp and kT in Eq. (5.91) are conjugate, namely, there is an element R in G such that Eq. (5.93) holds. Otherwise, the wave vectors in Eq. (5.91) have to be divided into at least two classes, where kp and kT in different classes are not conjugate, so that the representation matrix entry DTp(R, a) of every element g(R, a) in S is vanishing and the representation is reducible. One is able to collect the same diagonal entries in Eq. (5.91) together by a simple similarity transformation,
where k/l =J. kv if JL =J. v. The multiplicity of each k/l is the same d = mlq· Each irreducible representation of S corresponds to a star of wave vectors, ,vhere the dimension m of the representation is a multiple of the index q of the star. Arbitrarily choose one wave vector in the staT, say k j . There are some elements P in G satisfying (5.95)
k ] is said to be invariant in P. The set of P forms a subgroup H(kd of G. The left coset of H(kd is denoted by R/lH(kd, (5.96)
The nonvanishing vector of reciprocal crystal lattice appears in Eqs. (5.95) and (5.96) only when the star of wave vectors is located at the boundary of the Brillouin zone. The index of H(kd is equal to the index q of the star of wave vectors. Although the choice of R/l is not unique, R/l is assumed to be fixed, where RJ = E. Any element in G can be expressed as R/lP. The set of g(P, a), where P E H(kJ), forms a subgroup S(kr) of S, called the wave vectors group with respect to k j . The index of S(kr) in S
224
Chap . .5 Symmetry of Crystals
is q. T is also an invariant subgroup of S(kd, where the quotient gro S(ktJ/T is isomorphic onto H(k j ).
5.5.3
R epresentation Matrices of E lements in S
vVe discuss an arbitrary mdimensional unitary irreducible representati D( S) of S which corresponds to a star of wave vectors with index q. Divi the representation matrix D(R, a) of an arbitrary element g(R, a) of S in q2 square submatrices with the dimension d = m/q,
Dll(R, a) D12(R, a) '" D1q(R, a)) D(R, a) = ( D2d~, a ) D22(~' a) :.: D2q(~, a)
(5.9
D qj (R, a) D q2 (R, a) ... Dqq(R, a)
It will be shown that there is only one nonvanishing sub matrix in each c umn (or row) of D(R, a). First, the representation matrix of a translati T( £) is diagonal,
where Id is a ddimensional unit matrix. Second , discuss the representation matrix of g(RJ.<, tJ.<). Substituting E (5.98) into Eq. (5.92), one has exp { i27fkv . £} Dvl (R I", t l ,)
= D//d RIl> tJ.< )
exp { i27r (RJ.
Thus, in the first column only the submatrix in the p,th row is nonvanishin
(5.9
where DfJJ(RfJ , t J.<) is unitary. Up to now, only D(E, £) has been determined. The representation D ( is allowed to be made a unitary similarity transformation X, which is com mutabl e vvith D(E, £.). In order to simplify D(RI", tJ.<), one chooses: when
~l
=
1,
when p, =I 1.
(5.10
The representat.ion after the similarity transformation is still denoted the sa me symbol D. D(E, £.) remains the form given in Eq. (5.98), but
(5.10
§5. 5 [,,,neaT Representations of Space Groups
225
Since the representation is unitary, the submatrices in the J,Lth row are vanishing except for that in the first column,
T hird , through similar derivation, from Eq. representa t ion matrix of g(F, a)
(5.94), one obtains the
(5.103) At last, calculate the submatrix DPI,(R, t) in the J,Lth column of the rep resenta tion matrix of g(R, t). Since the product RRp. is an element of G, it can be ex pressed as RuP,
(5.104) For the gi ven Rand Rp., t , til' Rv, t u, P, and a in the space group 5 are all determined. Th e vth row of the J,Lth column of the representation matrix of g(R, t) is
D up.(R, t) =
L
DUT(Ru,t[l)Dd(P,a)Dp.>.(Rp., t p. l* = Dll(P,a). (5.105)
Since the representation is unitary, (5.106) Thus , in each column (or row) of D(R , t ) there is on ly one non vani shi ng submatrix, which is ex pressed as Dll (F, a). The representation matrix of a ny element of t he space group 5 is comp letely determined by the irred uci ble representation Dll (P , a) of the wave vector group.
5.5.4
Irreducible Representations of S(k 1 )
We first prove that the representation D(S) of the space gro up 5 is irreducible if and only if the representation Dll (P, a) of the wave vector group S(k d is irreducible. In fact, if Dll is reducible, th ere is a non constant matrix Y commutable with the matrices Dll (P, a). Then , X = lq x Y is commuta ble wi th all matrices D(R,t) and D(E,£), so that D(S) is reducible. Conversely, if D(S) is reducible, there is a nonconstant matrix X commutable with all represen tation matrices of D(S). Since X is commutable with D(E ,.f.) , X = EEl!l YI" Since X is commutable with D (RJ1' t l ,),
226
Chap. 5 Symmetry of Crystals
y,~
is independent of 11 and commutable with Dll(P, a). Thus, Dll (P, a is reducible because Yj is not a constant matrix. Second, make a correspondence from an element g(P, a ) of the wa vector group S(kd to the matrix Dll(P,a) = exp(i27Tk 1 · a)f(P),
(5.10
where f(H) is an irreducible representation of the point group H(kJ). O wants to show whether this correspondence remains invariant in the mul plication of the two elements of S(kl)' g(P, a)g(p l , a/) = g(PP I , a
+ Pa l) ,
namely, whether the following equation holds,
Dll (P, a)Dll (Pi , a/) = Dll (P pi, a
+ Pa').
(5.10
Note that f (P)f(PI) = f(P Pi). The exponential function on the lefthan side of Eq. (5.108) is exp [i27TkJ . (a
+ a/)] ,
and that on the righthand side of Eq. (5.108) is exp [i27Tkl . (a
+ Pa' )] = exp (i27Tkl
. a) exp [i27T(Pl kJ) . a'l
If the star of wave vectors is located inside the first Brillouin zon P1k 1 = k 1 , Eq. (5.108) holds. If S is a symmorphic space group , a l = even though the star of wave vectors is located on the boundary of the fir Brillouin zone, p 1 kl = kl + K, one has exp (i27T K £) = 1, and E (5.108) still holds. Therefore, for the two special cases, the set of D J } (P, a given in Eq. (5.107) forms an irreducible representation of the wave vect group. Remind th at the number of the irreducible representations in t form (5.107) is equal to the number of classes of H(k)). We will not discu the remaining cases where the star of wave vectors is located on the boun ary of the first Brillouin zone but the space group S is not symmorphic. In summary, an irreducible representation D(S) of a space group S characterized by the star of wave vectors and the irreducible representati f(H) of the point group H(kd. Its dimension is m = qd, where q is t index of the star of wave vectors and d is the dimension of the representati f. The representation matrix of an element of the translation subgro T is diagonal and given in Eq. (5.98) . The representation matrix of element g(R, t) in the coset of T is composed of q x q submatrices wi
227
§5.5 Linear Representations oj Space Groups
the dimension d as given in Eq. (5.97). In each row (or column) there is only one nonvanishing submatrix, which is related with the irreducible representation DlI (P, a) of the wave vector group S(kd (see Eqs. (5.106) and (5.107)). It can be proved that for a given star of wave vectors, the representation D(S) is independent of the choice of k J .
5.5.5
The Bloch Theorem
Let 1/JJ m (r) be the basis functions belonging to the irreduci ble representation Dll (P, a) of the wave vector group S(kd (5.109) m'
For a translation T(£), due to Eq. (5.98), Pg (E ,l)1/Jlm(r)
= 1/Jlm(r 
£)
= 1/Jlm(r) exp (i21fkl
. £) .
(5.1l0)
Thus , (5.ll1) Extending the basis functions, one has
'lj;J1m(r)
PR~ 'um(r 
£)
= PR~ 'lj;Jm(r) = 'lj;lm(R;:lr) = exp {i21fkJ . (R;:lr)} um(R;:J r ) = exp {i27f (RJ1kd . r} PR~ um(r), = Um(R;:lr  R;:l£) = um(R;:lr = Um(R;:lr) = PR~ um(r).
(5.ll2)
 £')
The basis functions 'lj;J1m (r) belong to the irreducible representation D(S) characterized by the star of wave vectors and the irreducible representation f(H) of the point group H(kIl. In fact, the definition (5.ll2) for 'lj;J1m(r) shows that Eq. (5.101) holds, so does Eq. (5.102). For the translation T(£), Pg (E ,.f.)VJ J1m(r) = 'lj;wm(r  £) = exp {i27f (RJ1k 1 ) . £} 1/J,'m(r)
= exp {i27fkJ1 . £} 'lj;J1m(r).
(5.113)
Thus, Eq. (5.98) as well as Eqs. (5.105) and (5.106) hold. The static wave function of an electron moving in the crystal belongs to an irreducible representation of the space group S of the crystal. If'tPlm(r)
I
228
Chap. 5 Symmetry of Crystals
is an eigenfunction of the Hamiltonian, its partners 1/J/ Ln1 (r) in an irreducib representation of S are the degenerate eigenfunctions of the Hamiltonia wi th the same energy. The Bloch theorem, which is a fundamental theorem in solidsta physics (see p.211 in [Dradley and Cracknell (1972)]), says that the stat wave function of an electron moving in a potential with the translatio symmetry satisfies
(5.11
Equat ion (5.112) gives an exp lanat ion of the Bloch theorem from grou theory.
5.5.6
Energy Band in a Crystal
In this subsection we will discuss the dependence of the energy of an electro in a crystal on the wave vector k by the approximation of free electrons. A electron in a crysta.l is moving in a potential V(r) with the symmetry of t space group S. V(r) varies around its average field Va, and the differen Vdr) = V(r)  Va plays the role of perturbation. The Hamil ton ian of th electron is
(5.11
where me is the mass of the electron. If the energy level is normal degene ate, the eigenfunction of the energy belongs to an irreducible representatio of thr space group S, namely, 1/J/lm(r) given in Eq . (5 .112) has the sam energy, independent of fJ, and m. For simplicity we denote kl by k an om it the subscripts J.1 and m in the wave function,
1jJ (r)
= exp (i2Jrk . r) l1(r),
l1(r 
l) = l1(r).
(5.11
can be made a Fourier expansion with respect to the vector of reciproc crystal lattice K n
U
u(r)
=L
l1n
ex p (  i2 Jr K n' r),
n
1jJ(r)
=L
(5.11 unexp {i21f (k  Kn)' r},
n
where
'U n
is a constant to be determined.
Each term in the expansi
229
Exercis es
(5.117) is an eigenfunction of Ho with the eigenvalue F Il 2
En
h = 2me
(k  K n)
2
+ Vo.
(5.118)
En varies continuously as th e wave vector k  Kn changes. Calculate the en ergy correction to the ground state with n = O. Since the first approximation is vanishing, the en ergy correction com es from the second approximation of the energy. When k is near the center of the first Brillouin zone, Eo depends on k 2 continuously. En  Eo is large and U n with n i 0 is much smaller than 'Uo. When k is in the boundary of t he first Brillouin zone, (k  Kn)2 may be closed to k 2 , and the energy interference o(:r1!rs. The result is that one energy level enhances a nd the other depresses , so that an energy gap appears. Th e condition for the boundary of the Brillouin zones is
k
2
= (k
2
 K) .
T he solution is 2k· K
= K2
(5.119)
It is the equation for the perpendicular bisector plane of the line from the origin to a point of reciprocal crystal lattice, called the Bragg plane in solidstate physics. The Brillouin zones are divided by the Bragg planes. k
o
Fi g. 5.2
5.6
K
Th e Bragg plane.
Exercises
1. III t.he rectangular coordinate frame , write th e doublevec tor forms and the mat.rix forms of the prop er and improper sixfold rotations a.round t he zaxis.
230
Chap. 5 Symmetry of Crystals
2. Express the directions of the proper and improper rotational axes in groups Td and Oh by the basis vectors ej of the rectangular coordin frame. Express the dou blevector forms of the generators of the pro rotational axes in Oh by ej.
3. Write the doublevector form of a rotation R(n,w) around the direct
n through the angle w
by the identity
f
and the unit vector
n.
4 . Let R be a rotation around the direction n = (e} + e2) /V2 throu 27r/3. Find the symmetric straight line for g(R, t) where (a) t = e3 , t = ej + e3. If the symmetric straight line is a screw line, find its glid vector. Simplify g(R, t) by moving the origin to the symmetric strai line for checking your result.
5 . Let R be a reflection with respect to the x y plane. Find the symmet plane of g(R, t) where (a) t = e3, (b) t = ej +e3. Hthe symmetric pl is a gliding plane, find its gliding vector. Simplify g(R, t) by moving origin to the symmetric plane for checking your result.
6. Analyze the symmetry property of the crystal ·w ith the following sp group: No. 52 [D~h' p±2(HO)2'(~H)], No. 161 [C~v' R32'(H~) and No. 199 [T5, 1312(~O~)2'(HO)1. Point (a) the general form of symmetric operation; (b) the relations of directions and lengths amo three lattice bases; (c) the symmetric straight line, symmetric pla and the gliding vector of the generator in each cyclic subgroup of space group, if they exist; (d) the equivalent point to an arbitrary po r = ajXj + a2x2 + a3X 3 in t.he crystal cell.
7. Calculate the 12 space groups related to the crystallographic point gro
D2d ·
8. Calculate the nine space groups related to the crystallographic po group D 2 .
Chapter 6
PERMUTATION GROUPS
A permutation group describps the permutation symmetry of a system of identical particles. We will establish the representation theory of the permutation groups by the Young operator method that is an elegant theory and is widely used in modern physics. From the viewpoint of mathematics, the theory of permutation groups is very interesting because each row in the group table of a finite group shows a permutation of group elements such that every finite group is a subgroup of a permutation group.
6 .1 6.1. 1
Multiplication of Permutations
Permutations
A rearrangement of n objects, denoted by digits I, 2, ... , n, is called a permutation. A permutation R which moves the jth object into the T'jth position is characterized by a 2 x n matrix,
R=(l
2 ..
n).
T'l
T'2···
Tn
(6.1)
In expression (6.1), the order of the columns does not matter, but the corresponding relation between two digits in each column is essential , for example,
R=
(134521 2 3 4 5) = (512345 4 1 2 3)
(6.2)
A di git j describ es an object located in the jth position. Hereafter, a permutation of ohjects is said as a permutation of digits for simplicity. Usually, this simplification does not cause any confusion. 231
232
Ch(lp. 6' PerrnutallOn Groups
The mult.iplication of two permuLations does not satisfy the multipli cation rule of matrices, a lthough a permutation is expressed by a matrix The multiplication SR of two permutations is defined as successive appli cations of R and then S. In other words, the jth object is transform e into the: Tjth position by R, and then, into the STj th position by S. Not that in the: second permu tation S, the j th 0 bj ec t in the new order after th permutation R is t ransformed into the 8jth position , no matter in whic position it was located b efore the first permutation R. From the definitio of multiplication , SR can be calculated as follows. First , change the orde of columns in S or R such that the first row of S is the same as the secon row of R. Then , combine the first row of R and the second row of S to new 2 x n matrix , which is the matrix form of SR . For example,
s=(12 ... n) 8] 82 ... Sn
Th e multiplication of tvvo permutations is a permutation. Being a mul tiplication of transformations, the multiplication of permutations sat isfie the associa tive law, (RS )T = R(ST), but it is generally not. commutable The identical p ermutation E is characterized by a matrix whose two row arc exactly the same, E
=
(11 22 33 ...... n) n
(6.4
From the multiplication rule (6.3), ER = Rand E is th e identical element The matrix of R  ] is obtained from that of R by switching two rows,
R] =
(T] T2 ... Tn) 1 2 ... n
(
T] T2 ... Tn)
1 2 ... n
(1 2..
(6.5
'
n )
T] T2 ... Tn
= E.
There are n l diffe:rent permutations among n objects. In the multiplicatio rule: (6.3 ) the set of n l permutations sat.isfies the four axioms for a group and forms the permut.at ion group of n objects, denoted by Sn . The: orde of Sn is 9 = nL Choosing arbitrarily m objects among n obj ects, on obtains a permutat.ion group Sm from the permuta.tions of m objects. Sr is a subgroup of Sn.
§ 6.1
233
Multiplication of Permutations
In the literature, there are different definitions for a permutation. For example, a permutation R given in Eq. (6.1) may be defined as a transformati on moving the Tjth object to the jth position. In fact, two definitions are mutually inverse. Attention must be paid to the definition when one begins to read a new book. 6.1. 2
Cycles
If a permutation S preserves (n  e) objects invariant and changes the remaining objects in order, S is called a cycle with length characterized by a onerow matrix:
e
e,
ae l Ut b1 ... bn  e ) ae al b1 ... b,,_[
a2 S = ( al a2 a3
= (al
a2
..
.
ae  t ae)
= (a2
a3
...
(6.6)
at al) .
In the onerow matrix for a cycle, the order of digits is essential, while the t ransformation of digits in sequence is permitted. A cycle with length 1 is t he identical element E. A cycle with length 2 is called a transposition,
(a b) = (b a) ,
(a b) (a b)
= E.
(6.7)
e e
The order of a cycle S with length is (see Prob. 3 of Chap. 6 in [Ma and Gu (2004)]), namely, e is the smallest power with Se = E. Two cycles are called independent if they do not contain any common digit (object). The multiplication of two independent cycles is commutable. Any permutation can be decomposed as a product of independent cycles uniquely up to the order. In fact, checking an arbitrary digit at in a given permutation R, one finds that after R, aj changes to a2, a2 changes to a3, an d so on. In this cha in there must exist an obj ect, say at, such that ae cha nges to aj. Thus, R contains a cycle (aj, a2, ... , ad whose length is e. In the remaining objects one checks another object, say bj , and finds an other chain of objects which gives another cycle independent of the first one. This process can be done until all objects are contained in the cycles. T hus, the permutation is decomposed into a product of independent cycles. For example,
R=
(~
s=
C
2 3 4 4 5 2 2 3 4 1 2 4
~) ~)
= (1 3 5) (2 4) = (24) (1 3 5) ,
(6.8) = (1 32) (4) (5)
= (1 32) .
Chap. 6 Per·mutation Groups
234
The set of lengths of the independent cycles in a permutation R is called its cycle structure. For example, the cycle structure of R in Eq. (6.8) i (3 2), and that of 5 is (3,1,1) == (3,1 2 ). The order of the cycle length £i in the cycle structure of a permutation is arbit.rary and the repetitive cycle lengths can be expressed as an exponent. It is a common sense tha the simplest form of a permutation is its decomposition into a product o independent cycles. In the calculation of the multiplication of two permutations one has to deal with the multiplication of two cycles with a few common objects. The multiplication of two cycles with one common object can be calculated as
(a bed) (d e f)
=
(~~ ~ : j ~)
f (abedef) ( abede ) abeefd bedefa
= (a bed e f) . Generally, two cycles with one common object are connected to be one,
(a
f) = (a ... bed ... 1).
.. b e)(e d
(6.9
Conversely, a cycle can be decomposed into a product of two cycles by cut ting it in any digit and repeating the digit. Generalizing this method, one can calculate the multiplication of two cycles with more than one common objects by the above "cutting" and "connecting", say (al ... ai e ai.+l ... aj d) (d b1 ... br c br +1 ... bs )
=
(al
...
ai e) (e ai+l ...
aj
d) (d bl
...
b( e) (c br+ 1
= (al ... ai e) (ai+l
aj de) (e d b1
= (al ... ai e) (ai+l
aj d) (d e) (e d) (d b1
= (a1 ... ai c) (aiH
aj d) (d b1
...
...
br) (e b,+1 ...
bs) bs )
br ) (e b(+1 ... bs)
br) (e br+1
=(al ... aieb,·+1 ... bs)(ai+J ... ajdh
6.1.3
...
bs )
.. br) .
Classes in a Permutation Group
The multiplication rule (6.3) of permutations can be understood from othe viewpoints. When a permutation R is leftmultiplied by a permutation 5 the digits (objects) in the second row of R make a 5 permutation, and when R is rightmultiplied by 5, the digits (objects) in the first row of R make an 5 1 permutation. Therefore, the conjugate permutation 5 R5of R can be calculated by transforming all digits in both rows of R by the permutation S. Especially, when R is a cycle, one has
§6'.1 Multiplication of Permutations
S (a be ... d) Sl
= (sa
Sb Sc
235
(6.10)
... Sd)·
Formula (6.10) for the multiplication of permutations is called the interchanging rule. From the rule, two permutations are conjugate to each other if and only if their cycle structures are the same. The class in a permutation group is characterized by the cycle st ructure of the permutations in the class. Equation (6.10) can be rewritten in another way,
S (a be . .. d)
= (sa
Sb Sc
. _. Sd)
(6.ll)
S.
W hen a permutation S is moved from the leftsid e of a permutation R to its rightside, the digits (objects) in R are made an S permutation. Conversely, when S is moved from the rightside of R to its leftside, the digits in Rare made an S1 permutation. A set of positive integers £j, whose sum is equal to a positive integer n, is called a partition of n. The cycle structure of a permutation R as well as a class is characterized by a partition of n, m
L
£i
= n.
(6.12)
i=l
T he number 9c(n) of classes in Sn is equal to the number of partitions of n. T here is no analytic formula for 9c(n) (see Chap. 1 in [Angrews (1976)]), but all partitions for a given positive integer n can be listed without dropping (see the next section). Some 9c(n) are give n as follows.
= I, 9c(2) = 2, 9c(5) = 7, 9c(6) = ll, .9c(9) = 30, 9c(10) = 42, 9c(100) = 190569292, 9c(1)
9c(4) = 5, = 3, 9c(8) = 22, 9c(7) = 15, 9c(20) = 627 , 9c(50) = 204226, 9c(200) = 3972999029388. 9c(3)
e
On the other hand, the number n(C\') of the elements in a class a of Sn can be expressed analytically. Denote by N[n, m] the number of elements in S" which contain m cycles with length £ and the length of the remaining cycles is 1,
e,
n! If the number of integers
e in the partition of Co is me, 1 .s; e .s; k, one has
236
Chap. 6' Perm.utation Groups
k
n(o:)
= II N[n 
= n! II (me!e m ,) !
e= ! where TJ
6.1.4
e l
k
T e, e, me]
,
Te
=L
e=!
am
Q ,
(6.13
a= l
= 0 and T k + 1 = n.
Alternating Subgro"ups
Each cycle can be decomposed into a product of t.ranspositions by cuttin the digits in the cycle, so can each permutation :
(a l! c
p q) = (a b) (b c) ... (p q) .
(6.14
This decomposition is not unique. However, it is fixed for a given permuta tion whether the number of the transpositions in its decomposition is eve or odd. This conclusion can be proved by making use of the Vandermond determ in a nt. The Vandermonde determinant is defined as 1
'11[
Xl
1
1
nI .nI X2 ... Xn
The determinant chan ges its sign when the variables x J are made a transpo sition. \Vhen the variabl es Xj are made a permutation R , it is completel determined by R whether the determinant changes its s ign or not. O the other hand, R can be decomposed into a product of tra nspositions Whether the determinant changes its sign or not depends on \vhether th numb er of transpositions in the decomposition is odd or even. Thus, th conclusion is proved. A permutation is called even (or odd) if it is decomposed into a produc of eve n (or odel) transpositions. The product of two even (or two ode permutations is an eve n pe rmutation. The product of an even permutatio and an odd permutation is an odd permutation. The identical element E is an even permutation. The set of all even permu tations in a p ermutatio group Sn forms an invariant subgroup of Sn with index 2, call ed the alter nat ing subgroup of Sn, and the set of all odd permutations in Sn forms it cose t. The quotient group is isomorphic onto the inversion group V 2 . Thus S", n > 1, has an ant isymmetric representation, where th e representatio ma trix of R is called the permutation parity of R, denoted by
§6.2
Young Patterns, Young Tableaux and Young Operators
o(R)
=
{
~l
when R is even,
6.1.5
(6.15)
when R is odd.
The permutation parity of a cycle with length
237
eis (_I)el
Transposition of Two Neighbored Objects
Denote by Pa = (a a + 1) the transposition of two neighbored objects a and (a + 1). From the interchanging rule Pa satisfies
P; = E,
(6.16)
Each transposition can be expressed as a product of the transpositions of neighbored objects, so can a permutation,
=
d
PdPd+l ··· Pa2PalPa2 ... Pd+lPd ,
< a.
(6.17)
Denote by W a cycle of length n,
W
=
(1 2 . .. n),
W 1
=
Wn
1.
(6.18 )
As shown in Prob. 4 of Chap. 6 of [Ma and Gu (2004)], Wand PI are the generators of a permutation group Sn. The rank of Sn is 2.
6. 2
6. 2.1
Young Patterns, Young Tableaux, and Young Operators
Young Patterns
A class in the permutation group Srt is characterized by a partition of n, (£) = (e l , e2 , ... , em) with Lj e) = n. Since the number of the inequivalent irreducible representations of a finite group is equal to the number of its cl asses , therefore, the irreducible representation of Sn can also be described by a partition of n , denoted by
[A ] = [AI, A2, ... , Am],
m
(6.19) j=1
Note that the irreducible representation [A] has no relation witi1 the class (e), no matter whether the partitions are the same or not. Based
238
Chap. 6 Permutation Groups
on a partition [A], a Young pattern [A] (or called a Young diagram) can be defined. We will show later that a Young pattern [A] characterizes an irred uci ble representation of Sn. A Young pattern [A] consists of n boxes lined up on the top and on the left, where the Jth row contains Aj boxes. For example, the Young pattern [3,2] is
Note that in a Young pattern, the number of boxes in the upper row is not less than that in the lower row, and the number of boxes in the left column is not less than that in the right column. In order to emphasize the above rules for a Young pattern, a Young pattern is sometimes called a regular Young pattern in the literature. In fact, we will not be interested in an irregular Young pattern. A Young pattern [A] is said to be larger than a Young pattern [X] ifthere is a row number j such that Ai = A~, 1 :s: i < j, and Aj > Aj. There is no analytic formula for the number of different Young patterns with n boxes. However, one can list all different Young patterns with n boxes from the largest to the smallest. Namely, one lists the Young patterns in the order that A1 decreases from n one by one, then for a given A], A2 decreases from the minimum of A1 and (n  AIl one by one, third for the given AI and A2, A3 decreases from the minimum of A2 and (n  A1  A2) one by one, and so on. For example, the Young patterns for n = 7 are listed as follows:
6.2.2
[7],
[6, 1],
[5,2]'
[5,1,1]'
[4,3] ,
[4,2,1]'
[4,1,1,1]'
[3,3,1]'
[3,2,2] ,
[3,2,1,1]'
[3,1,1,1,1],
[2,2,2,1]'
[2,2,1,1,1],
[2,1,1,1,1,1]'
[1,1,1,1,1,1,1].
Young Tableaux
Filling n digits 1, 2, ... , n arbitrarily into the Young pattern [A] with n boxes, one obtains a Young tableau. There are n l different Young tableaux for a given Young pattern with n boxes. A Young tableau is said to be standard if the digit on the left is smaller than the digit on the right in the same row, and the upper digit is smaller than the lower digit in the same column. It is proved that the number dl.\l of the standard Young tableaux
§ 6.2 Young Patterns, Young Tableaux and Young Operators
239
for the Young pattern [AJ with m rows is (6.20) where rj
= Aj + m
 j. The square sum of the numbers diAl is n'
L
d[A](Sn)2
= n'.
(6.21)
[A]
The proof can be found in Chap. IV in [Bo ern er (1963)]. The formula (6.20) for calculating d[A] is not the simplest because there are some common factors between the numerator and the denominator. For example, d[3,2,) ,1] (S7)
= 7!
5x4x2 3x2 1 1 , x  ,  x I" x I"
x
6.
4.
2.
1.
= 35.
A simpler method for calculating diAl is called the hook rule. The hook number h ij of the box at the jth column of the ith row in a Young pattern [A] is equal to the number of boxes at its right in the ith row, plus the number of boxes below it in the jth column, and plus l. y~A] is the product of the digits filled in a tableau of the Young pattern [A] where the box in the j t h column of the ith row is filled with its hook number h i j . The number diAl of stan dard Young tableaux for the Young pattern [AJ is (6.22)
F
For comparison , d[3,2 ,),1] is recalc ulated from Eq. (6.22), 3
y [3 ,2,]
h
,I]
=
4
1
71
1
2
d[3,2,1,1](S7) =
'
6 x 3 x 4 x 2 = 35.
1
From the calculation one finds that two formulas are the same because for each row i, _l ~i
II h j= J
m
ij
= mil
{
II j = l+ l
(ri  rj)
}
l
Chap. 6 Permutation Croups
240
Compare the filled digits in two standard Young tableaux of a given Young pattern from left to right of the first row, and then those of the second row, and so on. For the first different filled digits, a smaller filled digit corresponds to a smaller standard Young tableau. Enumerate the standard Young tableaux of a given Young pattern [A] by an integer J from 1 to d[>,j' This in creasing order is the socalled dictionary order. The readers are suggested to learn how to arrange the standard Young tableaux of a given Young pattern in the dictionary order. For example, the orde of the standard Young tableaux of the Young pattern [3,2] is
CillIIJ [J]ITI] IT1IJ5l ITIIIIJ
[]J]J}] ~ ~ ~ ~ ~
Two Young patterns related by a transpose are called the associated Young patterns. The corresponding standard Young tableaux of two asso ciated Young patterns are also related by a transpose, but the larger Young tableau of one Young pattern becomes t he smaller of its associated Youn g pattern. The numbers of t he standa rd Youn g tableaux of two associated Young patterns are the same. 6.2.3
Young Operators
A permutation of the digits in the jth row of a given Young tableau is called a horizontal permutation Pj of the Young tabl eau, and a permutation of th digits in the kth column is called its vertical permutation Qk· The produc of horizontal permutations is also a horizontal permutation, denoted by P The product of vertical permutations is a vertical permutation, denoted by Q. The sum of all hori zontal permutations of a given Young tableau is called its horizontal opera tor , and the sum of a ll vertical permutation multiplied by their permutation parities is call ed its vertical operator. Th Young operator of a given Young tableau is th e product of P and Q:
II P = II (L PJ) , Q = L i5(Q)Q = L II i5(Qk)Qk = II [L
P
=L
P
=L
j
j
j
i5(QdQk 1 '
(6.23
k
Y=PQ
P a nd Q are also said to be the hor izo ntal permutation an d the vertica permutation of the Young operator Y, respectively. A Young operator i said to be standard if its Young tableau is standard. Since for a given
§6.2
Young Patterns, Young Tableaux and Young Operator·s
241
Young operator y, its Young tableau and its Young pattern are fixed, the Young tableau and the Young pattern are usually respectively called the Young pattern Y and the Young tableau Y for convenience. The symbol Y denotes the Young operator itself. Except for the id entical element E , no permutation can be both the horizontal permuta tion and the vertical permutation belonging to the same Young tabl eau. The readers are suggested to be familiar with writing the expansion of a Young operator of a given Young tableau. The following is an example:
ITITJ3l
[I]}J Y
= {E + (1 2) + (13) + (23) + (1 23) + (3 2 I)} . {E + (4 5)}{E  (1 4)}{E  (2 5)} = {E + (1 2) + (13) + (23) + (1 23 ) + (3 2 1) + (45) + (1 2)(45) + (1 3)(45) + (2 3)(4 5) + (1 2 3)(4 5) + (32 1)(4 5)} {E  (1 4)  (2 5) + (14)(2 5)} = {E + (1 2) + (1 3) + (2 3) + (1 23) + (32 1) + (45) + (1 2)(45) + (13)(45) + (23)(4 5) + (1 23)(4 5) + (3 2 1)(4 5)}  {(I 4) + (2 1 4) + (3 1 4) + (2 3)(1 4) + (2 3 1 4) + (32 14) + (5 4 1) + (2 1 54) + (3 1 5 4) + (2 3)(5 4 1) + (2 3 154) + (3 2 1 54)}  {(2 5) + (1 25) + (1 3)(2 5) + (325) + (3 125) + (1 325) + (452) + (1 245) + (1 3)(4 5 2) + (3 245) + (3 1 2 4 5) + (1 3 2 4 5)} + {(I 4)(25) + (1 4 2 5) + (3 1 4)(2 5) + (14)(325) + (3 1 4 25) + (1 4325) + (4 1 5 2) + (4 2)(15) + (4 3 1 5 2) + (3 2 4 1 5) + (4 2)(3 1 5) + (4 3 2)(1 5)}.
From the definition of a Young operator or from the above exampl e on e sees that a Young operator is a vector in th e group algebr a of t.he permutation group Sn,
y =
L
F(R)R,
(6.24)
RES"
F (E)
= F(P) = o(Q)F(Q) = o(Q)F(PQ) = 1.
(6 .25)
The coefficients F(R) are taken to be 1, 1, or O. Th e permutat ion R with nonvani shing coefficient F(R) in Eq. (6.24) is ca lled a permutation belonging to th e Young operator Y (and the Young tableau Y), and th e remaining permutations do not belong to y. We will discuss the criterion whether a permutation belongs to th e Young operator Y or not..
Chap. 6 Permutation Gmups
242
It is easy to determine a permutation S uniquely which transforms Young tableau Y to another Young tableau Y' with the same Young patter In fact, the first row of S is filled with the digits of the Young tableau and the second row of S is filled with the corresponding digits of the You tableau Y' in the same order. For example,
Young tableau Y
= 11f3T5l ~
11f2T3l
Young tableau Y' = ~
S=(113524) 2 3 4 5
.
Then, one has from the interchanging rule (6.10)
= Y',
SYS l
6.2.4
SPSl
= pi,
SQSl = Ql
(6.2
Fundamental Property of Young Operators
For a given Young operator y, the set of its horizontal permutations form a group, and the horizontal operator P is the sum of elements in the grou Similarly, the set of its vertical permutations forms a group, and the ve tical operator Q is the algebraic sum of elements in the group with the permutation parities. Thus, from the rearrangement theorem (Theore 2.1), one has
pp
= pp = P,
(6.2
QQ = QQ = o(Q)Q.
The Young operator Y satisfies pY
= o(Q)YQ = y.
(6.2
It is the fundamental property of a Young operator. Except for the norma ization condition F(E) = 1, Eq. (6.25) can be derived from this propert In fact, from Eq. (6.24) one has p1y
=
YQl
=
L L
F(R)Pl R
RES "
L = L
=
=Y =
F(SQ)S
= o(Q)Y =
SES"
F(R)RQl
RES"
S ES"
L
F(PS)S
S ES "
F(S)S,
L
o(Q)F(S)S
S ES"
Thus, F(S)
= F(PS) = o(Q)F(SQ) = 5(Q)F(PSQ).
(6.2
§ 6.2
Young Patterns, Young Tableaux and Young Operators
243
Letting S = E and assuming F(E) = 1, one obtains Eq. (6.25). Fock found another important property of a Young operator.
The orem 6.1 (Fock conditions) Let A and N be th e numbers of boxes in the jth row and in the j'th row of a Young tableau y, respectively. Denote by ap' the digits filled in the jth row and by bv those in the j'th row. If A ~ N, then (6.30) Let 7 and 7 ' be the numbers of boxes in the kth column and in the kith column of a Young tableau y, respectively. Denote by cp. the digits fllled in the kth column and by d v those in the k' th column. If 7 ~ 7 ' , then (6.31) The following example shows the filling positions of ap" bv , cp., and d v in a Young tableau, where j = k = 2 and j' = k' = 3: I
a1
a2 bv
a3
a4
I
as
I
Cl
I
C2
dv
I
I
I
I
C3
C4 Cs
Proof The proofs for the two Fock conditions are similar. In the following we will prove the condition (6.30) as example. Denote by (L Pj ) and (L Pj ,) the sums of horizontal permutations in the jth row and in the j'th row, respectively. pi is the sum of horizontal permutations not containing the digits in the jth row and the j'th row. Hence,
(L Pj )
is the totally symmetric operator for the A objects in the jth row, because it is the sum of AI different permutations of those objects. LeftmUltiplying it with { E + Lp. (ap. bv ) }, one obtains a sum of (A + 1)1 different permutations of the object bv and A objects in the jth row. Hence , the sum is the totally symmetric operator for those (A + 1) objects, denoted by
244
Chap. 6 Permutation Croups
I: Pj (b v ). When each term Pj' in the sum (I: P j ,) moves from the r side of I: P j (b v ) to its leftside, the only change of I: Pj (b v ) is that bv be replaced 'with another digit bp in the j' row,
(
I: PJ (b p )
is commutable with pl Since A 2' A', there exists an object the jth row which is located at the same column as bp . The transpos (a p bp ) preserves [I: Pj (b p )] invariant but changes the sign of Q, so th
Thus, Eq. (6.30) is proved. The key in the proof is that if the box number A of the jth row is less than the box number A' of the j' th row, Y is annihilated from the by symmetrizing bv in the j'th row with all al" in the jth row. Note tha (6.30) holds if j < j', but when j > j', it holds only if A = A'. Simi if the box number T of the kth column is not less than the box numb of the kith column, Y is annihilated from the right by antisymmetrizin in the kith column with all ell in the kth column. Equation (6.31) ho k < lei, but when k > lei, it holds only if T = Tl
6.2.5
Products of YOlmg Operators
For two Young operators Y and Y', one cannot derive YY' = 0 from Y 0, and vice versa. Two Young operators are called orthogonal only whe two products vanish simultaneously. The method used in Eq. (6.33) i typical one for proving that t he product of two Young op era tors vanis
Theorem 6.2 If there exist two digits a and b in one row of a Y tableau Y which also occur in one column of a Young tableau Y', or eq lently, if To = (a b) is both the horizontal transposition of a Young ope Y = PQ and the vertical transposition of a Young op erat.or Y' = pi Q',
Q'P
=0
and Y'Y
= o.
(
Proof Q'P = Q'ToP = Q'P = O. The subscript 0 indicates To to be a transposition. In proving th lowing corollaries, the key is to find the pair of digit.s.
§6.2
Young Patterns, Young Tableaux and Young Operators
Corollary 6.2.1 then Y'Y = o.
2
If a Young pattern Y' is less than a Young pattern Y
P roof Denote by [).'] and [A] the partitions of the Young patterns Y' an Y, respectively. Since the Young pattern Y' is less than the Young patte y, there is a row number j such that A; = Ai when i < j and Aj < Aj. Check the digits filled in the first (j 1) rows of the Young tableau Y see whether there exist two digits in one row of the Young tableau Y whi also occur in one column of the Young tableau Y'. If yes, from Theore 6.2, Y'Y = O. If no, there is a vertical permu ta tion Q' of the Young table Y' which transforms the Young tableau Y' to the Young tableau Y" su
that each row among the first (j  1) rows of both the Young tableau and the Young tableau Y" contains the same digits. Note that
Y" = Q'Y'Q'l = <5(Q')Q'Y',
Y' = <5(Q')Q'1 y".
Since Aj < A), there exist at least two digits in the jth row of the You tableau Y, which occur in one column of the Young tableau Y". Thu Y" Y = 0, and then, Y'Y = O. In this corollary the Young operators a not necessary to be standard. C orollary 6.2.2 For a given Young pattern, if a standard Young table Y' is larger than a standard Young tableau Y, then Y'Y = O.
Pro of Compare the filled digits in two standard Young tableaux Y an Y' from left to right of the first row, and then those of the second row, an
so on. If the first different filled digits occur at the jth column of the i row where the digit in the Young tableau Y' is a and the digit in the You tableau Y is b, then a > b. Thus, b has to be filled in the j'th column of t i'th row of the standard Young tableau Y', where j' < j and i' > i. T digits filled in the j'th column of the ith row of both the Young tableaux a nd Y' are the same, say c. Thus, the pair of digits band c occurs both t he same row of the Young tableau Y and in the same column of the You ta bleau Y', so that from T heorem 6.2, Y' Y = O.
C orollary 6 .2.3 For a given Young pattern, if the digits which occur one column of the Young tableau Y' never occur in the same row of t Young tableau Y, the permutation R transforming the Young tableau Y the Young tableau Y' belongs to bot.h the Young tableaux Y and Y'.
P roof Remind that the condition of Corollary 6.2.3 is equivalent to th the digits which occur in one row of the Young tableau Y never occur
246
Chap. 6 Permutation Groups
the same column of the Young tableau y. Let P be a horizontal permutation of the Young tableau Y which trans forms the Young tableau Y to the Young tableau Y" such that each column of both the Young tableau Y' and the Young tableau Y" contains the sam digits. Thus, the permutation transforming the Young tableau Y" to th Young tableau Y' is a vertical permutation Q" of the Young tableau Y" Q" = PQP 1, where Q is a vertical permutation of the Young tableau y Therefore, R = Q"P = PQ and R = RRR 1 = (RPR1)(RQR 1). 0 The converse and negative theorem of Corollary 6.2.3 says that if R transforming the Young tableau Y to the Young tableau Y' does not belon to y, there at least exists a horizontal transposition Po of Y which is als a vertical one of Y', Po = Q~ = RQoRl. Namely,
R
= PoRQo.
(6.35
By the way, the converse theorem of Corollary 6.2.3 holds obviously. In fact if R transforming the Young tableau Y to the Young tableau Y' belongs t y, then, RYR 1 = Y', R = PQ = (RQR 1)P, and the Young tableau Y" i transformed both from the Young tableau Y by a horizontal permutation P of Y and from the Young tableau Y' by a vertical permutation (RQR 1)1 of Y'. Thus, the digits which occur in one column of the Young tableau Y never occur in the same row of the Young tableau y.
Corollary 6.2.4 A permutation R does not belong to Y if and only there is a horizontal transposition Po and a vertical transposition Qo of Y such that Eq. (6.35) holds.
Proof From the converse and negative theorem of Corollary 6.2.3, if R does not belong to the Young tableau y, Eq. (6.35) holds. Conversely, R = PoRQo, from Eq. (6.29), F(R) = F (PoRQo) = F(R) = O. 0 6.3
6.3.1
Irreducible Representations of Sn
Primitive Idempotents in the Group Algebra of Sn
In §3.7 we introduced the method of finding the standard irreducible base bLv in the group algebra 12 of a finite group G in terms of idempotents. Th idempotent e a is a projective operator in 12, satisfying e~ = ea. The idem potents e a are called mutually orthogonal if ebea = 6baea. There are thre main theorems for idempotents. Corollary 3.7.1 says that an idempoten e a is primitive if and only if
§ 6.3 Irreducible Representations of Sn
247
Vi E L,
(6.36)
where At is a constant depending on i and is allowed to be O. Theorem 3.7 says that two primitive idempotents ea and eb are equivalent if and only if I.here exists at least one element S E G satisfying (6.37) Theorem 3.8 says that the direct sum of n left ideals La generated by the orthogonal idempotents eo, respectively, is equal to the group algebra L if and only if the sum of ea is equal to the identical element E: n
n
a=]
0=1
(6.38) We are going to show that the Young operator Y is proportional to the primitive idempotent of the permutation group Sn, and then, to calculate the irreducible representations of Sn. T heorem 6.3
If a vector X in the group algebra L of Sn satisfies
PX = o(Q)XQ = X for all horizontal permutations P and vertical permutations Q of X is proportional to the Young operator Y
(6.39)
Y, then
X=AY. P roof
Let X
= 2..=
(6.40)
F] (R)R. Similar to the proof of Eq. (6.29), from
RES"
Sq. (6.39) one has
Taking S = E and Fl (E) = A, one obtains
If a permutation R does not belong to Y, then from Corollary 6.2.4
In comparison with (6.25), Eq. (6.40) is proved. Since ytY satisfies Eq. (6.39), Corollary 6.3.1 follows directly.
0
248
Chap. 6 Permutation Groups
Corollary 6.3.1
For any vector t in the group algebra L of Sn, one ha
(6.4 where At is a constant depending upon t and is allowed to be O. Corollary 6.3.2
The square of a Young operator Y is not vanishing:
YY = AY]i O.
(6.4
Proof We are going to calculate the constant A explicitly. The right ide generated by Y is denoted by R = Y L. R is not empty because it contai at least a nonvanishing vector y. Let f ]i 0 be the dimension of R. Take complete set of basis vectors xJ1. in the group algebra L of Sn such that th first f basis vectors xJ1., fL :::; f, constitute a set of bases in R, and the la (nl  f) basis vectors xJ1., fL > i, do not belong to R. Since R is generate by Y, any vector in R, including its basis vector x,)' can be expressed as product of Y and another vector in R:
(6.4
Now, calculate the product YX 1L from two viewpoints. On one han xJ1. is a basis vector in L, and Y is an operator applying to xJ1." Thus, th matrix form D(Y) of Y in the basis vectors xJ1., n!
YXJ1. =
L
XVDVIL(Y)'
(6.4
v=1
is t.he representation matrix of Y in the representation D(Sn), which equivalent to the regular representation of Sn,. Then, Tr D(Y) = Tr D(E) = n!.
(6.4
On the other hand, since YXJ1. E Y L = R, the summation in Eq. (6.4 only contains the terms with v :::; i, when
v> f.
(6.4
When {i :::; f, from Eq. (6.43) one has YxJ1. = YYyJ1. = AYY/1 = AxJ Thus, DvJ1.(Y) = OVJ1.A when fL :::; f, and Tr D(Y) = fA. In compariso with Eq. (6.45), one obtains A = n!/ f ]i 0,
where
i
]i O.
(6.4
§6.:1 Irreducible Representations of Sn
C orollary 6.3.3 tion group Sn.
a = (f /n!)Y is a primitive idempotent of the
p cnnlll il
C orollary 6.3.4 If the digits in one column of the Young tableau Y' never occur in the same row of the Young tableau y, then Y'Y oj::. O.
P roof From Corollary 6.2.3, the permutation R transforming the Young tableau Y to the Young tableau Y' belongs to the Young tableau y. Thus, Y' = RYR 1 , R = PQ, and Y'Y RYQ1p1y = o(Q)RYY = o(Q)ARY oj::. O. [J
C orollary 6.3.5 Two minimal left ideals generated by the Young operators Y and Y', respectively, are equivalent if and only if their Young patterns are the same as each other.
P roof If the Young patterns Y and Y' are the same, there exist a permutation R transforming the Young tableau Y to the Young tableau Y' such that Y' = RYR 1 , and Y'RY = RYY oj::. O. From Theorem 3.7, two left ideals are equivalent. Conversely, if two Young patterns Y and Y' are different, without loss of generality, the Young pattern Y is assumed to be larger than the Young pattern Y'. For any permutation R, the Young pattern Y", where Y" = RYR 1 , is the same as the Young pattern y. Then, due to Corollary 6.2.1 Y'Y" = 0 and y'RY = Y'Y"R = O. 0 Therefore, an irreducible representation of Sn can be characterized by a Young pattern [Aj. Two representations denoted by different Young patterns are not equivalent to each other. Thus, the following Corollary follows Theorem 3.7. Corollary 6.3.6. Two Young operat.ors corresponding to differen Young patterns Y and Y' are orthogonal to each other, YY' = Y'Y = O. The number of different Young patterns is equal to the number of partitions of n, which is equal to the number gc(n) of classes in Sn. Hence, the irreducible representations denoted by all different Young patterns with n boxes constitute a complete set of the inequivalent and irreducible representations of Sn.
6.3.2
Orthogonal Primitive Idempotents of Sn
Two Young operators corresponding to different Young patt.erns are orthogonal to each other. However, two standard Young operators corresponding to the same Young pattern are not necessary to be orthogonal. The non
250
Chap. 6 Permutation Groups
orthogonal standard Young operators occur only for Sn with n 2: 5. Fo n = 5 there are two Young patterns, [3,2] and [2,2,1]' where some standar Young operators are not orthogonal to each other. For example, list th standard Young tableaux for [3,2] from the smallest to the largest:
Due to Corollary 6.2.2, Y/LYII = 0 when f..L > v. Check the product YIIY/ with f..L > v one by one whether two digits in one row of the Young tablea Y/L occur in the same column of the Young t.ableau YII' If no , YIIY/L f::. The result is that only
(6.48
The permutation R IS transforming the Young tableau Y5 to the Youn tableau Yl is 13524) R 1S = ( 12345 =(3245)=(24)(453) = (2 4) (53) (34) =
= (4 5) (3
2) (2 5)
(6.49
h Q5
= Pj
Q 1,
where P5 = (24) (53), Q5 = (34), PI = (4 5)(3 2), and Q1 = (2 5). Th decomposition of R IS in Eq. (6.49) is a typical technique for the decompo sition of the permutation between two nonorthogonal Young operators. For a given Young pattern, we want to orthogonalize the standard Youn operators by leftmultiplying or rightmultiplying them with some vecto y" . in L. In the above example, there are two sets of orthogonal Youn operators. One set is
u> 1.
(6.50
Since YIPS = YI R I5 Qi 1 = 8(Qs)R1s Ys, Y;Y/L = YIY/L when f..L < = Y1(E  P s )Y5 = O. The other set is
Y; Ys
Y~'
= [E + Qr] Ys,
Y~ =Y/L'
f..L
< 5.
(6.51
Since QIYS = P I 1 R I5 YS = Yt R 1S , YIIY~' = YIIY5 when u > 1. YIY~' YdE + Ql)YS = O. Generally, for a given Young pattern [,\] where the standard Young op erators are not orthogonal completely, we want to choose some vecto
y1"j
§6. 3 Irreducible Representations of Sn
251
YL.\] such that the new set of YL.\JYL.\J or the new set of YL.\] YL.\J are mutually orthogonal. We will discuss the first set in detail and give the result for the second set. Since the Young pattern [AJ is fixed, in the following we will omit the superscript [AJ for simplicity. The problem is to find YJ1. such that
YL.\] or
1 ~ p, ~ d,
1
< v < d.
(6.52)
Then, YJ.l.YI1 Yv = 6J.1.vYJ.l.Yw Denote by RJ.l.v the permutation transforming the standard Young tableau Yv to the standard Young tableau YJ.I.'
RJ.l.vYv
= YJ.l.R/LV'
RJ1.vPv
Rl1pRpv = RJ.l.v,
= PJ.l.RJ.l.v ,
(6.53)
RJ.I." =E.
Due to Corollary 6.2.3, (6.54) where PS") and Q~") are the horizontal permutation and the vertical per(v)
mutation of the Young tableau Yv, respectively, and P J.I. those of Y w Let
Obviously, when YJ.l.Yv
when YJ.l.Y"
i: 0,
when YJ.l.Yv
= 0.
(v)
and Q J.I.
are
(6.55)
i: 0,
PJ.l.vQ" = RJ.l.I/Qv
(Q~)) 1
= QJ.l.PJ.l.I/'
(6.56)
PVPJ.l.V = PJ1. V P V = P v , YJ1.PJ1.V = RJ1.vYv Define Y/, one by one from p,
(Q~)) 1
=
=d
= 1,
to
P,
6(Q~))RJ1.vYv.
d
YJ1. = E 
L
PJ1.PYP'
Yd = E,
1~
P, ~
d.
(6.57)
p=J1.+1 It is easy to show by induction that Eq. (6.52) holds. As a matter of fact, Eq. (6.52) holds when p, = d owing to Corollary 6.2.2 . Suppose that Eq. (6.52) holds for p, > T. For p, = T, Eq. (6.52) also holds because
d
QTYTP V = QTP V
rl
L

= QTP
QTPTPYpP,/
V
L

PTPQpYPp"
p=T+I
when v < when v = when v>
T, T,
T.
Note that YI" is the algebraic sum of elements of Sn with the coefficients ±l, and due to Eq. (6.56) d
d
6(QSt'))R/1vYvYv =
L
tpYp,
(6.58)
P=I'
where the sum index v in the middle expression runs over from fJ. + 1 to d and in the condition YI"Yv # 0, and tp in the last expression is a vector in [. which is allowed to be zero. Similarly, let
# 0, YI"Y'/ = O.
when YI"Yv when
VI.' are defined one by one from v = 1 to v
(6.59)
= d,
vI
VI.'
=E
L
1 ~ v ~ d,
V/Jpv,
(6.60)
p=1
such that YI"VvY'/
= 6l"vYvYv.
Theorem 6.4 The following eL>'] const.itute a complete set of orthogonal primitive idempotents: e[>']
=
I"
d[>.] y[>']y[>']
n!
/1
(6.61)
1"'
and the identical element E can be decomposed as E
= ~!
diAl
L [>']
Proof
d[>.] L
yl>']yl>']·
(6.62)
/,=1
For a given Young pattern [AJ, there are d[>.] orthogonal primitive
idempotents e~;], given in (6.61) but replacing dl>'] with 1[>.], where 1[>.] is the dimension of the left ideal .c\;\] generated by eL>'] (see Theorem 3.10). Since the multiplicity of each irreducible representation in the reduction of
!i 1i.:J lr'TCI!w;ible IlcTirescntalions of S"
t.he regular representation is equal to the dimension of the representation, one has diAl ~ J[A]' On the other hand, the square sum of the dimensions of the inequivalent and irreducible representations of a finite group G is equal to its order, " " J[A] 2 ~
(6.63)
=n ,..
[A]
In comparison with (6.21) one obtains (6.64)
eh
Therefore, the direct sum of the left ideals L~] generated by A] is equal to the group algebra L, and Eq. (6.62) follows Theorem 3.8. 0 In the same reason, AI also constitute a complete set of orthogonal primitive idempotents:
eh
e[AI I"
= d[A]y[A]Y[AI n! IJ. IJ.'
(6.65)
and the identical element E can be decomposed as 1 E = 'I "" d n. ~ '["] [A]
6.3.3
diAl
"".y',. ~
~;] Y,[.,A] .
(6.66)
1"=1
Calculation of Representation Matrices for Sn
For a given Young pattern [>'J, we are going to choose a set of standard bases b~J and calculate the representation matrices. Since the Young pattern [>.J is fixed, we omit the superscript [>.J for simplicity. In terms of the permutations RIJ.II given in Eq. (6.53), where Rw transforms the standard Young tableau Yv to the standard Young tableau YIJ.' d 2 basis vectors bl1v can be defined: bl"V
= el"Rl1v ev = (din!)
2
YIJ.YI"R I11I YIIY,1
'J
= (dln!r YIJ.YIJ.YIJ. RIJ.IIY,I
= (din!) YI"RIJ.vYv = (din!) RIJ.IIYIIYII
(6.67)
Those basis vectors b,.LII are standard because they satisfy the condit.ions (3.130), (6.68)
Chap. 6 Permutation Groups
254
For a given lI, d basis vectors bJ1.v are the complete bases in the left idea Lv, and for a given f.l, d basis vectors bJ1.v are that in the right ideal RJ1 In the standard bases, the representations of both the left ideal Lv and th right ideal RJ1. are the same, d
d
SbJ1.'/ =
L
bpv DpJ1.(S),
bJ1.v S
=L
p=l
D vp (S)bJ1.p.
(6.69
p=l
Replacing 1I with T in the first equality of Eq. (6.69) and leftmultiplyin it with bTl" one obtains
(6.70
where T is arbitrary, 1 :::; T :::; d. Due to Corollary 3.7.1, the righthan side of Eq. (6.70) is proportional to e T . The representation matrix of an element S of Sn in the representation [t\] is calculated from Eq. (6.70). In calculating DV/i(S), one has to move out the quantity YvS betwee two Young operators in Eq. (6.70) such that two Young operators reduc to one Young operator. Yv is an algebraic sum of group elements with th coefficients ± 1 and can be expressed as follows formally,
Yv =
L
6k T k,
(6.71
k
where Tk is a permutation and 15 k = ± 1. Denote by Yvk the Young tablea transformed from the Young tableau Yv by (Tk)~l, and by YJ1.(S) the Youn tableau transformed from the Young tableau YJ1. by S. Hence, Eq. (6.70 with T = 1 becomes
DV/i (S)el =
L
15 k
(djn!)2 R1vTkYvkY/i (S)SR/ilYI .
(6.72
k
Now, we calculate the product of two Young operators. If two digits in on row of the Young tableau YJ1.(S) occur in the same column of the Youn tableau Yvk, the product YvkYJ1.(S) is vanishing. If the digits in one row of the Young tableau YJ1.(S) never occur in the same column of the Youn tableau Yvk. from Corollary 6.2.3, the permutation transforming the Youn tableau YJ1.(S) to the Young tableau Yvk belongs to YJ1.(S),
The quantity in the bracket, denoted by Qvk. is a vertical permutation o the Young tableau Yvk' (Qvk)~l transforms the Young tableau Yvk to th
§ 6. 3 Irreducible Representations of Sn
25[,
Young tableau Y' such that the digits in each row of the Young tableau also occur in the same row of the Young tableau YfJ,(S). Hence,
Y'
(din!) YVkY~L(S)
= (din!) Yvk6(Qvk)QvkPfJ,(S)YfJ,(S) = (din!) 6(Qvk)QvkPfJ,(S)YfJ,(S)YJL(S) = 6(Qvk)Qvk PfJ,(S)YfJ,(S),
Substituting it into Eq. (6.72), one obtains DVfJ,(S)el =
L
6k6(Qvk) (dln!){RlvTkQvkPfJ,(S)SRfJ,d YI Yl·
(6.73)
k
The product of permutations in the curve bracket of Eq. (6.73) has to be equal to the identical clement E because the righthand side of Eq. (6.73) is proportional to e 1 = (dln')YI Yl' In fact, RfJ,l first transforms the Young tableau Yl to the Young tableau YfJ,' Then,S transforms the Young tableau YfJ, to the Young tableau YfJ,(S). Third, QvkPfJ,(S) transforms the Young tableau YfJ,(S) to the Young tableau Yvk. Fourth, Tk transforms the . Young tableau Yvk to the Young tableau YV' At last Rlv transforms the Young tableau Yv to the original Young tableau Yl. Namely, the product of permutations preserves the Young tableau Yl invariant so that it is equal to the identical element E. Hence, DvfJ,(S) =
L
6k J(Qvd·
(6.74)
k
It means that in the standard basis vectors bfJ,v, the representation matrix entry DvfJ,(S) is always an integer, so that each irreducible representation of Sn is real and its characters are integers. Equation (6.74) provides a method for calculating DvfJ,(S), Due to Eq. (6.71) Jk is known. 6(Qvd can be calculated by comparing two Young tableaux Yvk and Y~l(S), If two digits in one row of the Young tableau Y!,(S) occur in the same column of the Young tableau Yvk, J(Qvk) = O. Otherwise, from Corollary 6.2.3, there is a vertical permutation Q;;/; of Yvk that transforms the Young tableau Yvk into the Young tableau Y' such that the digits in each row of the Young tableau Y' also occur in the same row of the Young tableau YfJ,(S). J(Qvd is the permutation parity of Q;;A1. The matrix entry DL~(S) of 5 in the irreducible representation [A] of Sn can be calculated by the tabular method as follows. Denote by YfJ,(S) the Young tableau transformed from the standard Young tableau yr'] by the permutation S. List YJL(S), 1 ::; f.1 ::; d[>,], on the first row of the table
256
Chap. 6 Permutation Groups
to designa te its columns. Let yU'1 = Lk OkTk· Denote by Yvk the Young t.ableau transformed from the standard Young tableau y!/l by the permutation T k l . List the sum of the Young tableaux Lk OkY,/k, 1 :::; v :::; d[A], on the first column of the table to designate its rows. The representation matrix entry DL~(5) is equal to Lk okA~k(5), which is filled in the Jith column of the vth row of the table. A~k (5) is calculated by comparing two Young tableaux Y"k and YJl(5). A~k(5) = 0 if there are two digits in one row of the Young tableau YJl(5) which also occur in the same column of the Young tableau Yvk. Otherwise, A~k (5) is the permutation parity of the vertical permut.ation of the Young tableau Y//'\;' which transforms the Young tableau Y"k to the Young tableau Y' such that the digits in each row of the Young tableau Y' also occur in the same row of the Young tableau Y1I (5). The sum of the diagonal entries in the table is t.he character Xl),] (5). The calculated irreducible representat.ion DI)'I(5) is generally not unitary. Table 6.1 Tabular method for calculating the irreducible representation matrix DL~ (5) of Sn
[AJ = [3,2]'
5
= (1
2345), k
T k 1 transforms the Young tableau y" to the Young t.ableau Yuk, 5 transforms t.he Young tableau YJl to the Young tableau YJl(S). Young tableau YI' (S)
LOk
{Young tableau Yvk}
A·
123 45 1 2 35 12 34 13 25 13 24
145 23 4 5 4 5
234 5 1
235 4 1
23 1 4 5
245 3 1
24 1 35
1  0
01
1 0
0+1
00
1
0
0
0
1
0
1
0
0
0
1
0
0
1
0
0
1
0
1
0
As example, in Table 6.1 the representation matrix DI 3 ,2](S) of S5, where 5 = (1 234 5), is calculated by the tabular method:
257
§6·.3 Irreducible Representations of SOn
Di 3 ,2j [(1 2345)]
[1 1 = o
1 0 1 1 0 o 1
1 1 0 0
0 0 0 1 0 1
1)
Di 3 ,2j [(1 2)] and D1 3 ,2j [(5 4 3 2 1)] can be similarly calculated
[ ~ n~: ~:) [H~:~: l) . o o
and
0 0 1 0 0 0 0 1
0 0 1 0 1 0 1 1 1 1
Then, the representation matrices for the clements in each class are calculated by the products
(2 3) = (1 2 3 4 5)(1 2)(5 4 3 2 1), (3 4) = (1 2 3 4 5)(2 3)(5 4 3 2 1), (45) = (1 2345)(34)(5 432 1),
(1 23) = (1 2)(2.3), (234 5) = (1 2)(1 2345), (1 2)(34), (1 23)(45).
The characters for the representation are listed in Table 6.2. Table 6.2.
6.3.4
Character table for representation [3,2] of S5
Calculation of Characters by Graphic Method
The representation matrices of 8 n , as well as the characters in the representation, can be calculated by the tabular method. However, there is a graphic method for calculating the character of a class (£) in the irreducible representation [A]. The integers ej of the partition (e) can be arranged in any order, but in the increasing order, e1 :S e2 :S ... em, will simplify the calculation. According to the following rule, we first fill £1 digits 1 into the Young pattern [AJ, then fill £2 digits 2, and so on, until filling em digits m.. The filling rule is as follows:
:s
(a) The boxes filled with each digit, say j, are connected such that from the lowest and the leftmost box one can go through all the boxes filled with j only upward and rightward.
258
Chap. 6 Permutation Groups
(b) Each time when ej digits j are filled, all the boxes filled with digits i ::; j form a standard Young pattern, namely, the boxes are lined up on the top and on the left such that the number of boxes in the upper row is not less than that in the lower row and there is no unfilled box embedding between two filled boxes.
It is said to be one regular application if all digits are filled into the Young pattern according to the rule. The filling parity for the digit j is defined to be 1 if the number of rows of the boxes filled with j is odd, and to be 1 if that is even. The filling parity of a regular application is defined to be the product of the filling parities of m digits. The character XIAI [(e)] of the class (e) in the representation [A] is equal to the sum of the filling parities of all regular applications . XIAI [(e)] = 0 if there is no regular applicat ion, namely, if m digits cannot be filled in the Young pattern, all according to the above rule. For the class (1 n) composed of only the identical element E, each regular application is just a standard Young tableau, so its character is nothing but the dimension of the representation. Usually, the character of the class (In) is calculated by the hook rule instead of the graphic method. In Table 6.3 all the regular applications of each class for the Young pattern [3,2] are listed, and their characters are calculated. The five regular applications of the class (1 5 ) are omitted . Table 6.3 The character table for the representation [3,2] of S5 calculated by the graphic method Class
(1 ")
Regular application
(1 s, 2)
(1, 2~)
W,3)
(2,3)
(1,4)
123
122 33
133 23
1 22 I 2
122 22
44
Filling parity
xI3 ,2] [(e)l
5
1
1
 1
1
1
1
1
1
1
1
(5)
0
If one changes the order of ej for (e), there may be more regular applications, but the calculated results of the characters are the same. For example, if one changes the order for the class (1,2,2) to be (2,2,1), there are three regular applications in the Young pattern [3,2]:
Filling parity
=
113 122 123 22 13 12 1 , (1) , 1
X(I,2 2 )
= 1  1 + 1 = 1.
§ 6. ,1 Irreducible Representations of Sn
259
The Young pattern [n] with one row has one standard Young tableau, and the corresponding Young operator is the sum of all elements in 5n . Thus, [n] characterizes the identical representation, where the representation ma trix of each element in 5n is l. The Young pattern [In] with on e column also has one standard Young tableau, and the corresponding Young operator is the sum of all elements in 5n , multiplied with their permutation parities. Thus, [In] characterizes the antisymmetric representation, where the representation matrix of each element R is its permutation parity c5(R). Denote by [5.] the associate Young pattern of [A]. According to the graphic method, the characters of a class (f) in two associate Young patterns differ only with a permutation parity c5[(£)] of the elements in the class, (6.75)
If [5.] = [A], the Young pattern [A] is called selfassociate, and the character of a class (f) with odd permutation parity is O. In fact, the transpose of each regular application of the Young pattern [A] is a regular application of the associate Young pattern [5.], where the positions of each digit, say j, are the same as each other except for the interchanging of rows and columns. The sum of the row number and the column number of boxes filled with the digits j in the Young patterns is (£j + I), so that the product of two filling parities for the digit j in the two Young patterns is equal to the permutation parity (_l)tj+l of a cycle with length f j .
6.3.5
The Permutation Group 8 3
As an example, we calculate the standard bases and the inequivalent irreducible representations of 53 by the Young operator method . 53 is isomorphic onto the symmetric group D3 of a regular triangle. There are six elements and three classes in 53. The class (1 3 ) contains only the identical element E. The class (2,1) contains three elements, A = (2 3), B = (3 I), and C = (1 2). The class (3) contains two elements, D = (3 2 1) and F = (1 2 3). There are three Young patterns for 53. The Young pattern [3] characterizes the identical representation, where the representation matrix of any group element is equal to l.
y131
I
1
I2 I3 I
bPI = e l31 = {E + (1 2) + (2 3) + (3 1) + (1 23) + (3 2 I)} /6. For the Young pattern [2,1]' there are two standard Young tableaux.
260
Chap. 6 Permu.tation Groups
The representation [2 , 1J is twodimensional.
ITI:IJ ~
[]]JJ
Y2[2,li
and
~
The idempotents and the standard basis vectors are
b~21,li
= e\2,IJ = {E +
b~~,lJ
= (2
3)e\2,1] = {(2 3) + (321)  (2 3 1)  (2 I)} /3,
b\2/ J = (2 3)e~2,1]
b~22,lJ
(12)  (13)  (2 1 3)} /3,
= {(2 3) +
= e~,l] = {E + Table 6.4.
(2 31)  (321)  (3 I)} /3,
(13)  (1 2)  (3 1 2)} /3.
The representation matrices of generators in [2, 1] of S3
Yv
5 = (1 2) 21 23 1 3
12 3 13 2
5' = (1 2 3) 23 2 1 1 3
1
1
1
1
0
1
1
0
The representation matrices of the generators (1 2) and (1 2 3) ar calculated by the tabular method (see Table 6.4): D[2,lJ [(1 2)J
= (~ =~),
D[2 ,li[(12 3)J = (1 1).
1 0
It is not a real orthogonal representation because the standard basis vector are not orthonormal in the group algebra. Through a similarity transfor mation X, the new basis vectors are orthonormal and the representatio becomes real orthogona.i (see Eq. (2.12)) .
1(33 V3) V3 '
XI D[2,li(R)X = D(R),
X=2
1 ( 1 D[(12)J=2 V3
V3)
cPl = (3/2) (b l l + b2t}
= {E +
cP2
= (V3/2) (b ll
1
'
D [(1 2 3)J
=~
(_~ ~),
(2 3)  (3 1)  (1 2 3)} /2,
+b21 )
= {E + (23) + (31)  2(12)  (123) + 2(32 I)} /(2V3).
§6. 3 Irreducible Representations of Sn
,/ 1
I
The Young pattern [1,1,1] characterizes the antisymmetric represellt.1I tion, where the representation matrix of any group element R is equal t.o its permutation parity r5(R).
bll,l,l]
6.3.6
=
ell,l,l]
= {E 
(1 2)  (23)  (3 1)
+ (1
23)
+ (32
I)} /6.
Inner Product of Irreducible Representations of Sn
The direct product of two irreducible representations of Sn is specially called the inner product, because there is another product called their outer product (see the last section in this chapter). The inner product is usually reducible and can be reduced to the ClebschGordan series by the character formula (3.54).
Xl>] (R)X[,L] (R)
a>pv =
~ n.
L
=
L
v
a>,.wX[v] (R),
(6.76)
v
X[>I(R)XI'~](R)Xlvl(R).
RES n
Since the characters in the irreducible representations of Sn are rcal, a>pv is totally symmetric with respect to three subscripts. This property can be used to simplify the calculation of the ClebschGordan series. Noting Eq. (6.75) , one has
[n] x [A]
"=
[A],
[A] x
[J.L] "= [).]
x [iLl.
(6.77)
Due to Eq. (6.76) one concludes that there is one identical representation [n] in the reduction of [A] x [J.L] if and only if [A] = [~i], and there is one antisymmetric representation [In] in the reduction of [A] x [J.L] if and only if [A] = [iLl. For the group S3 one has
[3] x [31
"=
[1 3 ] x [1 3 ]
"=
[3],
[3] x [1 3 ]
[1 3 ] x [2,1] "= [2,1]' [2,1] x [2,1] "= [3] EB [1 3 ] EB [2, 1]. [3] x [2,1]
"=
"=
[1 3 ], (6.78)
Some results for the ClebschGordan series can be found in Prob. 31 of Chap. 6 of [Ma and Gu (2004)].
262
Chap. 6 Permutation Groups
6.4
Real Orthogonal Representation of Sn
The merit of the tabular method for calculating the representations of Sn i that the basis vectors are well known and the representation matrix entrie are integers. Its shortcoming is that the calculated representation is not rea orthogonal. In this section we are going to show a method to combine the basis vectors such that the new representation [A] is real orthogonal [Tong et al. (1992)]. In this section we neglect the superscript [A] for simplicity because the representation [A] is fixed, In the group algebra L of Sn, a transposition (a d) is unitary and Her mitian. Introduce a set of Hermitian operators M" in L: aI
Ma =
L
d=l
aI
(a d) =
L
Pa J P a 2
···
Pd+IPdPd+I'" P,,2 Pal,
d=1
2 ::; a ::; n,
(6.79
MJ = 0,
where P" = (a a+ 1) is the transposition of two neighboring objects. From the definition one has
(6.80 It is easy to show from (6.15) that if b < a or b > a + l.
(6.81
Then, the Hermitian operators Ma are commutable with each other
(6.82)
Theorem 6.5 In the standard bases bvp (6.67), the matrix form D(Ma) of Ma is an upper triangular matrix with the known diagonal entries,
(6.83) when J1 > v,
where the rows and columns are enumerated by the standard Young tableaux Yv in the increasing order. If a is filled in the cv(a)th column of the Tv(a)th row of the Youn!?; tableau Yv, mv(a) = cv(a)  Tv(a) is called the content of the digit a in the standard Young tableau Yv.
§6·4 Real Orthogonal Representation of Sn
M~ll=L (a ai) , i
M~3) = 
Mfl=L (a bj
)
263
+L
j
L
M~4) =
(a dd,
L
(a d k )
,
k
(a if) ,
k
where ai denotes the digits filled in the boxes on the left of the box a at the Tv(a)th row of the standard Young tableau Yv, bj and dk denote the digits, respectively smaller and larger than a, filled in the boxes at the first [TV (a)  1] rows, and te denote the digits less than a and filled in the lower rows than the Tvth row. From the symmetric property (6.28) and the Fock condition (6.30) one has
Ma(2) bvp _
{
1  Tv(a) } bvp .
When applying each transposition in lvIi 3 ) and M~4l to the Young tableau Yv, a smaller digit in a lower row is interchanged with a larger digit in the upper row. Although the transformed Young tableau is generally no longer standard, it can be proved in terms of the similar method used in the proof for Corollary 6.3.2 that when f.l;::: v. Noting (6.58) one has (3)
YJly"Ma Yv (3)
= YJLYJlMa(4) Yv = 0, 1(4)
when f.l;::: v.
bpJl1I1 a bvp = bpJlM a bvp = 0,
o
Letting X be a similarity transformation which changes the representation D(Sn) to a real orthogonal representation D(Sn) such that the representation matrices of lvI" are diagonal,
DJlv(Ma) = [X 1 D(Ma)X]JlV = 6Jlv m v(a), D(Pa ) = X 1 D(Pa)X = D(Pa)* = D(Pa)T.
(6.84)
Because there are no two different standard Young tableaux YJl and Yv satisfying m!L(a) = mv(a) for every a, the eigenvalues mv(a) of M" are not degenerate. In other words, the set of Ma, 2 ::; a ::; n, is a complete set of the Hermitian operators in the group algebra L of Sn. Since D(J11a) are real upper triangular matrices and D(Ma) are real diagonal, X has to be
264
Chap. 6 Permutation Groups
a real upper triangular matrix. The new basis vectors, called orthogon bases, are calculated in terms of X matrix d
¢p.v =
d
L
bpvXpp. E Lv,
R¢p.v
¢ pvDpp.(R),
p=l
p=l d
¢p.v =
=L
(6.8
d
L
(X  l)vpbp. p E Rp.,
¢p.v R
=L
Dvp(R)¢p.p,
p= l
p=l
or d
Pp.v = Rpp.v ==
L
d
LL
(XI) VT bpTXP1L>
T=l PpvDpp.(R), p=l
P!W R ==
(6.8
L
Dvp(R) Pp.p.
p
p
D(Pa) as well as X can be calculated from Eqs. (6.80), (6.81), a (6.84). In fact , substituting Eq. (6 .84) into Eq. (6.81), one has if b < a or b > a + 1,
(6.8
namely, Dp.v(Pa) i= 0 only if mp.(b) = Tnv(b) for every b except for b = and b = a + l. If a and (a + 1) do not occur in the same row and in t same column of the standard Young Tableau Yv, another standard You tableau, denoted by Yv a, can be obtained from the Young tableau Yv interchanging a and (a + 1) ,
Tn,/ (a)
= Tnl/" (a + 1), ITnv(a)  mv(a
Tnv(a + 1)
+ 1)1 >
= Tnvo (a),
(6.8
1.
The rela tion between v a nd V(L is mutual. On the other hand, if a a (a+ 1) occur in the same row or in the same column of the standard You tableau Yv, the Young tableau obtained by the interchanging is no long standard. In these cases we will say that Va does not exist for the standa Young tableau Yv. Therefore, Dp.//(Pa ) i= 0 only if J.L == v or J.L = Va, so that D(P,,) is block matrix. The submatrix Dvv (P(L ) is onedimensional if Va does n exist and is twodimensional if Va. exists,
(6.8
§ 6·4 Real Orthogonal Repre se ntation of Sn
where, without loss of generality, we assume II < lIa, namely, a. occurs al. I.he right of and upper than (a. + 1) in the standard Young tableau YV. Substituting Eq . (6.84) into Eq. (6.80), one obt.ains
The nondiagonal entries of the twodimentlional submatrix can be calculat ed from P; = E,
It is proved that the square root can be positive by choosing the phase angles of the basis vectors and noting the condition (6 .16). Thus, the onedimensional sllbmatrix is
+ 1) occur (a + 1) occur
a and (a
in the same row ,
a and
in the same column,
(6.90)
an d the twodimensional submatrix (6.89) becomes m
= mv(o.)
 mv(o.
+ 1) >
I, (6.91)
where m is equal to the steps of going from a to (0.+ 1) in the Young tableau
Y,/ downward or leftward. Equations (6.90) and (6.91) give the calculation method for the real orthogonal representation matrix D(Pa ) in [A]. It is easy to calculate the similarity transformation matrix X from D(Pa ) and D(Pa ). In the following we calculate the similarity transformation matrix X for the representation [3,2] of 8 5 as an example. The standard Young tableau x with the Young pattern [3,2] are listed 1 from Y 1[32 to Y 5[321 as follows: ' ' 123
124 35
4 5
125 34
135 24
134 25
The representation matrices D(Pa ) can be calculated by the tabular method (see §6.3)
D(Pr)
=
[!
0 1 0 0 0
1)
Q1 0 1 0
1 0 1 01 0 o 0 1
D(P,)~ [!
0 0 0 1 0
0 0 0 0 1
0 1 0 0 0
!J.
Chap. 6 Permutation Groups
266
D(P3 )
=
[!
1 0 0 0 0
0 0 1 0 0
0 0 0 1 0
1) 1 1 1 1
,
~ [!
D(P,)
o 00) o0 100 1 0 0 0 000 1 o0 1 0
.
The orthogonal representation matrices D(Pa ) are calculated from E (6.90) and (6.91). For example, in the calculation of D(P2 ) one need check the positions of 2 and 3 in the Young tableaux Yv' In the intercha ing of 2 and 3, the Young tableau Y2 is changed to the Young tableau with m = 2, and the Young tableau Y3 is changed to the Young tableau with m = 2, so that D(P2 ) is a block matrix with one 1 x 1 submatrix two 2 x 2 submatrices.
o1 100000 0 0) D(Pd = [ 0 0 1 0 0 o 0 0 1 0 o 0 0 0 1
1 ,
1 VB 0 0 0) VB100 0 = 3" [ 0 0 3 0 0 , o 0 0 3 0 o 0 0 0 3
D (P2 )
=2
1
1
D(P3 )
2 0 0 0 1 0 [ 0 0 1 o J3 0 o 0 J3
D (P4 ) =
2
0 J3 0 1 0
0) 0
J3 , 0 1
2 0 0 0 0) 0 1 v'3 0 0 [ 0 v'3 1 0 0 o 0 0 1 v'3 000J31
The similarity transformation matrix X, D(Pa)X = X D(Pa), is upper triangular one, whose column matrices X!l are denoted by
(0) 1
one obtains al =
l/VS and
a2 =
3/VS.
= _
From
~ 3
(1) + VB (at) , 0
.3
a2
§ 6.4 Real Orthogonal Representation of Sn
one obtains d 1 = 3/VS, d 2 = 1//8, d 3 = d4 = last, the similarity transformation matrix X is
267
l/V2, and
d.5 =
V2.
At
3)
VS 1 V3 V3 103V3V31 X = /8 [ 0 0 2V3 0 2 8 0 0 0 2V3 2 o 0 0 0 4
(6.92)
.
It is easy to check that X satisfies (6.84). The orthogonal bases
/8el,
J3 {E + (3 4) + 2(35 4)} el,
similarity
transformation
matrices
X[A],
D[A](Pa)X[A]
=
X[A]D1A ](Pa), for the representations [3,1]' [2, I, 1], [2,1]' and [2,2] are calculated in Probs. 2224 of Chap. 6 of [Ma and Gu (2004)], X[2,1] = X[2,2]
=
(~
CJl),
268
Chap. 6 Permutation Croups
X [3,lj
=
J3/8) ' (1yT78 J378 0 3/.;8
o
6.5
0
/372
X[2,],] ]
=
(1
yfJJ3
0 2/.)3 0
0
Jf16) Jf16 fi72
Outer Product of Irreducible Representations of Sn
The outer product of representations of the permutation groups is relate to the induced representation from a representation of the subgroup Sn Sm with respect to the representations of the group Sn+m' In this sectio the graphic method for calculating the outer product of representations w be given.
6.5.1
Representations of Sn+m and Its Subgroup Sn® Sm
First of all, we introduce our notations for this problem. The group algeb £ of the permutation group Sn+m among (n + rn) objects is (n + rn dimensional The primitive idempotent e[w] in £ generates the minimal le ideal £[w] = £e[w], which corresponds to the d[w ]dimensional irreducib representation [w], where the Young pattern [w] contains (n + rn) boxes. Denote by Sn the permutation subgroup among the first n objects in th (n + 171) objects, and by Sm the subgroup among the last m objects. Th product of elements belonging to two subgroups Sn and Sm, respectivel is commutable because they permute different objects. The only commo element in two subgroups is the identical element E. Therefore, we are ab to define the direct product Sn® Sm of two subgroups, which is a subgrou of S'dm with the index N,
(n+rn)! n'rn'
(6.9
Its leftcoset. is denoted by 2
< 0: <
N.
Each To: moves the first n objects to n new positions, which are differe for different To;. Although those To: are not unique, it is assumed that the have been chosen. The group algebra of the subgroup is denoted by £nm £ is decomposed as follows:
§6.5 Outer Product of Irreducible Representations of Sn
269
(6.94) Denote by e[.\[[J.L1 the primitive idempotent in £nm. el.\IIJ.L[ generates the minimal left ideal of £nm, £ 1.\ II J.LI = £ nm e[.\IIJ.LI, corresponding to the irreducible representation [,\] x [IL] of 81'10 8 m with the dimension d[.\ld[J.LI ' The Young patterns [,\] and [IL] contain nand m boxes, respectively. Note that e[.\I['LI is an idempotent of £ but generally not primitive, and £IAIIJ.LI is a subalgebra of £ but not its left ideal. Now, we discuss two related problems. One problem is the subduced representation from an irreducible representation [w] of 8 n +m with respect to the subgroup 81'10 8 m , which is generally reducible and can be reduced as a direct sum of the irreducible representations [,\] x [IL] of 81'10 8 m (6.95)
a}.'L is the multiplicity of [,\] x [IL] in the subduced representation [w] and can be calculated by the character formula in the subgroup XI.\IIJ.L1 (H)X[w l (H) ,
(6.96)
where XI.\1I1LI(H) and Xlwl(H) are the characters of H in the representations [,\] x [IL] and [w], respectively. From the viewpoint of the group algebra, if a vector tj in £ satisfies (6.97) elAII'Lltjelwl provides a map from the minimal left ideal £1.\[[,,1 in £nm onto the left ideal £Iwl in £ : (6 .98) namely, [,\] x [IL] is contained in the subduced representation [w] . From Eq . (6 .95), the number of the linearly independent maps (6.97) is equal to a}.'L' In other words, there are a\J.L vectors tj, 1::; j ::; a').J.L such that el.\IIJ.L 1tje[w l are linearly independent of each other . Usually, the standard bases in £Iw[ are chosen to be tje 1wl one by one until a').'Lindependent el.\IIJ.Lltjelwl have been found. The ClebschGordan coefficients for the decomposition can also be calculated from the map (6.98) (see Prob. 35) . Theorem 3.10 says that the left ideal and the right ideal generated by the same primitive idempotent in a group algebra correspond to the same
270
Chap. 6 Permutation Groups
irreducible representation. Therefore, there are also that eW tj e[ '\][JL] are linearly independent of each other,
a~/"
vectors
tj
such
(6.99) The other problem is the induced representation from an irreducible representation [A] x [Jl] of the subgroup Sn0 Sm with respect to the group Sn+m. The induced representation is called the outer product of representations in the permutation group, denoted by [A] 0 [Jl]. The outer product is generally reducible and can be decomposed as a direct sum of the irreducible representations [w] of Sn+m, (6.100) Due to the Frobenius theorem (see Eq. (3.76)), b~JL = a~JL' (6.101) where X[.\ ]0[JL] (R) is the character of the element R of Sn+m in the outer product representation [A] 0 [Jl]. From the viewpoint of the group algebra, e [.\)[JLI is also an idempotent in .c, but generally not primitive. e[.\)[JLI generates a left ideal .c.\JL in .c: .c AJL
= .ce[.\)[JLI.
(6 .102)
corresponds a representa tion of Sn+m, which is [A] 0 [Jl]. Substituting (6.94) into (6.102), one has
.c.\JL
N
.c.\JL
=
EB ..,=1
N
T..,.cnme[>')[JLI
= EB
T..,.c[>'][JLI,
(6.103)
<>=1
Each term in the sum is a d[.\]d[JL] dimensional subspace in .c, and there is no common vector between any two subspaces. Hence, the dimension of the representation [A] 0 [Jl] is (n + m)! d[>'] 0 [JL ] = I I d[>.]d[JLI· (6.104) n .m . Since (6.105) provides a map from the minimal left ideal .c 1w ] in .c onto the left ideal .c.\JL in .c. This is just the Frobenius theorem that the multiplicity
e1w]tje[>')[JL]
§ 6. 5 Outer Product of Irreducible Representations of Sn
271
of [w] in [A] 0 [J.l] is equal to a~ll· The basis vectors in LAIl are calculated IJY leftmultiplying To to the standard bases in L[A][Il]. Then, from the map (6.105) the basis vectors of Le 1w ] are expressed by the basis vectors of fAil where the coefficients are the ClebschGordan coefficients for the decomposition (see Prob. 37). 6 .5.2
LittlewoodRichard8on Rule
There is a simple graphic method for calculating b~Il' called the LittlewoodRichardson rule. Choose one representation between [A] and [Ii], say [A]. Usually, choose [A] whose box number is not less than that of [J.l] for convenience. Fill the boxes in the jth row of the Young pattern [J.l] with the digit j. Attach the boxes of the Young pattern [J.l] row by row, beginning with the first row, into the Young pattern [A] in the following way. (a) Each time when the boxes in one row of the Young pattern
Y ll have been attached, the resultant diagram constitutes a Young pattern. Namely, it is lined up on the top and on the left, where the box number of the higher row is not less than that of the lower row. (b) The boxes filled with the same digit never occur in the same column of the resultant diagram. (c) After all the boxes of the Young pattern Y ll are attached, read the digits filled in the boxes of the resultant diagram from right to left, row by row beginning with the first row, such that at every step of the reading process the number of boxes filled by a smaller digit is never less than that filled by a larger digit. If the resultant diagram, when aU the boxes of the Young pattern Y ll were attached according to the above rule, is the Young pattern [w], the representation [w] of Sn+m appears in the reduction of the outer product [A] 0 [J.l]. The times of appearing of a Young pattern [w] in the different res ultant diagrams is the multiplicity of [w] in the reduction. Let us give two examples to show how to use the Littlewood  Richardson rule. Example 1. Calculate the reduction of [2,1]0 [2,1].
7 ffiTI 0
Chap. 6 Permutation Groups
272
Attaching the boxes filled by the digit 1 to the first Young pattern acco to the LittlewoodRichardson rule, one obtains
x xII
x
X
X 1
X
1
X
X
1
X X
X
X 1
1
1
Then, attach the box filled by the digit 2 to the above diagrams, re tively. Note that the box filled by the digit 2 cannot be attached i first row due to the rule (c). It also cannot be attached to the secon of the fourth diagram. The allowed diagrams are as follows:
x x 1 1 x 2 x X 1 x 2 1
X
x 1 1
x 2 X
x 1 2
x 1
x x 1 x 1 2 x x x 1 1 2
x x 1 x 1 2 x x X 1 1 2
The reduction of [2,1 ] ® [2, 1] and their dimensions are given as follow
[2, 1] 0 [2, 1] ::: [4,2] EB [4, 1, 1] EB [3,3] EB 2 [3,2,1] EB [3,1,1,1] EB [2,2,2] EB [2,2,1,1]' 20
X
2 x 2 = 9 + 10 + 5 + 2 x 16 + 10 + 5 + 9.
Due to the Frobenius theorem, the LittlewoodRichardson rule ca be used to calculate the reduction of the subduced representation [w] respect to the subgroup Sn0 Sm. In this reduction, [A] and [fl.] run ov representations of S" and Sm, respectively, but the resultant diagram to be the given [w]. Example 2. Reduction of the subduced representation [3,2,1] with respect to the subgroup S30S3.
xxx xxI xxI xxI xxI xlI 11 EB x1 EB x l EB x2 EBx2 EB x2 2 1 2 1 3 x
Exercises
Hence,
Check the dimensions in the reduction, 16 = 1 x 2 + 2 x 1 + 2 x 2 x 2 + 2 x 1 + 1 x 2.
6. 6
Exercises
l. Simplify the following permutations into the product of cycles without
any common object:
(a) (1 2)(23)(1 2),
(b) (1 23)(1 34)(3 2 1),
(c) (1234)1,
(d) (1245)(4326),
(e) (1 23)(426)(34 5 6).
2. Please show that every permutation can be decomposed as a product of the transpositions Pa of the neighbored objects. 3 . Prove that the order of a cycle R with length £ is £, namely, Rt 4. Prove that P j = (1 2) and W permutation group Sn.
= (1
= E.
2 3 ... n) are two generators of the
5. There are 52 pieces of playing cards in a set of poker. The order of cards is changed in each shuffle, namely, the cards are made a permutation. If the shuffle is "strictly" done in the following rule: first separate the cards into two parts in equal number, then pick up one card from each part in order. After the strict shuffle the first and the last cards do not change their positions, while the remaining cards are rearranged. Try to find this permutation, to decompose it into a product of cycles without any common object, to write the cycle structure of the permutation, and to explain at least how many times of the strict shuffles will make the order of cards into its original one.
274
Chap. 6' Permutation Groups
6 . Write all Young patterns of the permutation groups 8 5 , 8 6 , and 8 7 from the largest to the smallest, respectively.
7. Calculate the number n(e) of elements in each class (e) of the permuta tion groups 8 4 , 8 5 , 8 6 , and 8 7 , respectively.
8. Calculate the number d[AI (8 n ) of the standard Young tableaux for eac Young pattern [A] of the permutation groups 8n , 5 S n S 9.
9. Write the Young operators corresponding to the following Youn tableaux:
(a)
[IpITI
(b)
[III]
(c)
L2l3J
llJ
2 I 3 I 4
10. Write five standard Young tableaux Y" corresponding to the Youn pattern [3,2] of 8.5 from the smallest to the largest, and calcula the permutations R,w transforming the Young tableau Yv into Youn tableau Y",.
11. Calculate the permutation R12 transforming the following Youn tableau Y2 to Yl, and check the formulas P J R 12 = R l2 P2 , QJRl2 R 12 Q2, and Y 1 R J2 = R 12 Y2, where
IiT2T3l
~'
[pID.
Y2:
12. It can be seen that there is no pair of digits filled in the same row the Young tableau Y and in the same column of the Young t a bleau Y Calculate the permutation R transforming the Young tableau Y int the Young tableau Y', and express R as PQ belonging to the Youn tableau Y, and as P'Q' belonging to t.he Young tableau Y': The Young tableau Y 1
3 6
2 5 8
4 I7 9 I
I
Th e Young tableau 1 5
8
2 6 9
Y'
3 I 4 I 7 I
13. Try to expand all nonstandard Young operators Y of the Young patter [2,1] with respect to the standard Young operators Y"" Y = 2:", t",Y, where tJ1 is the vector in the group space of 8 3 ,
14 . Expand explicitly the identity of the 8 4 group with respect to tl Young operators.
Exercises
27[
15. Calculate the orthogonal primitive idempotents for the Young pattern [2,2,1] of the permutation group 8 5 .
16. Calculate the orthogonal primitive idempotents for the Young pattern [4,2] of the permutation group 8 6 .
17. Calculate the orthogonal primitive idempotents for the Young pattern [3,2,1] of the permutation group 8 6 .
18. Calculate the orthogonal primitive idempotents for the Young patterns [3,3] and [4,1,1] of the permutation group 8 6 , respectively.
19. Give an example to demonstrate that the primit.ive idempotent generating a minimum left ideal is not unique.
20. Calculate the ClebschGordan coefficients for the reduction of the selfproduct of the irreducible representation [2,1] of the 8 3 group in the standard bases and in the orthogonal bases, respectively.
2l. Calculate the standard bases for the irreducible representation [3,1] in the group space of 8 4 , and calculate the representation matrix of the transposition Pa for the neighbored objects in the standard bases by the tabular method. 22. Calculate the real orthogonal representation matrix of the transpositions Pa for the neighbored objects in the irreducible representation [3,1] of the 8 4 group in the orthogonal bases by the formulas (6.90) and (6.91). Calculate the similarity transformation matrix between two representations in the standard bases (Prob. 21) and in the orthogonal bases and write explicitly the orthogonal bases
23. Calculate the representation matrices of the transpositions Pa for the neighbored objects both in the standard bases and in the orthogona bases of the irreducible representation [2,1,1] of the 8 4 group. Calculate explicitly the orthogonal bases
24 . Calculate the standard bases and the orthogonal bases in the group space of 8 4 for the irreducible representation [2,2]' and calculate the representation matrices of the transpositions Pa for the neighbored objects in these two sets of bases.
25. Calculate the symmetric bases of the oscillatory wave function of a molecule with the symmetry of a regular tetrahedron, for example, the methane CH 4 .
276
Chap. 6 Permutation Groups
26. Calculate the representation matrices of the generators (1 2) and (1 2 3 4 5) of the 8 5 group in the irreducible representation [2,2,1] by the tabular method.
27. Respectively calculate the real orthogonal representation matrices of the transpositions Pa of the neighbored objects in two equivalent irreducible representations [2 3 ] ':::' [1 6 ] x [3,3] of 8 6 , and find the similarity transformation matrix X between them. 28. Calculate the character of each class of the permutation group 8 5 in the representation [2,2,1] by the graphic method. 29. Calculate the characters of each class of the permutation group 8 6 in the representations [3,2,1]' [3,3], and [2 3 ] by the graphic method, respectively. 30. Calculate and fill in the character tables of 8 3 ,8 4 ,8 5 ,8 6 , and 8 7 by the graphic method. 31. Calculate the Clebsch Gordan series of the inner product of each pair of the irreducible representations for the permutation groups 8 3 , 8 1 , 8 5 , 8 6 , and 8 7 by the character method.
32. Calculate the ClcbschGordan coefficients for the reduction of the selfproduct of the irreducible representation [3,2] of 8 5 . 33. Calculate the reduction of the following outer products of the representations in the permutation groups by the LittlewoodRichardson rule: (1) [3,2,1] 0 [3], (2) [3,2]0 [2, 1], (3) [2,1] 0 [4,2 3 ]. 34. Calculate the reduction of the subduced representations from the following irreducible representations of 8 6 with respect to the subgroup
8 3 08 3 : (1) [4,2],
(2) [2,2,1,1] ,
(3) [3,3] .
35. Calculate the similarity transformation matrix in the reduction of the subduced representation from the irreducible representation [3 , 3] of 8 6 with respect to the subgroup 8 3 0 8 3 . 36. Calculate the representation matrices of the generators of 8'1 in the induced representation from twodimensional irreducible representation [2,1] of 8 3 with respect to 8 4 . 37. Calculate the similarity transformation matrix in the reduction of the outer product representation [2,1] 0 [2] of the permutation group.
Chapter 7
LIE GROUPS AND LIE ALGEBRAS
In Chap. 4 the fundamental concepts on Lie groups have been introduced through the SO(3) group and its covering group SU(2). In this chapter we will study t.he general theory on Lie groups and Lie algebras, such as the property of simple and semisimple Lie algebras, the regular commutative relations of the Cartan Weyl bases of a semisimple Lie algebra, t he classification of the simple Lie algebras, the Chevalley bases of generators, the weights, the irreducible representations, and the decomposition of direct product representations. The SU(N) groups, the SO(N) groups, and the USp(2£) groups are studied related to the classical Lie algebras in this chapter and further studied in the succeeding chapters because they are widely used in physics.
7 .1
7 .1.1
Lie Algebras and its Structure Constants
The Global Property of a Lie Group
We will begin with a review of the concepts of Lie groups. An element R in a. Lie group G can be characterized by gindependent real parameters r A, 1 ::; A ::; g, varying continuously in a gdimensional region. The parameters CA of the identical element E are usually taken to be zero for convenience. 9 is called the order of the Lie group G and the region is called the group space of G. It is required that there is a onetoone correspondence between the set of parameters r A and the group element R at least in the region where the measure is not vanishing. The parameters tA of the product clement are the real analytical functions of the parameters of the factor dements
T=RS,
(7.1)
278
Chap. 7 Lie Groups and Lie Algebras
fA (r; s) are called the composition functions, characterizing the multipli tion rule of elements in G completely. The composition functions have satisfy some conditions to meet four axioms for a group. The topolog property of the group space describes the global property of G, lf the group space of G falls into several disjoint pieces, G is calle mixed Lie group. G contains an invariant Lie subgroup H whose gro space is a connected piece in which the identical element lies, The of elements related to the other connected piece is the coset of H. T property of the mixed Lie group G is characterized completely by Ha the representative elements, respectively belonging to each coset. Hereaf we will mainly study the connected Lie groups. A connected Lie group G is called multiply connected with the degree continuity n if the connected curves of any two points in the group space separated into n classes where any two curves in each class can be chan continuously from one to another in the group space, but two curves different classes cannot, The Lie group with the degree of continuity n is called simplyconnected. For a Lie group G with the degree of continu n, there exists a covering group G' which is simplyconnected and hom morphic onto G with an ntoone correspondence. A faithful representat of G' is an nvalued representation of G. A Lie group is compact if its group space is compact, If the group sp is an Euclidean space, a closed finite region (including the boundary) compact, and an open finite region (without the boundary) or an infin region is not compact, The group integral can be defined only for a comp Lie group so that the most properties of a finite group can be generali to a compact Lie group, such as Theorems 3.1, 3.3, 3.5, and 3.6, T representations of a noncom pact Lie group G can be obtained throu those of a compact Lie group which is "close" to G (see §7.4.3 and §9.5 A Lie group is called simple if it does not contain any nontrivial invari Lie subgroup. A Lie group is called semisimple if it does not contain a Abelian invariant Lie subgroup (including the whole Lie group). A group of order 1 is Abelian so that it is simple but not semisimple.
7.1.2
The Local Property of a Lie Group
An infinitesimal element R which is located at the neighborhood of identical element E in the group space is characterized by infinitesim parameters rA = etA In a representation D(G) of a Lie group G, representation matrix of the identical element E is the unit matrix, D(E)
§ 7.1 Lie Algebras and its Structure Constants
279
I, and that of an infinitesimal element R(a.) can be expanded as a Taylor series. Up to the first order one has
D(R) = 1  i
L
IA =i BD(R) I . BrA R=E
rAI A + ... ,
A
(7.2)
f.1 are called the generators in a representation D (G). If the representa
tion is faithful, the generators are linearly independent of each other. The characterize the property of the infinitesimal elements in a representation of a Lie group G, so that, as shown by the Lie Theorems, they can characterize the local property of the Lie group. The First Lie Theorem says that the representation of a Lie group G with a connected group space is completely determined by its generators. D(R) satisfies a differential equation with a given boundary condition ~enerators
BD(R) BrA
.
= 2 L IaSaA(r)D(R), B
D(R)IR=E = I,
(7.3)
where the real functions
SBA(r) = BfB(r; s) I BrA S=Rl
(7.4)
depend upon the choice of the parameters of the Lie group. They are independent of the given representation. The Second Lie Theorem says that the generators in any representation of a Lie group G satisfy the common commutative relations
[lA, Ia]
=iL
(7.5)
C ABD ID,
D
where the real numbers C ABD are called the structure constants of G. Conversely, if 9 matrices satisfy the commutative relations (7.5), they are the generators in a representation of G with order g. The structure constants are calculated from the generators in a known representation. Although the structure constants are independent of the representation, they depend on the choice of the group parameters. If the parameters are rechosen, from Eqs. (7.2) and (7.5), both the generators and the structure constants are changed,
r~ =
L B
rB (X1)BA'
I~
= L XABIB, B
' D = 'L" XAPXBQCpQR (X1)RD' CAB PQR
detX
i
0, (7.6)
280
Chap. 7 Lie
GTOUpS
and Lie Algebras
In fact,
[lA' Is]
=L
XAPXBQ [Ip,
101 = i
PQ
=
iL {L
L
XAPXBQCpQRIR
PQR XAPXBQCpl (X1)RD}
1'0.
PQR
D
The Third Lie Theorem says that a set of constants CABD can be th structure constants of a Lie group if and only if they satisfy
CABD = CBAD,
L
{CAlC PDQ + CBl'C PAQ + CDIC PBQ }
= o.
(7.
p
Based on this theorem the Lie groups can be classified by their structu constants. We will study the classification later in this chapter. The adjoint representation is very important for a Lie group. If I are the generators in a representation D(G) of a Lie group G, the adjoi representation Dad (G) satisfies
D(R)IBD(R)l =
L
IDDtdB(R).
(7.
D
When R is an infinitesimal element, one has
[lA, I B] =
L
ID (I~d) DB'
(7.
D
The structure constants characterize some important properties of a L group G. A Lie group is Abelian if and only if its structure constants a all vanishing. A Lie group is simple if and only if its adjoint represe tation is irreducible. A Lie group is semisimple if and only if its adjoi representation is completely reducible and does not contain the identic representation. We are going to show that if a Lie group G is compact, its param ters can be chosen such that each structure constant of G is totally an symmetric with respect to its three indices . The adjoint representation of Lie group is real for any set of real group parameters owing to its definitio (4.28). For a compact Lie group G, a real representation is equivalent to real orthogonal one. Let yT be the real similarity transformation changi the adjoint representation to be real orthogonal so that its generators 1 becomes antisymmetric
§ 7. 1 Lie Algebras and its Structure Constants
_ () _ {( Y T) 1 () fA aD fA DB
L
=
ad
fAY
281
T} BD
(7.10)
(yl) PB (It) PQ YDQ.
PQ
Now, change the parameters in G such that the combination matrix X in Eq. (7.6) is equal to Y in Eq. (7.10). Thus, each new structure constant of G is totally antisymmetric with respect to its three indices
C~rP
=
L
YAPYBQCpQR (y1)RD
PQR
= i
L {L YAP
P
(y1)RD (IP)RQ Ya Q }
QR
 )
.~
, a
=2L YAP(f p DB=C AD , P
The Lie Algebra
7 .1. 3
Let fA be the generators in a faithful representation of a Li e group C. fA sat.isfy t.he commu t.ative relations (7.5), or equivalent.ly, [(if A), (iI a )]
=L
(7.11)
CAaD(ifD).
D
For the given structure constants CABD, the map (7.11) of two generators ( if A) and (if B) onto a combination of generators (if D) is called the Lie product. The real linear space spanned by the basis vectors (if A) is closed for the Lie product of vectors in the space
Y A
[X , Y] =
=L
(ifs)ys,
B
L AS
XAYB [(if A), (iIs)] =
L
(if D)
{L
XAyaCAaO}.
AB
D
This real space is called the real Lie algebra of a Lie group G , denoted by Ln. Generalizing the real space LR to a complex space L under the same product rule of vectors, one obtains a complex Lie algebra, brieRy called a Lie algebra. Evidently, the Lie algebra is not an associate algebra because the Lie product satisfies the Jacobi identity [fA, [fa, Iell [[lA, fa], Ie]
= [fs,
[fA, fell·
(7.12)
282
Chap. 7 Lie Groups and Lie Algebras
C ABD are the common structure constants of the Lie group G and 1.1 corresponding Lie algebra L. A representation of a Lie group is also representation (or called "module" in mathematics) of the correspondill Lie algebra. The Lie algebra L is the complexification of the real Lie algeb] LR, and LR is a real form of L. In this textbook we only discuss the L algebra L which has its Lie group G and its real form LR. The real Lie algebra is important for manifesting the compactness of Lie group. For example, the SO(4) group is a real orthogonal group in til fourdimensional Euclidean space. If the fourth coordinate is changed t be imaginary, X4 = ici, the orthogonal group becomes the proper Lorent group Lp. Both SO(4) and Lp have the same Lie algebra, but differen real Lie algebras. If the imaginary parameters of a group are allowed to b used, the generators in the corresponding representations of SO( 4) and L are the same but some parameters are different by a factor i. A represen tation of SO( 4) becomes a representation of Lp if some real parameters o SO( 4) are changed to be imaginary. This is the standard method to fin the irreducible representation of a noncom pact Lie group from that of compact Lie group. We will discuss it in Chap. 9 in detail. The real Li algebra of a compact Lie group is called the compact real Lie algebra. Two Lie algebras are isomorphic if their structure constants are th same, but two Lie groups with the same structure constants are only locall isomorphic generally. There are two typical examples where two locally iso morphic Lie groups are not isomorphic. The SU(2) group is homomorphi onto the SO(3) group, but they have the same structure constants. Th U(2) group contains two subgroups SU(2) and U(1), but does not contain subgroup SU(2)0U(1) because two subgroups SU(2) and U(l) have a com mon element 1 other than the identical element 1. Therefore, the U (2 group is not isomorphic onto the group SU(2)0 U(l), although they ar locally isomorphic onto each other. The commonly used concepts for an algebra can also be used for a Lie a gebra and for a real Lie algebra, such as subalgebras, ideals, Abelian ideal the direct sum, the semidirect sum, the isomorphism, the homomorphism and so on. For a Lie algebra, one also can introduce some concepts rela ing to a Lie group, such as a simple Lie algebra, a semisimple Lie algebra representations, and so on. A nonnull subset L1 of a Lie algebra L, L1 C L, is a Lie subalgebra o L if it is closed with respect to the sum and the Lie product of the vector in the subset
§7.1 Lie Algebras and its Structure Constants
[X, Y]
E £1,
(7.13)
where X E £1, Y E £1, and Cj and C2 are two arbitrary complex numbers. I'\lrthermore, a Lie subalgebra £1 of a Lie algebra £ is an ideal of £ if 'V X E £1 and Y E £.
(7.14)
There is no difference between a leftideal and a rightideal of a Lie algebra L A Lie algebra is called simple if it does not contain any ideal except ror the whole algebra. A Lie algebra is called semisimple if it does not contain any Abelian ideal. A onedimensional Lie algebra is Abelian, so it is simple, but not semisimple. A simple Lie algebra with dimension higher than 1 must be semisimple. The Lie algebra of a simple or a semisimple Lie group is simple or semisimple, respectively. A Lie algebra £ is called the direct sum of two Lie subalgebras, £ = £1 EEl £2, if (7.15) Evidently, both £ 1 and £2 are ideals of the direct sum L. A Lie algebra £ is called the semidirect sum of two Lie subalgebras, £ = £1 EEls £2, if (7.16) Now, £1 is an ideal of £, but £2 is not. If a Lie algebra £ is homomorphic onto a Lie algebra £', £ can be decomposed into a semidirect sum £ = £1 EEls £2 such that £2 is isomorphic onto £'. £1 is an ideal of £ and maps to null vector in £'. £1 is called the kernel of the homomorphism between £ and £'. Define a series of Lie subalgebras £(n), n = 1, 2, ... , from a Lie algebra £(1)
= [£,
£] ,
(7.17)
A Lie algebra £ is called solvable if there exists an integer m in the series of subalgebras £(n) such that £(m) = 0. It is proved that any Lie algebra ca,1 be decomposed into a semidirect sum of a solvable Lie algebra £1 and a semisimple Lie algebra £2, £ = £1 EEls £2. Any irreducible representation with a finite dimension of a solvable Lie algebra is onedimensional.
7.1.4
The Killing Form and the Cartan Criteria
Based on the Third Lie Theorem, the Lie groups as well as the Lie algebras can be classified by their structure constants . However, the structure
284
Chap. 'l Lie Groups and Lie Algebras
constants do depend on the choice of the group parameters. We hope find what property of the structure constants reflects the essence of a L algebra. The Killing form gAB of a Lie algebra is defined as gAB
= 2..:
C ApQCnl
= Tr (lAd lEd) = gBA·
(7.1
PQ
is a symmetric matrix. Defined the structure constants three subscripts which is totally antisymmetric
gAB
CABO
= 2..:
CABPgPD =
P
2..: = 2..:
2..:
CABO
wi
CAItCpQRCDRQ
PQR
=
{CBlC PA R
+ CQ,/C PBR } CDRQ
(7.1
PQR {CBQPCApFl.CDRQ  CAlCBpRCf)RQ}
PQR
= CBAD = CADB.
In the nomenclature of tensors (see Appendix B), with respect to t rechoice (7.6) of the group parameters, the generators IA is a covaria vector, the structure constant C.1 Bf) is a mixed tensor of rank (2, 1), C AB is a covariant antisymmetric tensor of rank 3, and the Killing form is covariant symmetric tensor of rank 2,
g~B =
2..:
X
AP
X nQ9PQ,
(7.2
PQ
The Killing form is a real symmetric matrix. A real symmetric matr can be diagonalized through a real orthogonal similarity transformatio Then, the diagonal entries can be changed to be ± 1 or 0 through the re scale transformation. Namely, for a real Lie algebra, through the rechoi of the real group parameters, the Killing form can be transformed into diagonal matrix with the diagonal entries ±1 and O. For a Lie algebr the sign of the diagonal entry can be removed by a scale transformatio However, the number of the zero eigenvalues of gAB is essential for a L algebra and cannot be changed by the rechoice of the parameters. Theorem 7.1 (The Cartan Criteria) and only if its Killing form is nonsingular,
det 9
i
0,
A Lie algebra is semisimple
(7.2
§ 7. 2 The Regular Form of a Semisimple Lie Algebra
and a real semisimple Lie algebra is compact if and only if its Killing form is negative definite. We will not prove this theorem here. Since the Killing form gA fj of a. scmisimple Lie algebra is nonsingular, gAB has its inverse matrix gAB ""' ~
9 AD gDB
.\:A = uB·
(7.22)
D
and gAB can be used as the metric tensors. Define an operator Cn which is a homogeneous polynomial of order n with respect to generators,
.IJ /\8
""' ""' ""' C AIDIJ.)2 C A,DD3 ~ ~ ~ 2 (D)
(A)
···
C AnDn DI (7.23)
(B)
X
gAIBI ... gAn 13 nIBJB2 ... IBn'
which is commutable with any generators
IA
in the Lie algebra, (7.24)
en
The reader is encouraged to prove Eq. (7.24). is called the Casimir operator of order n. Remind that the Casimir operator of order 2 is
C2
= LgABJ."In .
(7.25)
AU
7 .2
7.2 .1
The Regular Form of a Semisimple Lie Algebra
The Inner Product in a Semisimple Lie Algebra
Express the basis vector I A in a semisimple Lie algebra L by the Dirac symbol which is commonly used in quantum mechanics [G eorgi (1982)] lSASg ·
(7.26)
The inner product of two vectors in L is defined with its Killing form,
(AlB)
= gAB = 1'r (I~d Irf) ,
(7.27)
which is bilinear with respect to two vectors (AI and IB)
+ C2D)) = C1 (AlB) + c2(AID), ((CIA + c2D)IB) = C] (AlB) + c2(DIB),
(AI(clB (AlB)
= (BIA),
(AIDIB) == (AI [D, BD
= ([A,
D]IB).
(7.28)
286
Chap. 1 Lie Groups and Lie Algebras
When the group parameters are changed, the inner product will be related to the new Killing form in the same way,
IX!')
= I~
(x!'IXv)
X!'AIA)
=
L
=~
x!'AIA), (7.29)
X!'AXvBg AB = gW'
AB
7.2.2
The Cartan Subalgebra
In a semisimple Lie algebra L , any vector is also a linear operator which transforms a vector to another by the Lie product (7.30)
Then, one is able to calculate the eigenvalue and the eigenvector of a vector X in L . Evidently, any operator is its own eigenvector with zero eigenvalue,
XIX)
= I [X,
Xl)
= O.
Denote by ex the multiplicity of zero eigenvalue of X. Among all vectors in L the minimal ex is denoted by e
e = min ex e
> O.
(7.31)
is called the rank of the Lie algebra L as well as the rank of the Lie group G. A vector X in L is called regular if = e.
ex
e
Theorem 7.2 There are linearly independent eigenvectors H j of a regular vector X of a semisimple Lie algebra L with rank H j are commutable with each other
e.
ls,js,e,
ls,ks,e,
(7.32)
and X is a combination of H j . In the remaining subspace of L, the (g  e) common eigenvectors Ecx of eoperators H j are nondegenerate, (7.33)
We will not prove this theorem here. The Abelian subalgebra 11. spanned by e generators H j is called the Cartan subalgebra of L. 11. is not an ideal of L. The set of H j is the largest set of the mutually commutable generators in L. The edimensional vector a is called a root, and the £dimensional space is called the root space. There is a onetoone correspondence between the
§7.2 The Regular Form of a Semisimple Lie Algebra
287
root 0 and the generator Eo.. Sometimes, H j is also called the generator corresponding to the zero root. Note that the "zero root" is degenerate with the multiplicity £. The choice of the Cartan subalgebra is not unique. In fact, for any group element REG, the set of Hj = RHjR r also spans a Cartan subalgebra, conjugate to the original one. 7.2.3
Regular Commutative Relations of Generators
The basis vectors H j and Eo. in a semisimple Lie algebra £ are called the Cartan Weyl bases, or the regular bases. The regular bases satisfy the commutative relations (7.32) and (7.33). We are going to study their remaining commutative relations. From Eq. (7.28) one has
if
0
i=
(3.
(7.34)
For the same reason, (7.35) For a given root 0, if  0 is not a root, go.B are all vanishing, which conflicts to that £ is semisimple. Thus, in a semisimple Lie algebra, the roots ±o have to appear in pairs. Choose the factor bO/. in EO/. such that (7.36) The factor be. in Eo. can still be chosen in the condition bO/.b_O/. = 1. The part gjk of the Killing form related to the Cartan subalgebra 1i can be expressed by the components of roots:
(7.37)
where,6. is the set of all roots in £. Due to Eq. (7.35) g)k is also nonsingular, det (gjd i= O. Define gjk and its inverse _gjk to be the metric tensor in the root space such that they can be used to raise a subscript and to
288
Chap. 7 Lie Groups and Lie Algebras
lower a superscript of the vector in the root space. The inner product o two vectors in the root space is defined as
V . U   """ L 9 jkVj Uk  """ L VkUk   """ L gjk VjU k jk k jk k = (ajVj) (akU ) = (0: ' V) (0:' U).
LL
L
(7.38
From the Jacobi identity,
0 = [Hj , [E"" Ef3ll
+ [E""
[Ef3, Hjll
[Hj , [E"" Ef3ll (3j [E""
+ [Ef3,
[Hj , E",ll
E{3] +aj [E{3, E", ] ,
one has [Hj , [E"" H j I[E""
E{3]]=(aj+{3j)[E"" E{3], E{3])
cc
(a J
+ (3j) I[E""
(7 .39
Ef3]).
Since the nonzero root is nondegenerate, one concludes t.hat [E"" E{3] proportionaJ to E"' +f3 if 0: + (3 is a root , and [E"" Ef3) = 0 if 0: + {3 I 0 not a root. When {3 = 0:, [E"" E~",] belongs to the Cartan subalgebr and can be expressed as L: j Aj H j , where
j
j
= ([Hk, ).,J = 
L
E",IIE~",)
gjkak =
= adE",IE~",)
=
ak·
(7.40
0:1.
k
is nothing but the contravariant component of the root 0:. In summary, the regulaT commutative relations among the Cartan~vVey bases in a semisimple Lie algebra .c are ).,J
[H j
,
HkJ
= 0,
[Hj , E",]
= ajE""
N"' ,f3E"'+f3 [E""
Ef3]
= {
L
aj Hj
=
when 0: '
H
== H",
0:
when {3
+ {3 =
is a root,
(7.41 0:,
J
0,
the remaining cases,
where the antisymmetric coefficients N",,{3 =  Nf3,'" are to be determined E", can be multiplied with a factor b", satisfying b",b~", = 1, namely, tlw
§ 7.2 The Regular Form of a Semisimple Lie Algebra
28
fa.tio bOt/b_ Ot can be chosen arbitrarily. H j as well as the components cv of a root 0: can be made an arbitrary nonsingular combination.
7.2 .4
The Inner Product of Roots
The orem 7.3 The inner product of any two nonzero roots 0: and semisimple Lie algebral satisfies
/3
i
il
r
,
~ 0: / /3)
==
20:'
/3
.
;373 = 111 teger,
(7.42
a nd 0:  r(o://3)/3 is a root in L.
Proof Construct a root chain from a root 0: by adding and subtractin another root. /3 successi vely
. . . , (0:  2/3), (0:  /3), 0:, (0:
+ /3), (0: + 2/3),
Since the number of roots in L is finite, t.he root chain has to break off a t.wo sides after a finite number of tcrms. Without loss of generality, one ha
+ n/3, 0:  (q + 1)/3 0:
q ::; n ::; p, and 0:
are all the roots,
+ (p + 1)/3
are not the roots,
where p and q are both nonnegative integers. From Eq. (7.41) one has N(Ot +pfJ),fJ = N(OtqfJl,fJ = O.
(7.43
I,ct.ting
(7.44 OIlC
obtains from the Jacobi identity 0= [EOt+ nfJ , [EfJ, EfJJJ
+ [EfJ'
= L fY
+ [EfJ,
[E_ fJ , EOt+nfJJJ
[EOt +nfJ , EfJlJ
[EOt +nfJ' HjJ  N(Ot+nfJl.fJ [EfJ, EOt+(nllfJ]
j
+ N(Ot+nfJl ,fJ [E_ fJ , EOt+(n+J)fJ] = {/3' (0: + n/3)  F n  1 + Fn} E Ot +nfJ · T hus, Fn satisfies the recursive relation
290
Chap. 'l Lie Groups and Lie Algebms
Fn= Fnl+,I3·{a+n,l3}
Fn 
2
+,13. {2a + (n + n  1),13} = ...
Fn(n+q+ll +,13 . {(n + q + l)a +
~ (n 
q) (n + q + 1),13 }
(7
1
2" (n + q + 1),13 . {2a + (n  q) ,13} . When n
= p,
one has 2a . ,13 = (q  p) (,13 . ,13) .
(7
If ,13 . ,13 = 0, ,13 is orthogonal to each root a in L so that Hf3 is commuta with each generator in L, (7
Thus, Hf3 spans an Abelian ideal in L which is in conflict with the fact t L is semisimple. Since,l3·,13 :j:. 0, from Eq. (7.46) one has 2a . ,13 f(a/,I3) = ~ = qp.
(7
~ (p  q) ~ p, a  f(a/,I3),13 is a root. At last, the theorem also holds if the root chain a + n,13 contain zero root. Without loss of generality, let a = m,l3. Defining the genera Ea.mf3 = Eo = Hf3 corresponding to the zero root, one has
It gives Eq. (7.42). Due to q
[Ea.(mllf3, E f3
l=
[Ea.mf3, E±f3J =
± (,13.,13) E±f3,
Hf3,
N(a.(mllf3J,f3 N(a.mf3l,±f3 =
= 1,
± (,13.,13),
N(a.( m+ llf3l,f3 = 1.
[Ea.(m+llf3, Ef3] = Hf3,
Thus, the above proof holds for this case.
Corollary 7.3.1 The number of linearly independent roots in a semis pIe Lie algebra L with rank f. is
e.
Proof Prove the corollary by reduction to absurdity. If the number is than f., there is at least a nonzero vector V orthogonal to each roots i so that V . H is commutable with each generator in L. This contrad that L is semisimple.
Corollary 7.3.2 In the root space of a semisimple Lie algebra, the in product of any two vectors which are the real combinations of roots is r
§'l.2 The Regular Form of a Semisimple Lie Algebra
2
and the inner selfproduct of a non vanishing real combination of roots positive real. First, due to Eqs. (7.38) and (7.46) one has
P roof
where qOt and Pc< are the integral parameters in the root chain construct from a by adding and subtracting a root 13 successively. Since 13 . 13 i= 0
13 . 13 = 4
{L
(qOt  pOt)2} 1
=
positive real.
(7 .4
OtELl.
Then, due to Eq. (7.46), the inner product a . 13 of any two roots is rea Second, introducing two real combinations of roots
one has
v .U = L
L
bj3Cy
(13 .,) = real.
j3ELl. yEll.
Due to Eq . (7.38),
v .V =
L
(a· V)2 2: O.
oELl.
V .V
> 0 only if V i= O.
Corollary 7.3.3
(7 .5 Proof
Letting n
NOt ,j3N(o+j3),j3
= 0 in
Eq. (7.45) one has 1
1
= Fa = "2(q + 1) (2a· 13  qf3. 13) = "2 P(q + 1) (13' 13
(7 .5 If,
= a + 13,
and three roots are all nonzero,
(EoIEj3IEy)
Due to Eq. (7.43),
= (E o [Ej3, EyJ) = N_j3,y(E o IEo ) = N_j3 ,y = ([E_ Ot , Ej3ll Ey) = N o ,j3(EyIEy) = N o ,_j3 . 1
292
Chap. 'l Lie Groups and Lie Algebras
= N_{3,"I
N_ Ot ,_{3
=
(7.
N(Ot + {3) , {3'
Equation (7.50) follows Eq. (7.51). Corollary 7.3.4
N Ot ,{3
i
0 if 0:, {3, and (0:
+ (3)
are all nonzero roo
Corollary 7.3.5 Except for zero roots, there are only two roots along the direction of 0:.
Proof Let to: be a nonzero root, then f(to:/o:) = 2t and f (o:/(to:) 2/t. From the conditions that 2t and 2/t are both integers, one obta t = ± 1, ±2, or ±1/2. Since NOt,Ot = 0, ±20: is not a root. Then, ± 0:/ not a root, otherwise 0: = 2(0:/2) is not a root.
7.2.5
Positive Roots and Simple Roots
Theorem 7.4 For a semisimple Lie algebra L , the bases H j in the Car subalgebra H of L can be chosen such that the roots are all real and root space is real Euclidean.
e
Proof First, there exist linearly independent roots in L , say (3 1 :::; r :::; where is the rank of L. Any root 0: can be expanded w respect to (3(r) , 0: = 2.:r x r {3(r). Taking the inner product of each term the equation with 2{3Cs) / ({3(s) . (3(S)) , one obtains
e,
e
e
f (0: / (3(S)) =
2:=
xrf ((3(r) / (3( S) ) .
r= 1
This is the coupled linear algebraic equation with respect to the varia X r . Since all the coefficients in the equation are integers and the solution do exist, Xr have to be all real. Namely, each root in L is a real combinat of (3(r) Seco nd, for the set of new basis vectors becomes real, 91'S =
2:= jk
j3(r)j j3(s)k 9 jk
=
2:=
Hr
in H,
Hr = (3(r) .
H,
(0: ' (3( r) ) (0: . (3(S)) = real.
OtE~
The new 978 can be diagonalized through a real orthogonal transformat of H r , and then become 61'S through a scale transformation. Third, denote by H j the transformed bases in H, where 9jk = 6 In the condition 9jk = 6jk' Hr as well as the components of root.s still be made an orthogonal transformation. Due to Corollary 7.3.2,
§ 7.2 The Regular Form of a Semisimple Lie Algebra
29
loot space is real Euclidean only if the basis roots f3(r 1 can be transforme into real t.hrough the orthogonal transformation. Assume that the real an imaginary parts of the jth component of f3(r) are a;r) and b;r) , respectively
1;]") = a;rl + ib;r). Make a real orthogonal transformation on a;l) such tha
a\ I) is nonnegative and the remaining components are vanishing. Sinc is real positive ' 1 all) > 0 l and b(l) = 0 . Then , make a rea 1
{:3 (1) ·13(1)
orthogonal transformation on b;l), where j > 1, such that b~l) is non negative and the remaining components are vanishing. The components o the transformed root 13(1) become ,BP) = = a and ,B~l) = ib~l) = ib, an t.he remaining components are vanishing. Since f3{l) .f3{l) > 0, a > b 2 O Making an orthogonal transformation on the first two components,
ail)
on e has that only the first component ,Bi l ) of the transformed root f3{ is positive real, and its remaining components are all vanishing. Sinc j3(1) .13(1') is real, the first component ,Bi r ) of each basis root f3(r) is rea Define new basis vectors in l by the real combinations, ,(1') = f3(T) (Bi r ) / ,B;1)) f3(I), where r > 1. Thus, the first components of ,(1') wit r
> 1 are vanishing, and its remaining components are equal to those o
Each ,(1') is not a null vector. The rest can be deduced by analogy. Preserving the first one dimensional subspace invariant, one makes the orthogonal transformatio on the remaining (e  I)dimensional subspace such that the first compo nents ,BiT) remain invariant, the second component a~2) of the transforme lOot 13(2) is positive real, and its remaining components are all vanishing Corollary 7.3.2 is used in the proof. Thus, the first and the second compo nents of all basis roots j3(T) are real. In the same way, one can prove tha all the basis roots are real. Therefore, all roots in l are real, 9jk = Ojk and the root space is real Euclidean. For a given order of H j in the Cartan subalgebra, a root 0: is calle positive if its first nonvanishing component is positive, and the root negative if the component is negative. A positive root is called a simpl lOot if it cannot be expressed as a nonnegative integral combination o ot.her positive roots. Therefore, any positive root is equal to a nonnegativ integral combination of the simple roots, and the sum of the coefficients i t he combination is called the level of the positive root. The negative root j3(r).
294
Chap. 7 Lie Groups and Lie Algebras
are similar. Obviously, the number of the simple roots in a semisimple Li algebra with rank £ is not less than £. It is equal to £ if all simple roots ar linearly independent of each other.
Theorem 7.5 The difference of two simple roots is not a root, the inn e product of two simple roots is not larger than 0, and the number of tl) simple roots in a semisimple Lie algebra with rank £ is equal to
e.
First, denote by 'Y the difference of two simple roots a and ( 'Y = 0  (3 . 'Y is not a positive root, otherwise 0 is a sum of two positiv roots (3 and 'Y . 'Y is not a negative root , otherwise (3 is a sum of two positiV roots a and I Thus, the difference of two simple roots is not a root . Second, from Theorem 7.3, the inner product of two simple roots is no larger than 0, Proof
r
(a/(3) =
20 · (3 7373 =q 
p = p ~ 0.
(7.5:
At last, prove by reduction to absurdity that simple roots are linearl independent. Assume that there is a real linear relation among simple root.
L
Cjo(j) 
L
d k (3(k) = 0,
Cj
> 0,
k
j
Since aU) i (3(k), a(j) . (3 ( k) ~ 0. Then, it is in contradiction that tl inner selfproduct of the nonvanishing vector V is not larger than zero,
j
V .V =
k
L
CjdkO(j) . (3 (k)
~ 0.
jk
Theorem 7.6 Up to isomorphism , any semisimple Lie algebra L has and only one compact real form.
lI
Proof We only sketch the proof. Since Eo< can still be multiplied willi factor bo< satisfying bOt L Ot = 1, the ratios bOt/La. can be chosen such t.ll i
(7. ;; In fact, from Eq. (7.41) one has
N~ ,{3
N'Ot , {3
~ bf3
b_ Ot  f3
Na.,f3
La. Lf3 bOt +f3 N_a.,f3·
§'l.3 Classification of Simple Lie Algebras
2
One is able to choose the ratios bex./b_ex. such that Eq. (7.54) for the positiv roots are satisfied one by one as their levels increase. Under the conditio (7 .54) the Killing form gAB becomes OAB if E±ex. are replaced with
(7 .55
It can be shown straightforwardly that the structure constants C ABD a all real. The real Lie algebra with the basis vectors ( iHj ), (iEex.t), an (iEex.2) is the compact real form of L.
C orollary 7.6.1 Any representation with finite dimension of a semisimp Lie algebra is completely reducible. Since the adjoint representation of a semisimple Lie algebra is com pletely reducible, the following Corollary follows.
C orollary 7.6.2 Any semisimple Lie algebra can be decomposed to direct sum of some nonAbelian simple Lie algebras. Now, the classification problem of semisimple Lie algebras reduces the classification of simple Lie algebras with dimension larger than l. I the classification of simple Lie algebras one will find that in each simp Lie algebra L, there is only one root w whose level is the highest among a roots in L. w is called the largest root of L, which is related to the adjoi representation of L.
7 .3
Classifi cation of Simple Lie Algebras
A simple Lie algebra of one dim ension is Abelian. In this section we a going to study the classification of simple Lie algebras with the dimensio larger than 1. In th e compact real form of a simple Lie algebra L, the ro space is real Euclidean. Hereafter , denote by TJ.1. the simple roots of L.
7. 3. 1
Angle between Two Simple Roots
T il e inner product of two simple roots T J.1 is not larger than 0 so that the ;\Il gle () is not less than 7r /2 . The cosine square of () is
(7.56
Without loss of generality, let the length of
TJ.1.
be not less than that of TI/
296
Chap. 7 Lie Groups and Lie Algebra.y
(7 There are only four solutions for 4 cos 2 8 listed in Table 7.1.
Table 7.1
The angles and the lengths of simple roots
e 57r/6 37r/4 27r/3 7r/2
7.3.2
(150°) (135°) (120°) (90°)
cos 2 3/4 1/2 1/4 0
e
r(rJL/rv)
r(rv/rJL)
3 2 1 0
1 1 1 0
IrJLI/lrvl v'3 v'2 1
arbitrary
Dynkin Diagrams
The Dynkin diagram for a simple Lie algebra is drawn by the follo rule. It will be shown that there are one or two different lengths am the simple roots in any simple Lie algebra L. Denote each longer si root by a white circle, and denote each shorter simple root, if it exists a black circle. Two circles denoting two simple roots are connected single link, a double link, or a triple link depending upon their ang be 21':/3, 31':/4, or 51':/6, respectively. The ratio of their square lengt 1, 2, or 3, respectively. Two circles are not connected if two simple r are orthogonal and the ratio of their lengths is not restricted. Now, we going to study what kinds of Dynkin diagrams of L are allowed base the property of a simple Lie algebra. 1. The Dynkin diagram of a simple Lie algebra is connected.
If a Dynkin diagram is divided into two unconnected parts, two si roots belonging to different parts, say rand r', are orthogonal to each o From Eq. (7.53) both the sum and the difference of those two simple r are not roots. Thus, the generators are divided into two classes, respecti related to the roots in two parts. The generators Er and r· H belongin the first class are commutable with the generators Er' and r' . H belon to the second one. Therefore, L is decomposed into the direct sum of ideals, which contradicts to that L is simple. 2. The Dynkin diagram contains no loop.
If a Dynkin diagram contains a smallest loop composed of n circles neighbored circles denoting the simple roots Uj and uJ+ 1 are conne by links and there is no link inside the loop. Assume Un+l = Ul
§ 7.3 Classification of Simple Lie Algebras
297
convenience. The sum a of the simple roots is not vanishing. n
a =
2..=
Uj oj: 0,
j=l n
n
j=1
j= 1
n
=
2..= IUjl2 {I + r(uHtfUj)}:::: O. j=1
It contradicts to that a is nonvanishing. 3. The number of links fetching out from one circle is less than four.
Let a simple root r connect with n simple roots Uj, 1 :::: j :::: n. The number of links fetching out from r is equal to 2: j r (r IUj) r (ujlr). Since t.here is no loop , any two simple roots Uj are orthogonal to each other, and r is linearly independent of n simple roots r j. Thus,
nil') n Irl2 > 2..= (r· Uj)2 IIUjl2 = r4  2..= r (rluj) r (ujlr). j= 1
(7.58)
j=1
T he conclusion follows Eq. (7.58) by removing a factor r2. From this property, one concludes that there is only one Dynkin diagram with a triple link:
o
•
1
2
Its algebra is called the Lie algebra G 2 . The following diagrams as well as I.hose by interchanging the white and black circles are not allowed:
>< It will be shown that the diagrams by replacing the circle on the center a circle chain connected by single links are also fl ot allowed:
or the above diagrams with
298
Chap. 7 Lie Croups and Lie Algebras
===0     
===0     
~
~
~  ~
Let the circle chain contain m circles, denoting m simple roots Vj with same length v , 1 ::; j ::; m. The length of their sum, v = L j Vj, is equal to v, m
ml
j=J
j=1
Replacing T with v in the above proof, one also shows tha t the numb e links fetching out from the circle chain is less than four. 4. The Dynkin diagrams with a double link
The general form of the Dynkin diagrams with a double link is 0  0    0       ~      iI.Iil.f UnI
where the length of
VmI
Vm
Un
V2
...
VI
is v and tha t of Uj is v'2v. Letting
Vk
m
n
U
=
2:
JUj,
V
= 2:
j=1
kUk,
k= l
one has m
Ivl 2 =
L
mJ
k 2 v2
+
k=1
L
k(k
+ 1) (_v 2 )
k= 1
V2 (m2 _~I k) n(n + 1)v 2 . Since
U
k=1
and v are not collinear, one obtains
(7
299
§ 7.3 Classification of Simple Lie Algebras
o < luI 2 1vl 2 
(U' V)2
~n(n + l)m(m + l)v 4 1
2nm (n
(mn)2

(Un' V m )2
+ m + 1  mn) v 4
Then, (m  l)(n  1) < 2. If m = 1, n is an arbitrarily positive integer denoted by e 1. The Dynkin diagram is 000      DO====: 2
3
= 1, m
Its algebra is called the Lie algebra Be . If n integer e 1, and the Dynkin diagram is
e
(e  1)
(I'  2)
is an arbitrarily positive
••<.__ •. ~
2
t • .==~O
3
(C  2)
(I'  1)
Its algebra is called the Lie algebra Ceo If n = m = 2, the Dynkin diagram is
2
1
3
4
Its algebra is called the Lie algebra F 4 .
5. The Dynkin diagrams with a bifurcation The general form of the Dynkin diagrams with a bifurcation is Vm
V m !
"~
     DO
00      0D(X
r
"()(y     
DO
T he lengths of all simple roots are the same and denoted by =
L j=]
jUj,
V
=
L k=!
Let
p
m
n
U
V.
kVk,
W
=
L
SW s ·
s=l
'I'hey are orthogonal to each other and linearly independent of r, so that
300
Chap. 7 Lie Groups and Lie Algebras
Due to Eq. (7 .59),
n 2 v 4 /4 n(n+1)v 2 /2
3 1(1+
'2  '2
n
1
nv 2 2(n+1)
v2 2
v2 2(n+1)'
11)
+ m + 1 + p + 1 < 1,
I I I
  ++>l. n+1
m+l
p+l
Without loss of generality, letting p ::; m ::; n, and replacing nand m w p, one has 3 >1
p+ 1
'
then, p =
l.
Replacing n with m, one has
2
1
>m +1 2'
then, m = 1 or 2.
If m = p = 1, n can be chosen arbitrarily. Letting n thr Dynkin diagram
This algebra is called the Lie algebra De. If p
1
1 6'
>n
+1
= £
= 1 and
then, 2::; n
m
3, one obt
= 2,
one has
< 5.
There are three Dynkin diagrams, denoting the Lie algebras E 6 , E 7 ,
E8 ,
1
2
3
4
5
6
§7.3 Classification of Simple Lie illgebms
2
1
4
3
5
6
\II
7
6. The Dynkin diagrams with only the single links denote the Lie
al!J(""'It ~
At· 000      00
1
2
3
C2
C1
C
In summary, for the simple Lie algebras with dimension larger than 1 there are four sets of the classical Lie algebras At, Be, C e, and De and fiv exceptional Lie algebras G 2 , F 4, E 6 , E 7 , and Es. Their Dynkin diagram are listed in Fig. 7.l. G2 0==. 1
2
000 000 I
2
3
e2lJ
e
000   0O====e I
C t. ,
De,
e~
2
e~
4
2
3
l2£1
••iII.'~••   • 1
2
3
2
Fig. 7.1
3
:)
l  2el
000  I
•
e
t
~e
l  I
88
e3e2
The Dynkin di agrams of simple Lie algebras
In the next section we are going to show that the Lie algebra of th SU(C + 1) group is Ae, that of SO(2C + 1) is Be, that of SO(2£) is De, and that of USp(2C) is Ceo From Fig. 7.1 one finds that some Dynkin diagram are the same such that the corresponding Lie groups are locally isomorphic
Bl B2 D2 D3
~
~ ~ ~
Al ~ C I , C2 , Al EB A I , A3
SO(3) ~ SU(2) ~ USp(2) , SO(5) ~ USp(4), SO(4) ~ SU(2) 13) SU(2)' , SO(6) '" SU(4).
(7.60
302
Chap. 7 Lie Croups and Lie Algebras
If all generators in a given simple Lie algebra are multiplied with 1 common factor A, the structure constants C ABD are multiplied with A awl the Killing form gAB is multiplied with A2. For convenience, the ifll It " product (7.38) in the root space is redefined to be Euclidean again, nam 'Iy the metric tensor in the root space is 6AB instead of gAB ' Denote by II,. the half of square length of a simple root rIJ.' 1
dIJ.
= "2
(7.6 1)
rIJ. . rlJ.'
Usually in mathematics, the half of the square length of the longer simpl!, root is normalized to be dIJ. = 1. But in physics, the length of each ro()t in Ae is normalized to be dIJ. = 1/2, namely the generators are multiplied with a factor 1/,;2 (see next section).
7.3.3
The Cartan Matrix
For a simple Lie algebra L of rank £, there are esimple roots r IJ.' Define al £dimensional matrix A, called the Cartan matrix of L, AIJ.V
=r
(
rv/rIJ.)
=
2rv . rIJ. Ir IJ.12
1
= d"
(
rv . rIJ.) .
(7.6~)
When the order of simple roots in L is chosen, the Dynkin diagram of [. completely determines its Cartan matrix, and vice versa. The diagonal entry in the Cartan matrix A is always 2 and the nondiagonal one may 1)1 0, 1, 2, and 3. AIJ.v = AvIJ. = 0 if two simple roots rIJ. a.nd rv a n disconnected in the Dynkin diagram. AIJ.v = 1 and AvIJ. = 1, 2, or  :1 if the length of Tp is not less than that of Tv, and those two simple roOt.1 are connected by a single, double, or triple link, respectively. 7.4
7.4.1
Classical Simple Lie Algebras
The SU(N)
GTOUp
and its Lie Algebra
The set of all N x N unimodular unitary matrices u, detu
= 1,
(7.6:1)
in the multiplication rule of matrices, constitutes a group, called the N dimensional special unitary matrix group, denoted by SU(N). An N dimensional complex matrix contains 2N2 real parameters. The columll matrices of a unitary matrix are normal and orthogonal to each oth'r
§'l.4 Classical Simple Lie Algebras
30:)
There are N real constraints for the normalization and N(N  1) real consl·,raints for the orthogonality. One constraint comes from the determinant. Thus, the number of independent real parameters needed for characteriziIlg the elements of SU(N) is 9 = 2N2  N  N(N  1)  1 = N 2  1. The group space is a connected closed region so that SU(N) is a simplyI'Onnected compact Lie group with order 9 = N 2  1. Any element u of SU(N) can be diagonalized through a unitary similarity transformation X,
u = X exp { i p} XI = exp ( iH) , (7.64)
p=diag{'P], 'P2, .. . , 'PN}, where the diagonal entry of XlUX is written as exp(i'Pa). The phase aIlgle 'Pa is determined up to a multiple of 27r, N]
'PN
=
L
'Pa,
IT :::;
'Pa :::; 7r,
1 :::;
Q. :::;
(N  I),
(7.65)
a= l
where the sum of the phase angles is chosen to be 0 owing to det u = l. Due to X E SU(N), Eq.(7.64) shows that u is conjugate to exp(iP) which is diagonal. Thus, the classes of SU(N) is characterized by (N  1) I,arameters 'Pa given in Eq. (7.65). The integral on the classes of SU(N) is proved to be
(7.66) The Hermi tian traceless matrix H in Eq. (7.64) characterizes the eleIll cnt u of SU(N) completely. H can be expanded with respect to the Hermitian traceless basis matrices. In physics, the basis matrices of Ndimensions ; 11 <; divided into three types, generalized from iJ a /2, resp ectively,
1 :::;
Q.
< b:::; N,
304
Chap. 7 Lie Groups and Lie Algebras
when c
< a, 2 ::; a ::; N,
when c = a,
c> a,
when
where the subscripts a and b are the ordinal indices of the generator c and d are the row and column indices of the matrix. The matrix
symmetric with respect to both ab and cd, but T;~) is antisymmetric is diagonal. = (T(1)) = (T(1)) ( T(1)) ab cd ba cd ab de'
( T(2)) ab cd (3)
1'2
= _ (1'(2)) = _ (1'(1)) ba cd Ob dc'
.
=dlag{l, 1,0, ... ,0}/2,
T?) = diag {l , 1,  2, 0, ... , O} /(2/3), TP) = diag {I , 1, 1,  3, 0, . .. , O} /(2\1'6),
T~3) = diag {I, ... , 1,  (a  1), 0, ... , O} / j2a(a  1).
There are N(N 1)/2 basis matrices T~~), N(N 1)/2 basis matrice
and (N  1) basis matrices T~3). Altogether, the number of basis m is (N 2  1). Any Hermitian t.raceless matrix of N dimensions can be expande respect to three types of basis matrices where the coefficients are rea expanded coefficients of H are denoted by w~~), w~t), and w~3), whi the parameters of the SU(N) group,
H
N
= '\' L
{wi 1 )T(1) ob
ob
+ w(2)T(2)} + '\' w(3lT(3) ab ab Lao'
a
u = exp (  iH
0=2
Three types of basis matrices T(~;), T~;), and T~3) are the generators selfrepresentation of SU(N). Usually, the generators in three typ enumerated uniformly in the following order: (I)
1'1 = TI2 , (2)
T5 = T 13
'
( 1)
1'9 = TH , (I)
T 13 = 1'34 ,
(2)
T2 = T12 ,
T 3 
T6 = TJ~),
T7 = T23 '
TlO
(2)
= T14
(2)
1'(3) 2
(I)
,
(2)
,
1'14 = 1'34 ,
(I)
TIl = T24 , l'15  T(3) 4 ,
T4 = 1'13 ' T 8  T(3) 3 , (2)
T12 = T24 ,
The generators TA satisfy the orthonormal condition:
§ 7.4 Classical Simple Lie Algebras
A, B ::; (N 2
30S

1).
(7.72
The orthonormal condition (7.72) guarantees that the structure constan C ABD is totally antisymmetric with respect to its three indices. In fact multiplying the commutative relation (7.5) with Tc and taking the trace one obtains (7.73
It shows that the SU((N) group is a compact Lie group. In physical lit eratures, the antisymmetric structure constants C ABD are usua.lly denoted by fABo . In order to write the commutative relations of generators in the selfrepresentation in a unified form, one introduces the diagonal matrices
(Taa(I)) = OacOcri., T(3) =
n
C(d
. 1
20 (a _ 1)
) 1/2 {aI """"' L
b=l
T(I)  T(1) II
an
aI
= """"'
L
(
2 b(bl)
T(
I) _ bb
)1/2
(a _ I)T( I) an
T(3) b
(
+ ~ aI
} ,
)1/2
(7.74 T(3)
a
b=2
Through a stra.ightforward calculation, one obtains the commutative rela tions of generators in the selfrepresentation as follows,
(7 .75
a < c < b.
306
Chap. 7 Lie Groups and Lie Algebras
Note that from the second formula in Eq. (7.75), the set of generators spans a subalgebra in the Lie algebra of SU(N) . There are N constant matrices in the SU(N) group,
w
o :S m :S
= exp{ i27rjN},
(N  1).
(7
They are commutable with any element in SU(N) and forms the cente SU(N), denoted by Zrv. ZN is an Abelian invariant subgroup of SU( The group space of the quotient group SU(N)jZN is connected with degree of continuity N, and SU(N) is its covering group. In order to combine the generators in the selfrepresentation of SU to be the Cartan Weyl bases, one chooses (N  1) commutable genera T~3) to span the Cartan subalgebra. In the mathematics convention, j = 1, 2, . .. ,
e,
(7
C:=Nl.
The factor 12 is introduced to make d,1 = 1 for the simple roots. normalized common eigenvectors of H j are
!
a
< b,
[(N  j)j(N  j + 1)]1/2 Ea.ob' N  j + 1
[H J,
E
QQb
1
[(Nj+1)(Nj)]1/2Ea.ab' [(N  j + l)j(N  j)]
1/2
= a,
a
E Ctob ,
0,
the remaining case (7 Introduce e+ 1 vectors Va, distributing equally in an edimensional sp 1 . V; _ !lab _
Va
b 
a=l
2
1
2(e+1)'
[(N  j + l)(N  j)r 1 / 2
12 (Va)j = { 
,
[(N  j)j(N  j + 1)]1/2 ,
0,
:S
j
:S
e,
1 :S a :S
e+ 1,
a < N  j + 1, a
=N
(7
 J + 1,
a> N 
J
+ 1,
I2~ = {V(e: 1)e'Ve(€~ 1),···,/i,!f}, I2 va = { J
(C: 1)e' J C(€
~ 1)' . . . ' J (a: l)a '  J _a _: _1, 0, ... , o
§1.4 Classical Simple Lie Algebras
Thus, the positive root
Qab
for the generator
EOCab
3
is
bl
Qab = ..J2[Va 
Vb] =
L
T Jl ,
a < b,
(7.8
w=a
where
r Jl
are the simple roots of SU (N)
(7.8 From Eq. (7.79), the inner product of two simple roots is
(7.8
Therefore, the lengths of the simple roots of SU(N) are the same and t angle of the neighbored simple roots is 21T /3, namely, the Lie algebra SU(£ + 1) is Ae. The largest root in Ae is
e
w =
..J2[Vi  Vf+l]
=
L
rJl'
(7.8
Jl=l
7.4.2
The SO(N) Group and its Lie Algebra
The set of all N x N real orthogonal matrices R, R* =R,
(7.8
in the multiplication rule of matrices, constitutes a group, called the dimensional real orthogonal matrix group, denoted by O(N). From E (7.84) one has det R = ±l. Thus, O(N) is a mixed Lie group. The subs of elements with det R = 1 forms an invariant subgroup of O(N), denot by SO(N). An Ndimensional real matrix contains N 2 real paramete The column matrices of a real orthogonal matrix are normal and orthog nal to each other. There are N real constraints for the normalization a N(N 1)/2 real constraints for the orthogonality. The constraint on the d terminant is a discontinuous condition, which does not decrease the para eters. Thus, the number of independent real parameters needed for chara terizing the elements of SO(N) is 9 = N 2  N  N(N 1)/2 = N(N 1)/ The group space is a doublyconnected closed region so that SO (N) is doublyconnected compact Lie group with order 9 = N(N  1)/2. SO(N) is a subgroup of SU(N). The elements of the subgroup SO(N can be obtained by taking the parameters w~~) and w~3) of SU(N) to
Chap. 7 Lie Croups and Lie Algebras
308
vanishing. In the convention, one takes 80(N), R = exp
{i t
Wab
= wi~) /2 and
Tab
= 2T~;)
wabTab} '
a
(7.8
(Tab) cd =
i (baebbd  badbbc) ,
[Tab, Ted]
=
i
{ObeTad
+ Oadne 
ObdTac 
Oaend} .
The generators in the selfrepresentation of 80(N) satisfy the orthon mal condition so that the structure constant is totally antisymmetric w respect to three indices,
(7.8 Two generators Tab and Ted of 80(N) are commutable with each othe their subscripts are all different. The commutable generators
(7.8
span the Cartan subalgebra of both the groups 80(2e) and 80(2£ + Combining the remaining generators to be the common eigenvectors Eo. H j , [Hi, En] = Uj Eo. , one obtains four types of generators Eo. for 80(2 (1)
Eab
= "21
[
 "21
[
,(2) _ Eab (3)
E(Lb
T(2a)(2bl)
~
.
. . + ZT(2al)(2bl) + ZT(2(L) (2b)
1 [
= "2
Ei!) =
.
T(2a)(2bl)  ZT(2al)(2bl)  ZT(2a)(2b)  T(2al)(2b)
T(2a)(2bJ) 
[T(2a)(2bl)
 T(2al)(2b)
1,
1,
1 ZT(2aJ)(2bl) + 2T(2a)(2b) + T(2al)(2b) , .
+ iT(2al)(2bl)
.
 iT(2a)(2b)
+ T(2aJ)(2b)]
a<
,
(7.8 with the eigenvalues (roots) rea  eb}j' {e a + eb}j' rea + eb}j' a { eQ  eb} j' respectively, where {e a } J = 6aj . Two more types of g erators for 80(2£ + 1) are (5)
E"
(6)
En
=
(l [T(2a)(2f+l) V"2
=
(l [T(2a)(2C+l) + 2T(2aJ)(2C+l) . 1, V"2

.
ZT(2aJ)(2C+l)
],
(7.8
with the eigenvalues {eo.} j and  {e a } j. The generators labelled by ( (3), and (5) correspond to the positive roots.
§ 1.;' Classical Simple Lie Algebras
The simple roots Tp.
=
ep' 
Tp.
and the largest root w for SO(2£) are Te
ep.+l,
= e e J + ee,
bl
ea

eb
=
L
1S;j.lS;£1,
e 2
bl
Tp.,
:10
ea
+ eb
=
L
Til
+ 2L
Tv
+ Tel + Te,
(7.90
p.=a
[ 2
W
= e] + e2 = Tl + 2 L
Tv
+ Tel + Te·
(7.91
v=2
The angle of two neighbored roots is 211' /3 except for Te, which is orthogona to Te  l, but with an angle 211'/3 to Te2. dp. = 1 for all simple roots. Thus the algebra of SO(2£) is De. The simple roots T and the largest root w for SO(2£ + 1) are "
TI"
=
ep' 
ea

eb
ep.+l,
Te
bJ
=
L
1 S; j.l S; £  1,
= ee,
e
bl
TiLl
ea+eb= L
T ,1
+2L
p.=a
p.=a
Tv,
v= b
1"= a
(7.92
e w =
eJ
+ e2
=
Tl
+ 2 LTv.
(7.93
v=2
Except for Te, the angle of two neighbored roots is 211'/3 and dl' = l. But dE = 1/2 and the angle between Te and T e J is 311'/4. Thus, the algebra o SO(2C + 1) is Be.
7.4.3
The USp(U) Group and its Lie Algebra
In a (2€)dimensional space, the vector index a is taken to be j or j S; £, in the following order:
a = 1, I, 2, 2, ... ,
e, e.
J,
1S
(7.94
Generalizing the real orthogonal matrix by replacing the unit matrix with real antisymmetric matrix J, one obtains the real pseudoorthogonal matri fl,
RTJR where
= J,
R*
= R,
(7.95
310
Chap. 7 Lie Groups and Lie Algebras
when a = j, b = ], when a=], b=j, the remaining cases, det J
(
= 1.
Study the set of all (2€)dimensional real pseudoorthogonal matric in the multiplication rule of matrices. The product of two real ps orthogonal matrices is a real pseudoorthogonal matrix. The matrix uct satisfies the associative law . The unit matrix is also a real ps orthogonal matrix and plays the role of the identical element. The in R 1 = JRT J and the transpose RT of R are also real pseudoorthog
RJR T
= J,
Thus, the set of all (2€)dimensional real pseudoorthogonal matric in the multiplication rule of matrices , const.itutes a group , called the dimensional real symplectic group, denoted by Sp(2€, R). We are going to show that det R = 1 so that Sp(2£, R) is a conn Lie group. Let R be a transformation matrix in a (2£)dimensiona space 1;a
~ x~ = L RabXbl
R E Sp(2£, R).
(
b
The pseudoinner product of two vectors x and y is defined as
{x, Y}J ==
L
e
XaJabYb
ab
=L
(XjyyxyYj) ,
(
j=l
which is invariant in a real pseudoorthogonal transformation R
{x, yb
= {Rx, Ryb ·
(7
The pseudoinner selfproduct of a vector is vanishing. In terms of the totally antisymmetric tensor one has
L
l'.a, ... a2iJala2··· Jau_ lau
= 2f e!.
(7
a} ... a2C
In fact, in the sum there is only 2 f t! nonvanishing terms, which come the transpose of each J matrix and the permutations among £ differe
:1
§7.4 Classical Simple Lie Algebras
Denote by X~ the column matrix at the bth column of a (2£) x (2£) matr X. The determinant of X is
(7.10 Thus,
L
detX= (2££1)1
(detX)fal ... a2tlala2.·.lo2E_,a2t
al···a2.e
(2.'£1)1
L Q, ...
(2 .'01)1 (.
""" ~
fb, ... bz l
L
fb, .. b2eX~: ... X~~: b, ... b21 {Xb 1, X b2} J... {X b2;' . , X b2t} J.
le"oz···lou_102t
02t
Since the column matrix in RX is obtained from the column matrix in by a real pseudoorthogonal transformation R, one has from Eq. (7.10 that det (RX) = det X so that det R = 1. From the definition (7.95), R in Sp(2£, R) is given as follows if it is diagonal matrix
(7.10
where the real parameters Wj can be taken to be infinitely large such th Sp(2£, R) is not a compact Lie group. Replace the real matrix R in Eq. (7.95) with the unitary matrix u,
(7.10
The set of all (2f)dimensional unitary pseudoorthogonal matrices u, in t multiplication rule of matrices, forms a group, called the (2£)dimension unitary symplectic group, denoted by USp(2f). It can be similarly show that det u = 1. Since the module of any entry in a unitary matrix is finit USp(2£) is a connected compact Lie group. When f = I, 1 = i0"2 an (0" . n)T (i0"2) =  (i0"2) (0" . n). Thus, the elements in SU(2) satisfy t definition (7.104) so that SU(2) = USp(2). Discuss the infinitesimal elements R E Sp(2f, R) and u EUSp(2£)
R = 1  iaX,
XT = lXl,
u = 1  ij3Y,
yT = lY l,
X· = X, yt = Y.
(7.10
Both the imaginary matrix X and the Hermitian matrix Y contain (4 f real parameters. The component forms of XT = 1 X land yT = lY 1 ar
312
Chap. 'l Lie Groups and Lie Algebras
X kj
=
X kj =
XTk'
Ykj = Yj'k'
Y kj
X jk ,
= Yjk'
xl' =
lXl gives f!2 +£(£1)/2+£(£1)/2 = £(2£1) real constra yT = lY 1 gives £2 + e( e  1) £(2£  1) real constraints. Thus, the or of both Sp(2£, R) and USp(2£) are 9 = 4£2  £(2£  1) = £(2£ + 1). In the selfrepresentation of USp(2£) the generators can be expresse terms of the generators T}~) (r = 1,2) in SU(£) and the Pauli matrices From Eq. (7.105) they are
TR)
x
T)J)
CJd,
x
CJd/I2,
(7.
l::;j
1 ::; d ::; 3,
They satisfied the orthonormal condition Tr(TATa) = DAa so that structure constant C A aD of USp(2£) is totally antisymmetric and the Ki form gAa is a negative constant matrix. These coincide wit.h the fact USp(2e) is compact . The generators in the selfrepresentation of Sp(2e are pure imaginary and can be obtained from Eq. (7.106) by replacin and a3 with T) = iCJj and T3 = iCJ3, respectively. If one moves the additi fador i from the generators of Sp(2e, R) to its parameters, the generato the corresponding representations of Sp(2£, R) and USp(2£) are the s This is the standard method for calculating the irreducible representat of a noncom pact Lie group. We will calcula.te the representations o Lorentz group by this method in Chap. 9. The diagonal matrices in the generators (7.106) span the Cartan su gebra of the USp(2£) group,
1::;j::;f.
(7.
The remall1ll1g generators are combined to be the eigenvectors of
[Hj, E",]
=
O:jE",:
()) = {(!) x
Ejk
E;~)
= {
X
12
iTR)
X
12}
CJ3
T)~)
X CJ3 
(3) E jk =
T(l) ( jk X CJl
E;~)
TR)
=
. (2) + 1Tjk
Tjk
+ 1CJ2
X (CJl 
.) /
iCJ2)
M2 V L:,
/12,
} /
M V 2,
/12,
1 ::; j
<
k ::; £,
§7. 5 Represent.ations of a Simple Lie Algebra
E?)
=
TL1) x (CI]
+ iCl2) /2,
E)6)
=
TB)
~ iCl2)
X (Cll
/2,
3
(7.10
with the eigenvalues Jl72(ej ~ek), ~Jl72(ej ~ek), Jl72(ej +ek ~Jl72(ej + ek), J2ej, and ~J2ej, respectively. The positive roots are kJ
Jl72 (ej ~ ek) = L
rJ1.'
J1.=j
e ]
kl
Jl72(ej+eJJ=L rJ1.+ 2 L r,,+re, ,l=j J1.=k eJ J2ej = 2L r,,+re, J1.=j where rJ1. are the simple roots
rJ1. =
Jl72 (e,., ~ e,,+d ,
re
J2ee,
=
de
1 ::; J..L::; € ~ 1 ,
(7 .10
= 1.
The angle of two neighbored simple roots is 21f / 3 except for the angle rel and re which is 31T / 4. Thus , the algebra of USp(2€) is Ce The large root w in Ce is
el w =
7 .5
he]
= 2
L rJ1. ,,=1
+ reo
(7.11
Representations of a Simple Lie Algebra
A representation of a Lie group G is also a representation of the Lie algeb L of G. In a representatio n of a Lie algebra, one only studies its generator A simple Lie algebra has its compact real form where its representation equivalent to a unitary one. 7.5. 1
Representations and Weights
[n a unitary representation of a simple Lie algebra, the Cartan~ Wcyl bas of generators, denoted by D(Hj ) and D(Ea.) for convenience, satisfy
314
Chap. 'l Lie Groups and Lie Algebras
The basis vectors in the representation space, called the basis states and denoted by 1m), are usually chosen t.o be the common eigenvectors of the commutable Hermitian operators H j ,
H j 1m)
= mj
(7.111)
1m).
The edimensional vector m = (ml, ... , me), whose components are the eigenvalues mj, is called the weight of the basis state 1m). The edimensional space is called the weight space. A weight m is said to be multiple if the number n of the linearly independent basis states 1m) with the weight m in the representation space is larger than 1, and n is called the multiplicity of the weight m. A weight is called single if its multiplicity is l. For the adjoint representation, the representation space is the Lie algebra itself, a weight is a root, and the weight space coincides with the root space. Except for the identical representation, any irreducible representation of a simple Lie algebra with dimension larger than 1 is faithful and the generators are linearly independent of each other. Thus, there are elinearly independent weights in a representation of L with rank e. From the regular commutative relations, [Hj , Eo:l = (XjEo: and [Eo:, Eo:] = a· H, one has Tr D(Eo:)
= 0,
Tr D(Hj
)
= 0,
(7.112)
namely, the sum of weights of all basis states in an irreducible representation is vanishing,
L
(7.113)
m=O.
A weight m is said to be higher than a weight m' if the first non vanishing component of m  m' is positive. For a positive root a, one has
Hj (E±o: 1m)) =
[Hj , E±o:l 1m) (mj
+ E ±o:Hj
1m)
± (Xj) (E±or 1m)).
(7.114)
Thus, the action of Eo: on 1m) raises its weight m by a positive root a, and that of E_ OI lowers by a , so that Eo: is called the raising operator and E_ or the lowering operator. In a representation space with a finite dimension, there is a basis state 1M) whose weight M is higher than the weight of any other basis state 1m). lV! is called the highest weight of the representation and IM) is the highest weight state. Any generators Eo: with the positive root a annihilates the highest weight state 1M) , Eo: 1M)
= 0,
'1/ positive root a.
(7.115)
§ 7. 5 Representations of a Simple Lie Algebra
This is the main method to calculate the highest weight in a r
p l"l·~
hil i
Theorem 7.7 The highest weight M in an irreducible represelll.;I,I,;,"I , 01 a simple Lie algebra L is single. Two irreducible representations or I ; \I " equivalent if and only if their highest weights are the same. Proof Let 1M) be the highest weight state. We are going to prove' that if there is another state 1M)' with the highest weight M, it has to hC\ proportional to 1M). Since the representation is irreducible, 1M)' can be expressed as a combination of the following states each of which has Lhe highest weight M,
(7.116) If a raising operator, say Ep, is contained in Eq. (7.116), move Ep rightward according to EpEr = ErEp + [Ep, Er]. The additional term in the move is Np.rEp+r if p + T is a root, p' H if P + T = 0, and 0 if P + T is not a root. In each case, the additional term contains less operators than the original one. When the raising operator Ep moves to the rightmost position, the term is annihilated owing to Eq. (7.115), and the number of operators contained in each remaining term decreases. Making those moves of the raising operators in each term until it does not contain any raising operator, one finds that in the same time the lowering operators also disappear in the result terms because the weight of the term is M. Namely, each t.erm is proportional to 1M). The highest weights of two equivalent representations are obviously the same. Conversely, if the highest weights of two representations are the same, define a correspondence between the basis states of two representations
1M)
IJ) ==
E),. ... EfJEa. 1M)
ft ft
1M)'
IJ)' == E),. ... EfJEa. 1M)'.
We are going to show that the correspondence is onetoone, namely, any linear relation which holds in one representation space also holds in another representation space, and vice versa, \;j
I: j
Cj
Ij) = 0,
:J
I:
Cj
Ij)' == Iw)' = O.
j
By reduction to absurdity, if Iw)' i 0, it corresponds to the null state in the first representation space, so does the state Iw)' applied by the generators. Then, the set of those states constitutes an invariant subspace in the second
316
Chap. 1 Lie Groups and Lie Algebras
representation space, which is not equa.J to the whole space because it d not contain at least 1M)' . It is in contradiction to that the representa is irreducible. Hereafter, an irreducible representation of a simple Lie algebra L is called the highest weight representation. The basis states in the adjoint resentation space of a simple Lie algebra are the generators, and the wei are the roots, so that the highest weight of the adjoint representation is largest root w of £. 7.5.2
Weight Chain and Weyl Reflections
Theorem 7.8 If a is a nonvanishing root in a simple Lie algebra L m is a weight in an irreducible representation space of L , then,
2m·a
and m 
r (m/a) a

a·a
== r (m/a) = integer,
(7.1
is also a weight with the same multiplicity as m.
Proof Although Theorem 7.8 looks similar to Theorem 7.3, ther an important difference between them that the weight m is always on numerator and the multiplicity of a weight may be larger than 1. F multiple weight, any combination of the basis states with the same we is also a state in the representa.tion spa.ce. Before proving the theorem has to find a convention for choosing the basis states. Without loss of generality, let
(7.1
m·a 2':0.
Arbitrarily choose a basis state 1m) with the weight m and apply to it w EO'. successively until the state is annihilated,
0:::; n :::;
E;;' 1m) i= 0, 1m
+ pa) ==
Ef, 1m),
p,
Ef,+1 1m) = O.
Forget the choice of 1m). One defines a chain of basis states by appl EO'. to 1m + pa) successively,
+ na) == E~~n 1m + pa), E~~qH 1m + pa) = 0, 1m
q:::;
n:::;
p,
p + q + 1 > 0,
(7.1
where q is an integer larger than (p + 1). We do not care whether or the state E~O'. 1m +pa) coincides with the original state 1m). The prob
§ 7.5 Representations of a Simple Lie Algebra
3
is whether the subspace spanned by the basis states 1m + nO') is closed I.he application of E±O'., namely to show
EO'. 1m
+ nO') = En 1m + (n + 1)0'),
Prove it by induction. Equation (7.120) holds for n
EO'. 1m + PO')
(7.12
q ::; n ::; p.
= p and n = p 
= 0,
+ (p  1)0') = [EO'., EaJ 1m + PO') = 0" H 1m + PO') = (m· 0' + p10'12) 1m + PO'),
Ea. 1m
Ep
=
(7.12
0,
If Eq. (7.120) holds for n > k, when n = k,
Ea. 1m + kO')
EaEa. 1m + (k
+ 1)0') + 1)0') + EO'.EO'. 1m + (k + 1)0') = {m· 0' + (k + 1)10'1 2 + E k +!} 1m + (k + 1)0').
=
[Ea., EoJ 1m + (k
Thus, Eq. (7.120) is proved, and En satisfies the recursive relation
En = E n+1
= 8n+2
+ m . 0' + (n + 1)10'1 2 + 2m· 0' + {(n + 1) + (n + 2)} 10'1 2
=
Bn+(pn)
=
~(p 
+ (p 
n)m . 0'
n) {2m· 0'
1
E_q = 2(P + q) {2m· 0'
1
+ 2(P 
n)(n
+ p + 1)10'1
+ (n + p + 1)10'12}, + (q + P + 1)10'12}.
On t.he other hand,
o=
EoEa. 1m  qO')
+ EoEo} 1m  qO') {m· 0'  ql0'12 + B_q} 1m  qO'),
= {[Eo, EoJ =
E_q = m . 0'
+ q10'12.
In comparison one obtains
(p
+ q + 1) {2m . 0'  (q  p)IO'n
=
o.
'2
318
Chap. 7 Lie Groups and Lie Algebras
Since (p+q+ 1) > 0 and calculated to be q  p,
lal 2 > 0,
Eq. (7.117) is proved and the integ
2m·a
r (mla) = ~
= q p
=
integer.
(7.
Due to our convention (7.118), q '2: p '2: O. In the state chain 1m + given in Eq. (7.119), there are both states 1m) and 1m') with the we m and m', respectively, where
m' == m 
r
(mla) a = m  (q  p)a.
(7.
In the subspace orthogonal to the states given in Eq. (7.119) one f another state with the weight m and repeats the steps to obtain ano state chain containing two states with the weight m and m', respectiv If the multiplicity of the weight m is d, one is able to find d state ch so that the multiplicity d' of the weight m' is not less than d. Conver if the state chain is calculated from the state 1m'), one obtains d 2 Thus, two weights m and m' have the same multiplicity and are called equivalent weight
m·a m'
t
m al2a
o
Fig. 7.2
a
A Weyl reflection
Two weights m and m' are the mirror images with respect to the p perpendicular to a and across the origin. This reflection is called a W reflection in the weight space. The product of two Weyl reflections is fined as their successive applications. The set of all Weyl reflections their products for a representation forms the Weyl group W. The wei related by the elements of the Weyl group are equivalent, and the num of equivalent weights is called the size of the Weyl orbit of the weights A weight M satisfying
r
(M ITJ1.) = nonnegative integer,
'V simple root TJ1.,
(7.
is called a dominant weight. In an irreducible representation of a sim Lie algebra .c there are a few dominant weights, single or multiple. E
§1. 5 Representations of a Simple Lie Algebra
3
weight m in the representation is equivalent to one dominant weight. Th dimension of the representation is equal to the sum of products of th multiplicity of each dominant weight and its size of "Veyl orbit. The highe weight of an irreducible representation is a dominant weight and simp because of Eqs. (7.122), (7.115), and Theorem 7.7. Dynkin proved that space constructed by applying the lowering operators ETp successively a state 1M) with a dominant weight M is finite and corresponds to a irreducible representation of £. This is the foundation for the method the block weight diagram, which will be discussed in the next section.
Any dominant weight M is the highest weight of o irreducible representation of a simple Lie algebra £ with a finite dimensio
Theorem 7.9
7.5.3
Mathematical Property of Representations
The highest weight M gives the full property of an irreducible represe tation of a simple Lie algebra £. In this subsection we only quote som mathematical results of a highest weight representation. Let G be a compact simple Lie group with the Lie algebra £, and H the Abelian Lie subgroup produced from the Cartan subalgebra 1l. T elements in H are characterized by P. parameters <.p == (
R
= exp {i
{i t
}
E H.
(7.12
J=1
Each element in G is conjugate to an element in H so that the class in is characterized by t he parameters <.p. In an irreducible representation D the character X(M, <.p) of the class <.p is
x( M, '1')
~ 1, exp {i
t;
",jD(Hj) }
~ ~ him) exp {i
m} .
(7.12 b(m) is the multiplicity of the weight m, and in fact, the sum in E (7.126) runs over all basis states in the representation space. The charact X(M,<.p) can be calculated by the girdle ~(K,<.p) introduced by Weyl,
~(K,<.p)= ~ osexp{i t SEW
(SK)j
(7.12
j=1
where S is the element in the Weyl group and the reflection parity Os is
320
Chap. 'l Lie Groups and Lie Algebras
if S is a product of even Weyl reflections, and 1 if S is a product Weyl reflections. K = M + p, where p is the half sum of all positive in l... Then,
X
(M
,i.p
) = ~(K, i.p)
(
~(p,i.p)'
~ (p,
i.p) is independent of the representation and is related to the integral of the Lie group. The orthonormal relations of the charac two irreducible representations are expressed as
J
X(M, i.p)*X(M', i.p)I~(p, i.p)12(di.p) =
J~(K, i.p)*~(K',
i.p)(di.p) =
6
( When i.p goes to zero, dimension d( M),
xO\II, i.p)
d(M)
=
is an indefinite form and tends
M.O' .}  , II {1+ P·O'.
(
aEL'>.+
where .0.+ is the set of the positive roots in l... 7.5.4
Fundamental Dominant Weights
In the weight space of a simple Lie algebra L with rank fl., there dominant weights wJ1. satisfying
(
WiJ. is called the fundamental dominant weight. An irreducible repre tion whose highest weight is wiJ. is called the fundamental representa L. Comparing Eq. (7.131) with Eq. (7.62), one has
T"
=
L v=l
t
w,/A v,,,
Wv =
L
TiJ. (A1)iJ. v ·
(
J.L=!
Av" is the component of the simple root T" with respect to Wv. equal to an integer 2, 1, 2, 3, or O. The fundamental dominant are usually chosen to be the basis vectors in the weight space and root space such that the components of both the weight and the ro integers. In fact, M = LJ1. MJ1.wiJ. where MJ.L = r (M /TiJ.) is a nonne integer. rn = L" mJ1.wiJ. where miJ. = r (rn/TJ1.) is an integer. The
§'l.5 Representations of a Simple Lie Algebra
:l2
refl ection relation of two equivalent weights also becomes simpler,
(7.133 By making use of the symbol dJ.J.' one has from Eq. (7.131)
(7.134 dJ.J.A,.1V is usually called the symmetrized Cartan matrix. Furthermore ,
(7.135
T he shortage of the fundamental dominant weights is that they are no orthonormal.
7.5 .5
The Casimir Operator of Order 2
T he Casimir operator C 2 of order 2 is given in Eq. (7.25) which is com mutable with any generator lB. When the Killing form is a constant ma trix, C 2 = ~J\ I AlA. Due to the Schur theorem, C 2 is a constant matrix IC 2 (M) in an irreducible representation DM of a simple Lie algebra. The constant C 2 (M) is called the Casimir invariant of order 2 in the represen tation D M , or briefly the Casimir invariant. In the Cartan Weyl bases of generators,
C2
=L A
e
IAI/\
=L
L
HjHj +
{E.:.. E _ex
+ E  exEo.}.
(7.136
j= l
Applying C 2 to the highest weight state 1M), one has from Eq. (7.115),
C2 1M) = M2 1M) +
L L
[Eo.E_<> + Eo.Eo.] 1M)
<>Ef'.+
= M2 1M)
+
[Ee" Ea] 1M)
(7.137
e
= M· (M
+
2p) = L
MJ.J.dJ1. (A 1 )J1.V (Mv + 2),
J.J. ,v=1
where p is the half sum of the positive roots in the simple Lie algebra L and it is equal to the sum of the fundamental dominant weights
2p= L o.Ef'. +
e
a =
2L wJ.J.' J.J. = l
(7.138
322
Chap. 'l Lie Groups and Lie Algebras
In fact, it can be shown as follows that the inner product of 2p and simple root rl' is 2d w From any positive root a, except for the simple r 1" one is able to construct a root chain in ~+,
(a  qrl') , ... , (a  rl')' a, (a
+ rl')'
... , (a
+ prl').
The sum of the roots in the root chain is (p + q + 1) [a + r I' (p  q) /2], due to Eq. (7.46) it is orthogonal to rw The inner selfproduct of r 2dl'. Thus, Eq. (7.138) is proved. There is another way to calculate the Casimir invariant. The ma form of the generator IA in the representation DM is denoted by I;;:. fine a gdimensional matrix T(M), whose entry is TAB(M) =Tr[I;;: I For a compact simple Lie algebra, the adjoint representation is irreduc and the structure constant C ABD is totally antisymmetric with respec its three indices. Then, it can be shown from Eq. (7.9) that T(M commutable with any generator I~d in the adjoint representation,
[T(M), lAd] BD
=
L {Tr (I,W I{1) (I~d)
PD 
(I~d) BP Tr
(I{1 If!)
P
= iTr L {I,W I {1 CAt + cAl I{1 If!} P
= Tr
{1ft (I;;: 1[1  If! I;;:) + (I;;: I,W  1ft I;;:) If!}
Thus, T(M) is a constant matrix , denoted by TAB(M) Taking the trace, one has
=
=
o.
OABT2(
where d(M) is the dimension of the representation DM . The highest we M of the adjoint representation is the la rgest root w,
(7.
7.6
Main Data of Simple Lie Algebras
In this section we list the Dynkin diagram, the Cartan matrix A and inverse A I, the simple roots r 1" the positive roots, the largest root w, fundamental dominant weights wI" some Casimir invariants C 2 (M), the formulas for the dimensions d(M) of most simple Lie algebras. is the selfrepresentation. MI' is the component of the highest weight
§'l.6· Main Data of Simple Lie Algebras
32
in the basis vectors wJ1.' The Killing form is gAB = O AB C 2 (W). eJ1. ar the Euclidean basis vectors. The basis vectors Vj are given in Eq. (7.79 Due to the mathematical conventions, the half of square length of a longe simple root is dJ1. = l. For the Lie algebra At, dJ1. = 1/2 in the physic convention such that C 2 (M) as well as the Killing form shortens by a facto 2.
7.6.1
Lie Algebra Ai and Lie Group SU(l
+ 1)
The order of Ae is £(£+2) . There are £(£+1) roots, denoted by The Dynkin diagram of At is
J2 (Va
 Vb
000      00
1
A=
3
2 1 0
1 0 2 1 1 2
0 0
AI
= _1_
£2
2
0 0
£1
0 0 1
0
0 0 0
0 0
0
2 1  1 2
0 0
0
£
1.£ 1·(£1) 1 . (£  2) ... 1·(£1) 2·(£1) 2·(£2) ... 1·(£2) 2· (e  2) 3·(£2) ...
1·2 2·2 3·2
1.1 2·1 3·1
£+1
2·2 2·1
1·2 1.1
dJ1.
3·2 3·1
= I,
(£1) · 1
1 ::; Jt ::; £,
= W = WI + We, d(Madj) = £(e + 2),
MO=WI,
d(Mo)
(£  1) . 2 (£  1) . 1
Madj
= £+ 1 ,
C 2 (Mo) =
£2
+ 2£
e+l'
C 2 (Madj )
= 2(£ + 1).
,,=1
d(M)
=
} 11HI {bI + ?; b~J1.a 1
.
£·1
Cha.p. 'l Lie Groups a.nd Lie Algebms
324
7.6.2
Lie Algebra Bl and Lie Group 80(U
+ 1)
Being a Lie algebra, Bl ~ Al and B2 ~ C 2 . Here we list the data of algebra Be with £ ;::: 3. The order of Be is £(2£ + 1). There are 2 posi roots, denoted by eJ.t and eJ.t ± e//. The Dynkin diagram of Be is
e
000      00
•
1
e1
e
1 1 1 ... 1 2 2 ... 1 2 3 ...
1 2 3
2
2  I 0 . .. 0 I 2 1 .. _ 0 0 1 2 0
A
0 0
dJ.t =
0 0
3
0 0 0
e2
, A l
1 2 3 ... e  1 (£  1)/2 1 2 3 ... e1 e/2
2  1 0 0 ... 2 2
1
1,
~ JL ~
de
P.  1,
MO=Wl,
Madj=W=W2'
deMo) = 2£ + 1,
d( Madj)
C 2 (lWo)
= 2e,
= 2(2£ 
L,.;
v,
1 ~ JL
<
e,
= £(2e + 1)/4
C2 (Ms)
1),
J.t
J.t 
= 1/2,
Ms = we, d(Ms) = 2e,
= £(2£ + 1),
C 2 (Madj )
W'" e
1/2 1 3/2
Te
= ee,
1
e
We =:2
v=1
Lev. v=1
£1
e
d(M)
= IT
NIt + 2 1+
//1
IT
Mp
1+2(e..\)
'\=1
e
L
p= '\
1+
£ 1
Me + L Mp + L 2Mp __~p=___J.t,_ _P,=___v_ _ 1 + 2e  JL  v
vI
L
1+
Mp
,p=,J.t_ _
VJL
3
§'l.6 Main Data of Simple Lie Algebras
7 .6 .3
Lie Algebra Ce and Lie Group USp(U)
Iking a Lie algebra, C l ~ AI . Here we list the data of Lie algebra C e w P:::: 2. The order of Ce is €(2€ + 1). There are €2 positive roots, denoted Jl72 (ell ± e v ) and V2e,l . The Dynkin diagram of Ce is
0 0
0 0
e2
3
2 1 0 0 1 2 1 ... 0 0 1 2 0
o o
•

2
1
A
. .
•
•
0 0 0
, AI
1 2 3
1 2 3
1 2 3 €1 el 1/21 3/2 ... (e  1)/2 £/2
= I,
dlL = 1/2 , 1:::; f.L :::; (£  I),
de
Mo =
Maclj
WI,
=W
= 2W1,
= £(2e + I), C2 (Madj ) = 2(£ + 1).
d(Mo) = 2£, r ll =
€
1 1 1 1 2 2 1 2 3
22 . 1 2
C 2 (M o)
0
€1
d(Maclj)
= e+ 1/2,
Vlf2 {elL 
elL+d,
1:::; f.L:::; (e  I),
r e = V2ef,
1/=1
e
f
d(M) =
II
L 1+
Mp
p= ..\
£+1'\
,,\=1
e
vl
e
II 7.6.4
1+
L
Mp+
p=/"
L
vl
L
2Mp
p=v
2£+2f.Lv
1+
Mp
I '_ _ ,P=...c...
vJ.L
Lie Algebra De and Lie Group SO(U)
Being a Lie algebra, D2 ~ AL EE! AI an d D3 ~ A 3 · Here we list the data Lie algebra De with e :::: 4. The order of De is £(2£  1). There are e(£ 
Chap. 'l Lie Groups and Lie Algebras
positive roots, denoted by eJ1
± e v . The Dynkin dia.gra.m of De is ~£1
~~:3~£ 2 1 0 0 ... 0 1 2 1 0 ... 0 o 1 2 1 ... 0
0 0 0
0 0 0
A 0 0 0
1 2
AI
0 0 0
2 2 2 ... 2 4 4 ... 2 4 6 ...
2 4 6
1 2 3
1 2 3
2 4 6 ... 2(£  2) JI.2 JI.2 1 2 3 ... £2 (£2)/2 £/2 1 2 3 ... 1!2 (I!  2)/2 1!/2
1:S/l:S£,
= Wl,
MSJ =
2 1 1 0 0 ... 1 2 0 0 ... 1 0 2

dJ1=l,
Mo
0 0 0
Madj
= W = W2,
MS2 =
Wel,
= 21!, d(Msd = d(Ms2) = 2el, C2 (Madj ) = 4(1!  1),
We,
= JI.(2J1.  1), C2 (Mo) = 21!  1, C2 (Ms I ) = C 2 (Ms 2 ) = 1!(21! 
d(Mo)
d(Madj)
Te
= eel + ee,
v
Wv =
L
ep ,
1 :S
1/
:S I!  2,
p=1
1
WeJ
1 e
iI
="2 L
p=1
We
="2 L
p=l
ep .
1)
§ 7. 6 Main Data 0/ Simple Lie Algebras
t
vj
d( M ) =
vj
I: Mp + I: 2Mp 
e
II
1+
:3
Me  l  Me
LMp 1+
'P='J.1._ _'p:=:::v_ _ _ _ _ __
,P=,J.1._ _
2€  It  v
vIt
Lie Algebra G 2
7 .6 .5
====••
eEl 1 dl
= 1,
d(Mo)
= 1/3
d2
= 7,
AI
A=(21) 3 2 '
2
d(Madj)
= W2, Cz(Mo ) = 4,
,
Mo
= 14,
=
(2
1)
3 2
'
= W = WI, CZ(Madj ) = 8.
Madj
The order of G 2 is 14. There are six positive roots, rj = rj rl WI
r2
+ rz = J1i6 el + .J1fi ez, + 3r2 = J372 el  .J1fi e2 =
Jl16 el  .J1fi e2, rl + 2rz = V2f3 el, 2rl + 3r2 = J372 €I + jlf2 €z, W2 = V2f3 €l·
V2 e2, ,
J372 ej + .J1fi ez,
d(M)
=
=
+ M 1 )(1 + Mz)(4 + 3M1 + lvh)(5 + 3MI + 2M2 ) . (2 + MI + l\.fz)(3 + 2M + M 2 )/120 . (1
j
7.6.6
Lie Algebra F 4
•.   •• 3 4
O~C~I====~
1
2
1 0 0)
1 2 1 0 A= ( 2 o 2 2 1 o 0 1 2 dj
= d2 = 1,
Mo =
d3
W4,
Madj =
W
=
WI,
'
AI
=
( 3' 4 2
2 4 6 3 8 6 4 3
~)
= d4 = 1/2,
d(Mo) = 26,
C 2 (M o) = 12,
d(Madj) = 52,
C 2 (Madj) = 18.
Chap. 7 Lie Groups and Lie Algebras
328
T1
= eZ
T3
==
WI
W3
== e3  e4, T4 == (el  e2  e3 W2 == 2el + e2 + e3, W4 == el'
 e3,
T2
e4,
= el + e2, == (3el + e2 + e3 + e4) 12,
e4)
12,
The order of F4 is 52. B4 is a subalgebra of F 4 . F4 contains 48 roo including 32 roots in B4 and 16 other roots denoted by ae] +be2 +ce3 +d where four parameters are taken to be ±I independently.
d(M) = {l + (2M] + 4M2 + 3M3 + 2M4) Ill} {l + Md {I + M2} {l + M 3 } {I + M4} {l + (2MI + 2Jvh + M3) 15} {I + (2M2 + M3) 13} {I + (2M] + 31Vh + 2M3 + M4) 18} {I + (M2 + M3 + M 4 ) 13} {l + (MI + 3M2 + 2M3 + M4) 17} {I + (Ml + M2 + M3 + M 4 ) 14} {l + (Ml + 2Jvh + 2M3 + M 4 ) 16 {I + (M] + 2M2 + M3 + M 4 ) 15} {I + (M] + 2M2 + M 3 ) 14} {I + (MI + M2 + M3) 13} {I + (Ml + M z) 12} {I + (M2 + M3) 12 {I + (2M] + 4M2 + 3M3 + M 4 ) IIO} {I + (2M2 + 2Jvh + M4) 15} {I + (2M] + 2M2 + 2M3 + M 4 ) 17} {I + (2Ml + 4M2 + 2M3 + M 4 ) 19} {l + (M3 + M 4 ) 12} {I + (2M2 + M3 + M 4 ) 14} {I + (2MI + 2M2 + M3 + M 4 ) 16}
7.6.7
Lie Algebra E6
~ A==
2 1 0 0 0 0
 1 2 1 0 0
0
1 2 1
0 0 1
1 0 0 1 2 1 0
0 0
2 0
0 0 1 1 0 2 0 0 2
dfJ. == I,
1 :::;
d(Mo) == 27 ,
d(Madj) == 78,
f.1 :::;
6,
4
3
AI =
~
3
Mo
5 4 5 6 4 2 3
5 10 12 8 4 6
= WI,
C 2 (Mo) == 52/3 ,
6 12 18 12 6 9
4 2 3 4 6 6 9 5 6 4 3 3 6
8 12 10 5 6
Madj ==
== W6 C2 (Madj ) == 24. W
The order of E6 is 78. A5 is a subalgebra of E 6. E6 contains 72 roo including all 30 roots in A5 , V2 (Ua  Ub), and 42 other roots denoted ±e6 and V2 (Ua + U b + U c ) ± e6/V2, where 1 :::; a < b < c :::; 6. Va
329
§'l.6 Main Data of Simple Lie Algebras
I.lte basis vectors for As (see Eq. (7.79)). The first five components of Va are the components of Va but the sixth component is O.
Tl T3 TS Wl
J2e6, = J2 (V2 = J2 (V4
Ws =
7 .6.8
 V 3) ,
T4
 Vs) ,
7'6
+ e6/ J2, J2 (V) + V 2  2V6) , J2 (Vs + V 6) ,
e6/J2,
 V 4),  V2)
,
J2 (V) + V 2 + V3
W4 =
 V6),
W6=J2(V j V6 ).
Lie Algebra E7
4
1
2
A=
1 2 1 0 0 0 0 0 1 2 1 0 0 1 0 0 1 2 1 0 0
3
2 1 0
0 0 0
AI
dll
= J2(V3 = J2 (VI

W2 = 2J2V6 ,
= J2V6
W3 =
J2 (V3 + V 4 + Vs)
T2 =
=
= 1,
d(Mo)
1
= 2
= J2(Vs
0
0 1 2 1 0 0 0 1 2 0 0 1 0 0 0 2 4 6 8 6 4 2 4
6 12 16 12 8 4 8
8 16 24 18 12 6 12
6 12 18 15 10 5 9
4 8 12 10 8 4 6
2 4 6 5 4 3 3
4 8 12 9 6 3 7
= W6, C 2 (Mo) = 57/2,
Mo
 V 2) ,
T3=J2(V3 V4), TS
0
0
= 56, d(Madj) = 133, = J2 (Vi
0
6
0
1 ::::; JL ::::; 7,
Tl
0
5
 V 6 ),
T2 T4 T6
= J2 (V2 = J2 (V4 = J2(V6 
= W = Wj, C 2 (Madj ) = 36 . Madj
V 3) , Vs ) , V 7 ),
Chap. '7 Lie Groups and Lie Algebras
:330
T7
W2 W4
= V2 (V4 + V5 + V6 + V 7 ) ,
Wi
= V2 (Vi + V 2  2V8 ) , W3 = V2 (V5 + V6 + V 7 + 3V8 ), W5
w6 = V2 (V7
+ V 8) ,
= V2(Vl  Va), = V2 (Vr + V2 + V3  3V8 ), = V2 (V6 + V7 + 2V8 ),
W7 = 2V2Vs ,
where Va are the basis vectors for A7 (see Eq. (7.79)). The order o E7 is 133. A7 is a subalgebra of E 7. E7 conta.ins 126 roots, includ ing all 56 roots in A7 , V2 (Va  Vb), and 70 other roots denoted b V2 (Va + Vb + Vc + V d ), where 1 ~ a < b < c < d ~ 8.
7.6.9
Lie Algebra Es
8
2
A =
d/<=I,
d(Mo)
3
4
5
6
7
2 1 0 0 0 0 0 0 1 2 1 0 0 0 0 0 o 1 2 1 0 0 0 0 0 0 1 2 1 0 0 0 0 0 0 1 2 1 0 1 0 0 0 0 1 2 1 0 0 0 0 0 o  1 2 0 0 0 0 o 1 0 0 2 2 3 4 5 6 4 2 3 3 6 8 10 12 8 4 6 4 8 12 15 18 12 6 9 5 10 15 20 24 16 8 12 6 12 18 24 30 20 10 15 4 8 12 16 20 14 7 10 2 4 6 8 10 7 4 5 3 6 9 12 15 10 5 8 1~J.i~8,
= 248,
Mo=Madj=W=Wi'
332
Chap. 7 Lie Groups and Lie Algebras
where A lLv = (rv)1L is the Cartan matrix. Recall that the difference of simple roots is not a root, and the root chain r lL + nr v , 0 :::; n :::; p, bre off at p = A vlL . Therefore, in addition to Eq. (7.142) the Chevalley ba satisfy the Serre relations
[Ev, [Ev, ... [Ev, ElL] ... ] = 0, ,
v
I
IA""
JFv, [Fv~ ... [Fv,' F IL ] ... ]
(7 .1
= O.
IA""
As discussed in Chap. 4, the basis states in an irreducible representat space of the SU(2) group (AI algebra) are denoted by Ij,m), j :::; m::: The actions of the generators Ja on the basis state are
.hlj, m) = mlj, m),
(JI  iLl Ij, m)
= )(j + m)(j
 m
+ 1)1j, m 
The Chevalley bases for the AI algebra are
H=21J ,
E = J I +iJ2 ,
[H, E] = 2E,
[H, F]
= 2F,
F = J1

iJ2 ,
[E, F] =H.
(7.1
Changing the notations for the basis states, one has
Ij,j) = IM,M),
Ij,m) = IM,M  2n),
H IM , M  2n) = (M  2n) IM,M  2n),
F 1M, M  2n)
=
+ 1) 1M, M  2n  2) , n)(n + 1) 1M, M  2n),
(7.1
)(M  n)(n
ElM , M  2n  2) = )(M 
where 0 :::; n :::; M = 2j and m = j  n. The matrix form of H is diago and the form of F is the transpose of that of E. Now, three generators of Chevalley bases with a given subscript f.t sat the same commutative relations as Eq. (7.144),
(7.1
Thus , in th e simple Lie algebra .c, three generators ElL' FIL , and HIL s an Al subalgebra, denoted by AIL' In an irreducible representation sp of .c, som e states span an irreducible representation space of AIL' ca AILmult.iplet . Those AILmultiplets embed in t he representation space o in a compli cated way. The method of the block weight diagram will anal the complica ted overlapping of the AILmultiplets and calculate the ma forms of the generators.
3:
§1. 1 Block Weight Diagrams
7 .7 .2
Orthonormal Basis States
In an irreducible representation space of .L, the common eigenstates of H are chosen to be the basis states Hf'
1M, m)
= mf'
1M, m).
(7.147
The weight m is expanded with respect to the fundamental dominan weight w" so that the components mf' all are integers. Due to Eq. (7.142 t he action of EJl raises the weight m of the state 1M, m) by a simple roo T" and the action of FJl lowers m by r!1' The states with different weigh are orthogonal to each other. For convenience, the basis states are require to be orthonormal. In addition, when the weight is nmultiple, the ba sis states can be made any arbitrary unitary transformation. When n = only the phase of the basis state can be chosen. The phases and the unitar transformations should be chosen carefully as will be discussed later. In a block weight diagram of an irreducible representation of .L, eac orthonormal basis state with weight m is denoted by a block filled with i components mJl' When a component is negative, it is denoted by a numbe with a bar. For example, a state with a single weight m = 2Wl + W is denoted by When the weight is multiple, an ordinal number
ITIJ.
added as a subscript such as 1 (2,1 h I· A suitable combination of states 1M, m) with a given weight m in a irreducible representation DM can be obtained by applications of a few lowering operators FJl to the highest weight state 1M, M), so that th difference M  m is equal to the integral sum of the simple roots r Jl ,
e M  m = ~ Gf'rJl'
Jl=!
e h(M, m)
=~
GJl'
(7.148
Jl=!
h(M, m) is called the height of the weight m in the representation DM T he height of the highest weight is O. In the block weight diagram, t.he block for the highest weight is locate in the first row. The remaining blocks are arranged downward as the heigh increases. The blocks for the weights with the same height are put in th same row. Two blocks in the neighbored rows are connected by a specia line for the index p. if their states are related by the lowering operator FJ The line is indicated by the corresponding matrix entry of FJl except fo t.he entry 1 which may be neglected for convenience. The line is remove when the matrix entry is O.
334
Chap. 7 Lie Groups and Lie Algebras
There are two main problems in the calculations of the basis states alld the generators of the highest weight representation DM. One is how to pi("\< up the Allmuitiplets embedding in the representation space. The other i: how to determine the multiplicities of the weights. First, pick up the Allmuitiplets. For a normalized basis state 1M, m) which has a positive component m" . of m and is annihilated by Ell' Ell IM,m) =0,
(7.149)
one is able to construct an Allmultiplet, which contains (mil + 1) orthonormal basis stat.es, by applications of FIl successively, FIl 1M, m  nr ll )
=
J(mJ.l.  n)(n
+ 1) 1M, m
 (n
+ l)r'L)'
+ 1) 1M, m
 nr,,),
(7.150)
It can be shown by induction that
E,L 1M, m  (n
+ l)r ll ) =
J(m}L  n)(n
(7.151)
namely, the matrix form of EJ.l. is the transpose of that of F,L' The highest weight state 1M, M) satisfies the condition (7.149) so that one first constructs the AJ.l.multiplets from the highest weight state by Eq. (7.150). If there is a basis state in the Allmultiplet again satisfying the condition (7.149) but with another positive component, say ml/, one is able to construct an AI/multiplet by Eq. (7.150). New Apmultiplets can be constructed one by one from the basis states satisfying the condition (7.149). The connected multiplets beginning from the highest weight state are called a path. The state on the path is said to have a path connecting the highest weight. If the weights in an Allmultiplet all are single, 1M, mnr,,), including their phases, have been chosen to be the orthonormal basis states in the representation space. The difficulty comes from the intersection of, say, an A,"multiplet and an AI/multiplet at a weight mi . Generally, those two states with the weight m' are not orthogonal to each other. One can define that one basis state 1M, (m'h) belongs to, say, the Allmultiplet, and th other basis state 1M, (m'h) is orthogonal to the first one. The state with the weight m', belonging to the AI/multiplet, is a combination of above two basis states, and the matrix entries can be determined by the formula
(7.152) The detail will be explained through examples later.
§'l. 'l Block Weight Diagrams
335
Second, determine the multiplicities of the weights. It is well known t.hat the multiplicities of equivalent weights are the same. Any weight is equivalent to a dominant weight, so that the key is to determine the multiplicities of the dominant weights in the representation. The highest weight M is a dominant weight and simple. A weight equivalent to M is simple. A weight belongs to the same Allmultiplet as M is simple. A weight connecting to the highest weight through only one path is simple. Beginning with the highest weight state, one constructs Apmultiplets one by one from the basis states satisfying the condition (7.149) until a dominant weight Ml appears. Assume that the multiplicity of the dominant weight Ml is equal to the number n of paths through which M l connects to the highest weight M. Define n orthonormal basis states 1M , (MdJ, 1 ~ a .:::: n. According to the formula (7.152), one alculates the matrix entry of each F).L which lowers a state to the states 1M, (M1 ) J. If t.hen? is a state 1M, (Md a ) to which the matrix entry of each F).L is vanishing, it is a null state and the multiplicity of the dominant weight Ml decreases by 1.
7.7.3
Method of Block Weight Diagram
For a given highest weight representation DM of a simple Lie algebra L, draw its block weight diagram according to the following steps: (a) Write the expansions of AI'" is the Cartan matrix of L.
Tv
=
I:).L WW4,LV
(see Eq. (7.132)) where
(b) From Eq . (7.133), calcula te the weights equivalent to the highest weight. Those wei ghts are all single. (c) In the first row of the block weight diagram, draw a block filled with the components of the highest weight M. For each positive component Mil' one constructs an AILmultiplet by Eq. (7.150). Draw the blocks for the states in the multiplet downward one by one and connect the neighbored blocks with a special line attached with the matrix entry of F" as given in Eq. (7.150). Use different line (solid line, dotted line, and doubleline) for different 11. The matrix entry is usually neglected if it is equal to 1. The line is removed if the matrix entry is O.
(d) In the multiplets, find the weight m satisfying the condition (7.149) and construct new Afl.multiplet from the state with the weight m until an intersection between multiplets appears. If the weight m' at the intersection
336
Chap. 7 Lie Groups and Lie Algebras
is not a dominant weight, its multiplicity is known. If m' is a domi weight, check the number n of paths through which m' connects to highest weight M, and assume the multiplicity of m' to be n. Let us explain the calculation for the case where the multiplicit m' is 2 for simplicity. Assume that the intersecting multiplets are multiplet and AI/multiplet. Define one state 1M, (m'h) belonging to Altmuitiplet and the other 1M, (m'h) as the highest weight state of a Altmultiplet, so that EItIM, (m'h) is known and EItIM, (m'h) = O. matrix entries of FII as well as Ell can be calculated by Eq. (7.152 EIIIM, (m'h) is also equal to 0, 1M, (m'h) = 0 and m' is single.
(e) If there is a dominant weight appearing in an above multiplet its multiplicity has been calculated, calculate its equivalent weights by (7.133). They have the same multiplicity.
(f) Continue this method until a state with a weight  M' app where M' has no positive component (see Theorem 7.9). M' is lowest weight in the irreducible representation space DM. If M' = M representation DM is selfconjugate. Otherwise, two representations and DM' are conjugate to each other. Denote by D M ' the conjugate representat ion of DM
DM (R) = 1  i DM' (R)
=1
i
L A
L
wAD M (fA),
A
wAD M' (fA)
=1 
i
L
WA { _DM (fA)'} ,
A
DM'(fA) = _DM(fA) * = _DM(fA)T. The representation matrix of the Chevalley bases in D M
'
are
D M' (Hit) = _DM (Hit) , D M' (E It ) = _DM (Fit), D M' (Fit) = _ DM (EIt).
The minus sign can be attracted into the basis states in the conju representa tion D M '
IM' , m) = (_l)h(M .,.,.,,) IM, m)* .
(7.
Thus, (7
§'l. 'l Block Weight Diagrams
33
T he block weight diagrams of two conjugate represent.ations are upsid down of each other, and the corresponding states are related by Eq. (7.153) In drawing a block weight diagram one should check step by step. Chec whether the difference of two weights related by FJ.t is TJ.t, whether Eq (7.152) holds when it applies to each basis state, and whether the multi plicities of the equivalent weights are the same. In the completed bloc weight diagram, the number of blocks with the same height first increase and then decreases as the height increases, symmetric up and down like spindle. An excellent table book [Bremner et at. (1985)] is recommende where the useful data of the highest weight representations of all simpl Lie algebras with the rank less than 13 are listed such as their dimensions the number of different heights, and the multiplicities and orbit sizes of th dominant weights in the representation. In the succeeding subsections som examples will be given to explain the method of the block weight diagrams 7 .7.4
Some Representations of A2
(a) (1,0)
(b) (0,1)
(d) (3,0)
(e) (0,3)
(c) (1,1)
Fig. 7.3
111
The block weight diagrams of some representations of A 2 .
The Lie algebra of the SU(3) group is A2 . It has wide applicatio particle physics. The representations of A2 are easy to study by th
338
Chap. 7 Lie Groups and Lie Algebras
method of block weight diagrams. The calculations are left to be exerc for readers (see Prob. 9 of Chap. 7 in [Ma and Gu (2004)]). The re are given in Fig. 7.3. From Fig. 7.3 one sees that the represent (1,0) is the selfrepresentation of SU(3), and the representation (0,1) conjugate one. The representation (0,3) is conjugate to the representa (3,0). The representation (1,1) is the adjoint representation of SU(3). matrix entries of generators are indicated. From Fig. 7.3 (c) one sees there is a double weight (0,0) in the representation (1,1). The ma entries of generators related to the double weight are listed as follows
= J21(1,1),(O,O)l), F]I(I, I), (0, 0lI) = J21(1, 1), (2, 1)), F2 1(1, 1), (1,2)) = Ji721(I, 1), (0, Oh) + J3721(1, 1), (0, Oh), F21 (1, 1), (0, h) = Ji721 (1, 1), (1, 2)) , F21 (1,1), (0, 0h) = J3721 (1,1), (1,2)). F 1 1(1,1),(2,I))
°
(7
Three positive roots can be read from Fig. 7.3 (c), T] = 2Wl  W2, WI + 2W2, and a = WI + W2. From Eq. (7.79) and Sec. 7.6 one ha
(7
7.7.5
Some RepTesentations of C3
The Lie algebra of group USp(6) is C 3 . The Dynkin diagram and Cartan matrix of C3 are ••4'~=::::(O 123
Thus, the simple roots of C 3 are T3
= 2W2+2w3· (7
We are going t.o study three fundamental representations of C 3 . First, the block weight diagram of M(1) = (1,0,0). The weights equivale M(l) are
(1,0, 0) ~ (1,1, 0) ~ (0, I, 1) ~ (0, I, I) ~ (1, I, 0) ~ (1,0,0
§1. 'l Block Weight Diagrams
(a)
(1,0,0)
Fig. 7.4
(b)
(0,1,0)
33
(c)
(0,0,1)
The block weight diagrams of some representations of C 3 .
Since the first component of M(I) is 1, one constructs an AIdoublet (1,0,0) and (1,1,0), where the matrix entry of Fl is 1 and the differenc between the two weights is TJ = (2, 1,0). Then, from (1, 1,0), one con structs an A 2 doublet with (0, 1,1), where the matrix entry of F2 is 1 an the difference between the two weights is T2 = (1,2, I). From (0, 1,1), a A 3 doublet is obtained with (0,1, I), where the matrix entry of F3 is 1 an the difference between the two weights is T3 = (0,2,2). From (0,1,1), on has an A 2 doublet with (1, 1,0), and the matrix entry of F2 is 1. At las from (1, I, 0), one has an Atdoublet with (1,0,0), and the matrix entry o FI is 1. (1,0,0) is the lowest weight because it contains no positive compo nent. Since (1,0,0) =  M( I), the representation (1,0,0) is a selfconjugat representation. In fact, it is the selfrepresentation of USp(6). The bloc weight diagram is given in Fig . 7.4 (a). In the representation (1,0,0), ther is only one dominant. weight, and six weights are all single and equivalen
340
Chap. 'l Lie Groups and Lie Algebras
to each other. Since the Ai'multiplets in the representation are all blets, the nonvanishing matrix entries of Fi' are always 1, which can be shown in formulas Fl 1(1,0,0)) = 1(1,1,0)),
F2 1(1,1,0)) = 1(0, 1,1)),
F3 1(0, 1,1)) = 1(0,1, I)), FI 1(1, I, 0)) = 1(I, 0, 0)) .
F2 1(0,1, I)) = 1(1, 1,0)),
(7
Here and after, the representation symbol (1,0,0) in the basis stat neglected for convenience if without confusion. Second, draw the block weight diagram of M(2) = (0,1,0). The we equivalent to M(2) are (0, 1, 0) , (1, I, 1), (I, 0, 1), (1, 1, I) , (I, 2, I), (2, I, 0) , (1, 2, 1), (2, 1, 0) , (1, 0, I), (I, I, 1), (I, 1, I) , (0, I, 0) .
(7
From (0,1,0) one constructs an A 2 doublet with (1, 1,1). The w (1, 1,1) has two positive components so that from it two doublets are structed, an AIdoublet with (1,0,1) and an A 3 doublet with (1,1, I). (1,0,1) an A 3 doublet is constructed with (1,2, I). From (1,1, I), an doublet is constructed with (1,2, I) and an A 2 doublet is constructed (2, 1,0). Now, the weight (1,2, I) appears twice, one in an A 3 doublet (1,0, 1) and the other in an A I doublet with (1, 1, I). Since (1,2, I) is e alent to the highest weight M(2), it is single and the two states in two blets have to coincide with each other. In fact, if F31(I,0, 1)) = 1(1,2 one has from Eq. (7.152)
1(1,1, I)) = F31(1, 1,1)) = F3Ed(I , 0, 1)) = EIF31(I,0, 1)) = EJ 1(1, 2,
From (1,2, I), an A 2 triplet is constructed with (0,0,0) and (1, From (2,1,0), an AItriplet is constructed with (0,0,0) and (2, (0,0,0) is a dominant weight and has two paths connecting to the hig weight M(2) so that one has to assume temporarily that its multiplicity be 2. Define two basis states I(O,O,O)[) and I(O,O,Oh) such that 1(0,0 belongs to the A 1 triplet with (2, I, 0) and (2, 1,0), and 1(0,0, Oh) beI to an A [singlet, FI 1(2, I, 0)) =
V2 1(0, 0, 0 h ),
E[I(O, 0, Oh) = F11(0, 0, Oh) = 0,
F21(I,2,I)) =al 1(0,0,Oh)+a21(0,0,Oh), F21(0,0,0)[) = a3 1(1,2,1)),
aT+a~ =2,
F21(0,0,Oh) = a4 1(1,2,1)).
§ 7. 7 Block Weight Diagrams
3
Applying Eq. (7.152) to 1(1,2, I)), one has
E}F2 1(I, 2, I)) = an/2 1(2, I , 0)) = F2E11(I, 2, I)) = 1(2, 1,0)).
Jl72.
Thus, al = Choosing the phase of 1(0,0, Oh) such that a2 is = J3fi. Applying Eq. (7.152) j10sitive number, one has a2 = )2 1(0,0,0)1)' one has
E 2 F2 1(O,O,O)J)
at
= a~
1(0,0,0)1) + a3a4 I(O,O,Oh) = (1/2) 1(0,0, Oh) + J374 1(0,0,0)2).
= (F2E2 + H 2 ) 1(0,0, Oh)
Choosing the phase of 1(1,2,1)) such that a3 is a positive number, o has a3 = and then, a4 = J3j2. The lower half of the diagram symmetric to the upper half. Similarly, one obtains
Jl72,
F I I(l,2, 1))
=
1(1,1,1)),
= 1(1,1,1)), F I I(l, 0,1)) = 1(1, I, I)), F21(2,I,O))
F31 (1, 2, 1)) = 1(1, 0, I)) , F;ll (I, 1,1))
=
1(1, 1,1)),
F 21(I, I, T))
=
1(0,1,0)).
The representation (0, 1,0) is selfconjugate with dimension 14. In additi to M(Z), it contains another dominant weight (0,0,0) with multiplicity At last, draw the block weight diagram of M(3) = (0,0,1). The weigh equivalent to M(3) are (0,0, I),
(0,2 , I),
(2,2,1) ,
(2,0, I),
(2,0 , 1) ,
(2,2, I),
(0,2,1),
(0,0,1).
From (0,0, I), an A3doublet is constructed with (0,2,1). From (0,2, I an A2triplet is constructed with (1,0,0) and (2,2, I),
F 31(O, 2, I))
=
h
1(1,0,0)),
F31(I, 0, 0))
=
h
1(2,2,1)).
1(1,0,0)) is a dominant weight but has only one path connecting to t highest weight M(3) so t.hat it is single. The weights equivalent to (1,0, have been calculated. From (1,0,0) an AIdoublet is constructed wi (1 ,1,0). From (1,1,0) an Azdoublet is constructed with (0,1, 1). Fro (2,2,1) an AItriplet is constructed with (0, 1,1) and (2,0,1). Since (0, 1, is equivalent to (1,0,0), it is single and two states in the A2doublet wi (I, 1,0) and in the AItriplet with (2,2,1) and (2,0,1) have to coincide. fact, if F2 I(1, 1,0)) = 1(0, 1,1)), one has
/21(2 , 2,1)) = F21(l, 0, 0)) = FzE11(I, 1,0)) = EIF21(I, 1,0)) = E11(0, 1,1)
342
Chap. 'l Lie Groups and Lie Algebras
The remaining part of the block weight diagram for (0,0,1) can be d similarly. The matrix entries of FIJ. are listed as follows:
J2
F3 12,2,1) = 12,0, I), F] 10,1, I) = ,j2 12,2, I),
F] 12 , 0, I) =
F3 12,0,1) = 12,2, I), F] 11,1,0)= 11,0,0),
F2 10,1, I) = II, I, 0),
F2 11,0,0) = ,j2 10,2,1),
F3 10,2,1) = 10,0, I).
10, I, I),
F3 10, 1,1) = 10, I, I), F2 12,2, I) = ,j2 11,0,0),
The representation (0,0,1) is selfconjugate with dimension 14. It con two dominant weights (0,0,1) and (1,0,0) which are both single.
7.7.6
Planar Weight Diagrams
For a tworank Lie algebra, each weight has two components so tha weights and their multiplicities in a representation DM can be drawn dently in a planar rectangular coordinate system, called the planar w diagram. Since the fundamental dominant weights wlJ. are generally orthonormal and not necessary to be along the coordinate axes, it is convenient to express the weights and roots in the unit vectors e a a the coordinate axes. The planar weight diagram of an irreducible repre tation is the inversion of that of its conjugate representation with res to the origin. The planar weight diagram of a selfconjugate irreduc representation is symmetric with respect to the inversion.
e2
(r,O)
(2,2)
e2
(1,2)
(T,l)
(0,0) (1,0)
•
•
el
(1,2)
• (0, I)
•
(0, 1)
• (1, I)
(0,0) I
el
(
e2 T2
(1,2)
W
(
(1,0) (0,0)2
•
(1,2)
(
(0,2)
(a)
Fig. 7.5
(1 , 0)
(b)
(0,1)
(c)
(0,2)
The planar weight diagrams of some representations of SO
p.8
34
Clebsch Gordan Coefficients
For the Lie group SU(3) (the Lie algebra A2), the expressions of th l'undamental dominant weights wp. and the simple roots Tv with respect t e a are given in Eq. (7.156). The planar weight diagrams of some repre sentations in A2 are shown in Fig. 7.3 of [Ma and Gu (2004)]. For the L group SO(5) (the Lie algebra B2 ), which is locally isomorphic to the grou USp(4) (the Lie algebra C 2 ), the expressions of wp. and Tv are
The planar weight diagrams of some representations of B2 are given in Fig 7.5. The readers are encouraged to compare the planar weight diagrams o B2 with those of C 2 which are given in Prob. 10 of Chap. 7 of [Ma an Gu (2004)]. The planar weight diagrams of some representations of G 2 ar shown in Prob. 11 of Chap. 7 of [Ma and Gu (2004)].
7.8
ClebschGordan Coefficients .
M(l)
M(2)
. .
.
M(
The dIrect product D xD of two lrt'educlble representatIons D and D M (2) of a simple Lie algebra L is generally reducible and can b "1 anty ' . e M(1 )M I 2) , . re d uce d by a umtary SImI trans f ormatIOn
(e
M11 )M(2»)  I
(DM(I) x D M (2»)
e
M(1 )M(2)
= EB
aMDM.
(7.161
M
aM is the multiplicity of the representation DM in the reduction. Deno by d(M) the dimension of DM. The dimension of the space of dire product of two representations is d(M(1))d(M(2)), so is the dimension e M i l ) M(2). There are two sets of basis vectors in the product space. Befo transformation, the basis states are the products of two basis states of th representations DM(I) and D M(2 ), denoted by IM(1), m(1))IM(2), m(2) or briefly by Im(l))lm(2)). After transformation, the basis states belong t the representation DM, denoted by 11M, (r), m), where the ordinal inde r is used when the multiplicity aM of DM is larger than 1. Two sets . . M(I)M(2) basIs states are related by the ClebschGordan (CG) matnx e
11M, (r), m)
=
l::
Im (I) ) Im (2) )
e
M(I)M(2) 771 (I),771(2),M,(r),771 '
(7.162
771{l) ,771(2)
where the sum runs over all states, especially when the weights m(l) an are multiple. The series on the righthand side in Eq. (7.161) is calle
m(2)
344
Chap. 7 Lie Groups and Lie Algebras
the CG series, and the matrix entries of CM(J) M(2) are the CG coeffici Applying the generator fA to the two sides of Eq. (7.162), one has
'L
fAIIM,m) =
{(fAlm(1))) Im(2))
+ Im(I))
(f Alm(2)))}
=(1) ,771.'2)
.
C M (t)M(2) m(J),m(2),M,(r),1TI .
(7. When fA is taken to be HIl' one obtains M(l)M(2) m Cm(1),m.(2),M,(r),m. =
((I)
m
+m
(2)) cM(l)M(2) Tn(J),m.(2),M,(1'),m. ,
namely,
(7.
The weights before and after the transformation are the same. There are two main tasks in reducing the direct product of two re sentations. One is to determine the CG series. The other is to find expansions (7.162) for the highest weight states 11M, (1'), M)
11M , (1'),M)
"'" 1m (1) )IM  m(1))
= ~
M(1)M('2) Cm.(l),(M=(l)),M,(r) , M
(7 The expansions for the remaining states can be calculated by applying lowering operators PIl to Eq. (7.165) (see Eq. (7.163)). In this section method of the dominant weight diagrams will be introduced to calc the CG series and CG coefficients.
7.8.1
Representations in the CG Series
What kind of representations M will appear in the CG series? If in Eq. (7.165) is not the highest weight M(!) 1 there must exist a ra operator E/1 such that
The suspension points denote the possible states with the multiple we Since Ell annihilates the highest weight state 11M, (1'), M) (see (7.115)), the expansion (7.165) has to contain another term Im(l)+r/1m(l)  r l ,) such that. Ell1M  m(I)  rJ1.) = blM  mOl) + ... , and
Clebsch Gordan Coeffic" ni,
:3
Thus, if till: contains a term ) it has contain another higher weight, say other word the expansion (7 has Lo contain a term ll) NI(: I) and, f the same reason, contain another term 1M  M(2))IM(2)),
(7.16
This is the necessary condition for a highest weight representation O appearing in the CG series (7.161), namely, it is easy to write tht: C series (7161) from the block weight diagram of DM(I} or D M (2} based t.he condition (7166) The problem is how to cak1l1ate the multiplicity aM If aM = 0, DM disappears In the product DM(I} x D weight single and
Mo Due to the
=
, the highest weigh t
(7.16
repnlsentat.i
(7.16
C = L/L c~ is called the level of M in the CG series. Hereafter, M wi level c is denoted by Me. Sometimes, there are a few highest weights wi the same level c, then, they can be denoted by Me, M~, and so on.
7.8.2 There are two represenLs M(l)+m(
OS(Me)
Dominant
Weight Diagram.
vectors in the space m(l))IM(2), m(2)) from the block weight
(1),
Me The si
is calculated through ber ne of independent basis states I with m.tl) + 'm(2) = NIe In the product space can be counted. The Clebsch Gordan coefficients are shown through the expansio (7.162). The expansion for the highest weight state of DMo is given Eq. (7.167). Applying the lowering operators FIJ to Eq. (7.167), one o tains the expansions of the basis states in DMo, especially those expansio with the weight Me, if exist. Denote by nOe the multiplicity of Me in DM Thus, the multiplicity aJ of DMI is equal to the difference nl  nO]'
346
Chap. 7 Lie Groups and Lie Algebras
There are two ways to calculate the expansion of the highest weigh state in DMI. One is based on that the highest weight state is orthogon to those states with the same weight in the representation DMo. The oth is based on the property (7.115) that every raising operator EJ" annihilate the highest weight state. Then, by the lowering operators FJ"' calcula the expa.nsions of the basis states with the weight Me in DMl as well its multiplicity n1e. The multiplicity a2 of DM2 is equal to the differenc n2  n02  a1n12. Generally, the multiplicity a e of DMe is equal to e1
ae = ne 
L
(7.16
ae,ne'e'
c'=o
The dimension of the representation space
d(Me) =
L
c
OS(Me' )nce"
DM(l)
x
D M
(2)
is
(7.17
e'
When M(l) = M(2), the expansion (7.162) of the basis state in each re resentation DMc is symmetric or antisymmetric with respect to the inte change between ImP)) and Im(2)) such that the representations Me in th CG series are divided into symmetric ones and antisymmetric ones. Th sum of dimensions of the symmetric representations and antisymmctric on is d(M(1l)2, and their difference is d(MPl). The method of dominant weight dia.gram consists of two tables and th calculations of the expansions of the highest weight states of Me· On table lists the dominant weights Me based on the condition (7.166), the multiplicities n e , and the linearly independent basis states Im(1))IMc m(1l). The other table consists of a few columns. In the leftmost colum the products OS(Mc) x n e , where OS(Mc) is the size of Weyl orbit of M are listed and their sum is the dimension d(M(l))d(M(2l) of the produc space. Then, on the right of the table, each representation DMe correspond to one column and is arranged column by column in the increasing ord of c. For each representation D Me, one calculates its multiplicity a e in th CG series, the expansions of its highest weight states, the multiplicities ne of Me' contained in DMc, and its dimension d(Me) = Le' OS(Mc' )n ec In the bottom of the table, the dimension of the product space is calculate to be the sum of aed(Me), and the ClebschGordan series is listed. Th method will be explained in detail in the next subsection.
§ 7. 8 Clebsch Gordan Coefficients
7.8.3
347
Reductions of Direct Product Representations in Az
The algebra of SU(3) is A 2 . The block weight diagrams of some representations of A2 are listed in Fig. 7.3 . We first calculate the reduction of the direct product of two fundamental representations, (1,0) x (0, 1), where the highest weight directly denotes the representation for simplification. From Fig. 7.3 (b), in the product space there are two dominant weights Me, (1,1) = (1,0) + (0, 1) and (0,0) = (1,0) + (I, 0), with the multiplicities 1 and 3, respectively (see the upper table of Fig. 7.6). The expansion for the highest weight state (1,1) is given in Eq. (7.167). The expansions of other basis states in the representation (1,1) are calculated by applying the lowering operators FfJ to the highest weight state where Eq. (7.163) and the block weight diagrams in Fig. 7.3 are used. 11(1,1), (I, 1)) = 1(1,0))1(0,1)), II (1, 1), (I, 2))
=
II (1, 1), (2, I))
= F2 11(1, 1), (1, 1)) = I(1, 0)) 1(1, I)) ,
Fl II (1, 1), (1, 1))
=
1(1, 1)) I(0, 1)),
11(1, I), (0, Oh) = VlfiF1 11(l, 1), (2, I))
= J1fi {1(l, 0))1(1, 0)) + 1(1,1))1(1, I))} , II(l,l),(O,Oh) = J273{F211(l,l),(I,2))  VlfiII(l,l),(O,O)])} =
J176 {2 [1(1,1))1(1, I)) + 1(0,1))1(0,1))] + 1(1,1))1(1, I))]} J176 {1(1, 0))1(1, 0)) + 1(1,1))1(1, I)) + 2
 [1(1,0))1(1,0)) =
11(1,1)'(2,1)) II (1,1), (1,2))
I(O,I))I(O,l))}
= VlfiF1 11(1, I), (0,0)1) = I(I, 1))1(1,0)), = J2F211 (1, I), (0, O)d = I(0, I)) 1(1, I)),
11(1, I), (I, I)) = F211(1, 1), (2, 1)) = 1(0,1))1(1,0)). (7.171) The representation (1,1) contains a single dominant weight (1,1) and a double dominant weight (0,0) . Thus, the multiplicity of the representation (0,0) is aj = n]  n01 = 3  2 = l. Since the highest weight state II (0,0), (0,0)) is orthogonal to II (1,1), (0, Oh) and II (1,1), (0, Oh), one has
II (0, 0), (0, 0))
=
Ji73 {I (I, 0)) I(I, 0)) 
1(1, 1)) I(1, I))
+ 1(0, I)) I(0, 1)) } .
(7.172) The dominant weight diagram for the reduction of (1,0) x (0,1) of A2 is given in Fig. 7.6.
348
Chap. 7 Lie Groups and Lie Algebras
Me n c
Independent basis states
(1,1) 1
1(1,0))1(0,1))
(0,0) 3 1(1,0))1(1,0)),1(1,1))1(1,1)),1(0,1))1(0,1))
D+D 1x 3
9
8
(1,0) x (0,1)::: Fig. 7.6
~2~
(1,1)
+
1
+ (0,0)
The dominant weight diagram for (1,0) x (0,1) of A 2 .
Now, we calculate the reduction of the direct product of two a representations, (1,1) x (1,1). From Fig. 7.3 (c), in the product there are five dominant weights, (2,2), (3,0), (0,3), (1,1), and (0,0). multiplicities and the sizes of Weyl orbits are I, 2, 2, 6, 10 and 6, 3, 3 respectively. The independent basis states are listed in the upper ta Fig. 7.7 where the states by interchanging Im(l») and Im(2») are negl Except for (2,2), the block weight diagrams of the representations M given in Fig. 7.3. The block weight diagram of (2,2) is left to readers ercise. The representation (2,2) contains the dominant weights (2,2), (0,3), (1,1), and (0,0) with the multiplicities I, I, 1, 2, and 3, respec The results can also be calculated by the method of the Young ope (see Chap. 8). Based on those data, one calculates the multiplicit representations in the CG series to be a(2.2) = a(3,O) = a(O,3) = a(D.O and a(1,l) = 2. In Fig. 7.7, the number on the right shoulder of the denotes the multiplicity of the dominant weight in the representation the number above the the highest weight denotes the multiplicity representation in the CG series. The number is neglected if it is eq 1. The expansion for the highest weight state (2,2) is given in Eq. (7 and those of the remaining representations contain the coefficients determined:
11(2,2), (2, 2)) 11(3,0), (3,0)) 11(0,3), (0, 3))
= 1(1,1))1(1,1)), = at 1(1,1))1(2,1)) + a2 1(2,1))1(1,1)), = bl 1(1,1))1(1,2)) + b2 1(1,2)) 1(1,1)),
:
§ 7. 8 Clebsch Gordan Coefficients
11(1,1), (1, 1))5 =
CI
{1(1, 1))1(0, 0)1) + 1(0,0)1)1(1, I))}
+
C2
{I(l, 1)) 1(0, Oh) + 1(0, Oh)I(I , I))}
+
C3
{I(I, 2))1(2, I)) + 1(2,I))I(I,2))},
11(1 , 1), (1, 1))A = d l {1(1, 1))1(0,0)1) 1(0, 0)1)1(1, I))}
+ d 2 {1(1, 1))1(0, Oh) 1(0, Oh)l(l, Il)} +d3 {1(I,2))1(2,I)) 1(2,I))I(I,2))},
11(0,0), (0,0)) =
CI
(7.17
1(1,1))1(1, I)) + e2 1(1,2))1(1,2))
+ e3 1(2,1))1(2,1)) +
e4
1(0,0)1)1(0,0)1)
+e51(0,Oh)I(0,Oh) +e6 I(O,Oh)I(O,Oh)
+ e7 1(0, Oh)I(O, Oh) + es 1(2,1))1(2, I)) + eg I(1, 2)) I(I, 2)) + e to I(I, I)) I(1, 1)) .
Me nc
Independen t basis states
(2,2) 1
1(1,1))1(1,1)),
(3, 0) 2
I(1, 1)) I(2, I)),
(0,3) 2
1(1,1))1(1,2)) ,
(1, 1) 6 I(1, 1)) 1(0, 0) 1) , I(1, 1)) 1(0, 0h) , 1(1, 2)) I(2, I)) , (0,0) 10 1(1 , 1))1(1, I)), 1(2,1))1(2,1)), 1(1,2))1(1,2)), I(O,O)r1)I(O,O)b) , a,b
= 1,2.
6x1 3x2 3x2 6x6 1 x 10
2 3
i
1,1
0,0
2
~
0,0 2
[IQJ
= 27 + 10 + 10 + 2 x 8 + 1 (1,1) x (1,1):::(2, 2)s+(.3, O)A +(O,3)A+(I, 1)s + (1,I)A+(O, 0)5 64
Fig. 7.7
The dominant weight diagram for (1,1) x (1,1) of A2
Chap. 'l Lie Groups and Lie Algebras
'li d )
Since E2 annihilates 11(3,0) ,(3,0)), one has a1 = a2. Due to norm ization, aj = Jf72. Similarly, bl = b 2 = Jf72. Applying E1 and E2 11(1,1), (1, 1))5, one obtains
Elil (1, 1), (1, 1)) 5 = '.i2C1 {I (1, 1)) 1(2, I))
+ 1(2, I)) I(1, 1)) }
+ C3 {I (1, 1)) I(2, I)) + 1(2, I)) I(1, I))} = 0, E211 (1,1), (1, 1))5 = Jf72C1 {1(1, 1))1 (1,2)) + I(1,2)) I(1,1)) } + V3fiC2 {I (1,1)) I(1, 2)) + I(1,2))1 (1,1)) } + C3 {1(1, 2))1(1,1)) + 1(1,1))I(1,2))} = o.
Thus, '.i2Cj + C3 = 0 and C1 + V3C2 + '.i2C3 = O. The normalized solut is C1 = )3/20, C2 = )1/20, and C3 =  )3/10. The formulas for ap cations of E1 and E2 to II (1, 1), (1, 1)) A are similar except for replacing with d" and changing a sign in each curve bracket. From the formulas obtains '.i2d 1 + d3 = 0 and d1 + V3d 2  '.i2d 3 = O. The normalized solut is d1 = d2 = 1/2, and d3 = Applying E1 and E2 11(0,0),(0,0)), one obtains
JI7T2,
J1!6.
= (e1 + e2) 1(1,1)) I(1,2)) + '.i2 (e3 + e4) I(2, I)) I(0,0)1 + '.i2 (e4 + es) 1(0,0)1)1(2, I)) + '.i2 e61(2, 1))1(0, Oh) + '.i2 e71(0, Oh)I(2, I)) + (eg + elO) 1(1,2))1(1,1)) = 0, E 2 11(0, 0), (0,0)) = (ej + e3) 1(1,1))1(2,1)) + Jf72 (e2 + e4) I(1,2)) I(0,0)1) + V3fi (e2 + e5) I(1,2)) I(0, Oh) + Jf72 (e4 + eg) 1(0, Oh)I(1, 2)) + J372 (e5 + eg) 1(0, Oh)I(I, 2)) + Jf72 e61(1, 2))1(0, 0)2) + V3fi e61(0, O)dl(I, 2)) + Jf72 e71(0, 0)2)1(1, 2)) + J372 e71(1, 2))1(0, Oh) + (es + elO) 1(2,1))1(1,1)) = O. E111 (0,0), (0,0))
The normalized solution is e1 = e2 = e3 = e4 = es = eg = eg elO = ,jff8 and e6 = e7 = O. From the expansions of the highest wei states, one obtains the symmetries of the representations Me in the in change between 1'171(1)) and 1'171(2)), denoted in the bottom of Fig. 7.7.
7.9
Exercises
+ qc
1. Prove that the number Pc
+n
Exercises
;\
2. Calculate the Cartan matrix of the Lie algebra E 6 .
3. Draw the Dynkin diagram of a simple Lie algebra where its Carta matrix is as follows, and indicate the enumeration for the simple root
A=
( ~1o ~1 ~1 o
~
)
2 2 1 J 0 1 2 /
4 . Calculate all simple roots and positive roots in the G 2 Lie algebra.
5. Calculate all simple roots and positive roots in the F 4 Lie algebra . 6. Calculate all simple roots and positive roots in the C 2 Lie algebra. 7. Calculate all simple roots and positive roots in the B3 Lie algebra. 8. Calculate all simple roots and positive roots in the D4 Lie algebra.
9. Draw the block weight diagrams and the planar weight diagrams of th representations (1,0), (0,1), (1,1), (3,0), and (0 , 3) of the A2 Lie algeb (the SU(3) group).
10. Draw the block weight diagrams and the planar weight diagrams two fundamental representations and the adjoint representation (2,0 of the C 2 Lie algebra.
11. Draw the block weight diagrams and the planar weight diagrams three representations (0,1), (1,0), and (0,2) of the exceptional L algebra G 2 .
12. Calculate the ClebschGordan series and the ClebschGordan coeff cients for the direct product representation (1,0) x (1,0) of the C 2 L algebra.
13. Calculate the ClebschGordan series and the ClebschGordan coeff cients for the direct product representa tion (0,1) x (0,1) of the C 2 L algebra. 14. Calculate the ClebschGordan series for the direct product represent
tion (1,0) x (0,1) of the C2 Lie algebra and the expansion for the highe weight state of each irreducible representation in the ClebschGorda series.
352
Chap. 'l Lie Groups and Lie Algebras
15. Calculate the ClebschGordan series for the following direct prod ll, representations of the G 2 Lie algebra and the expansion for the higlll': weight state of each irreducible representation in the CG series:
(a) (0,1) x (0,1),
(b) (0,1) x (1,0),
(c) (1,0) x (1,0) ,
where the dimensions d(M) of some representations lVI, the Wey l oi bital sizes OS(M) of M, and the multiplicities of the dominant weiglII in the representation M are listed in the following table. M
(0,0)
d(M) 1
O S (M) 1
The multiplicity of the dominant weight (0, 1) (1,0) (0,2) (1, 1) (0,3)
(0 , 0) 1
(0, 1)
7
6
1
(1,0)
14
6
2
1
(0,2)
27
6
3
2
1
1
(1, 1)
64
12
4
4
2
2
(2, 01
1
1 1
(0,3)
77
6
5
4
3
2
1
1
(2,0)
77
6
5
3
3
2
1
1
1
16. Calculate the Clebsch Gordan series for the direct product represen tation (0,0,0,1) x (0,0,0,1) of the F4 Lie algebra and the expansio for the highest weight state of each irreducible representation in th ClebschGordan series, where the dimensions d(M) of some represen tations M, the Weyl orbital sizes OS(M) of M, and the multiplicitie of the dominant weights in the representation M are listed in the fo lowing table. The multiplicity of the dominan t weight (0,0,0,0) (0,0,0,1) (1,0,0,0) (0,0,1,0) (0,0,0,2) 1
M (0,0,0,0)
d(M) 1
1
(0,0,0,1)
26
24
2
1
(1,0,0,0)
52
24
4
1
1
(0,0,1,0)
273
96
9
5
2
1
(0,0,0 , 2)
324
24
12
5
3
1
OS
1
Chapter 8
UNITARY GROUPS
In this chapter the reduction of the tensor space is studied by the metho of Young operators. The symmetry of the tensor subs paces is characterize by the Young pattern. The independent and complete irreducible bas tensors in the tensor subspace are described by the tensor Young tableaux Combining with the method of block weight diagrams, one orthogonalize the irreducible basis tensors which are the orthonormal basis functions b longing to the given irreducible representation. Some applications of th SU (3) group to the particle physics are briefly introduced.
8.1
Irreducible Representations of SU(N)
The concept of tensor is related to the transformation group (see Appendi B). In this chapter, we study the tensor with respect to the SU(N) grou The element of SU(N) is an N x N unitary matrix, which transforms vector V in a compl ex space of dimension N, N
Va ~ V~ == (OuV)a =
L
1:::; a:::; N.
uabh,
(8.1
b=l
A tensor Tal ,.,a" of rank n contains n indices and N n components. In th SU(N) transformation u, each index plays the role of a vector index, T al . a" ~ (OuT)al ... a" =
L
Ualb l
···
Ua"b" T bt .. bn ,
(8.2
bl ... b"
namely, the transformation matrix of a tensor of rank n is the direct produ of n matrices u. The direct product of n selfrepresentations is general reducible. For example, the direct product selfrepresentations of SU(2) a
354
Chap. 8 Unitary Groups
Dl/2
X
D 1/ 2
D 1/ 2
X
D
X
1/ 2 X
Dl/2 :::: D3/2 ffi 2 Dl/2, D 1/ 2
X
Dl/2 :::: D2 ffi 3 Dl ffi 2 DO.
We are going to reduce the direct product representation of SU(N) a find the irreducible basis tensors. 8.1.1
Reduction of a Tensor Space
The tensor space T is an Nndimensional linear space which is invari in the SU(N) transformations, namely, the transformed tensor still belon to T The u matrices appear in the transformation (8.2) in the form the matrix entries Uab. The product of Uab is commutable. Therefore, t permutation symmetry of the tensor indices is invariant in the SU(N) tran formation 011.' The subset of tensors with the Bame permutation symme of indices constitutes an invariant subspace of T in 011.' It is easy to def the totally symmetric tensors and the totally antisymmetric tensors. H to describe the invariant subspaces with other permutation symmetrie For example, one cannot define a tensor of rank 3 whose first two indi are symmetric and the last two indices are antisymmetric. We begin with the definition of a permutation R on a tensor, RT = T which has to satisfy the group property. The definition for a transpositi (j k) contains no confusion because it is selfinverse,
[(j k )T]a,.aj ... ak.an = T~, ... aj ... ak ..
an = Tal .. ak .. OJ ... a n .
(8
A permutation R on a tensor can be defined based on the property tha permutation R can be decomposed into a product of transpositions. Fo simple example, one has
R= (123) = (312) =(123)=(12)(23) 231 123 '
(8
Generally,
R =
(1 2... n ) (Tl1 T22 ...... Tn) , n =
(8
Tl T2 ... Tn
(8
The permutation R moves the Tjth index aTj , NOT the Tjth index a to the jth position. Equivalently, R moves the jth index aj to the T
§8.1 Irreducible Representations of SU(N)
position, NOT to the Tjth position. The set of permutations of tenso forms the permutation group Sn and the tensor space T is invariant in S It is well known that a tensor of rank 2 can be decomposed into the su of symmetric and antisymmetric tensors,
Tab
1
1
= "2 {Tab + T ba } + "2
{Tab  na} .
The decomposition can be written in terms of the Young operators, 1
"2 {E + (1
2)}Tab
1
+"2 {E 
(1 2)}Tab
(8.
~ {y[21 + y[l,ll} Tab = ETab ,
namely, the decomposition can be achieved by the expansion (6.62) of t identical element with respect to the Young operators which are the pr jective operators,
A[>'1
Tala~ = ETa1 .. an = n. L
d[>,1
L
yl>'l yL>'ITa1 .a n ·
(8.
!1.
For example, a tensor of rank 3 is decomposed as _ 1 [31 Tabe  6Y Tabe
1 [2,11
+ 3" Y l
Tabc
1 [2,11
+ 3"Y2
Tabe
1 [1,1,11
+ 6Y
(8.
Tabc·
The first term is a totally symmetric tensor, the last term is a totally a tisymmetric tensor, and the remaining terms are tensors with mixed sym metry. The Young operators decompose the tensor space T into the dire sum of tensor subspaces 7;1>'1,
T
= E T = ~! EB
EB
d[>'1 [>.I!1.
yl>'l yL>'JT
= EB EB [>'1
(8.1
7j>.I.
!1.
Let us study the property of the tensor subspace 7;1>'1. First, there is common tensor between two subspaces 7;1>'1 and TJwJ because the You operators are orthogonal to each other. Thus, the decomposition (8.1 is in the form of direct sum. Second, the constant factor d[>.l/n! and t operator yL>'1 do not make any change with the subspace 7;1>'1. In fact, d to yL>']r c T and y l>'lT C T , one has
y(>'IT !1. Thus,
= y[>'ly[>'] {d[>'1 Y[>.]r} !1.!1. n!!1.
C y[>'ly[>'IT
!1.
!1.
.
356
Chap. 8 Unitary Groups
(8.11) For the same reason,
RT=T,
Y IJ.[>') RT = T!>') IJ. '
(8.12)
Third, as shown in Theorem 8.1, the subspace 7j>') is invariant in Ou.
Theorem 8.1 (Weyl reciprocity) The permutation R and the SU(N) transformation Ou for a tensor is commutable with each other.
Proof The key of the proof is that the matrix entries Uab in Eq. (8.2 are commutable:
(OuRT)a, .. n Q
=
L
= (OuTR)a,
.a n
=
L
Ua,b, · · ,Uanbn (TR)b,.b n
b, ... b"
Uar,br,' .. uarnbrnTbr, ... brn
= (OuT)a,., ... arn = (ROuT)a, .. an
b, .. . b n
(8.13 Thus, 0" {yi>')T} = yi>') {OuT} C yi>'IT
= 7j>') .
o
Remind that R on the left of yi>']T may change the subspace
R lIJ.L y J..1.[>'IT = Y[>']R T = T[>'] v Vf.i. V
l
(8.14
where Rvf.J. is the permutation transforming the standard Young tableau yi>'] to the standard Young tableau y];').
8.1.2
Basis Tensors in the Tensor Subspace
First of all, we review the property of the basis vectors (see §4.7). A basi vector (}d is a special vector with only one nonvanishing component, which is equal to 1, «(}d)a = Oda. Any vector V can be expanded with respect to the basis vectors N
(V)a =
L
Vd «(}d) a = Va·
(8.15
d==l
Note that (V)a and Va are different in the SU(N) transformation although they are equal in value. A basis tensor (}d, ... d n is a special tensor with only one nonvanishin component which is equal to 1,
«(}d, . dn )a,fl n = Od,,,,Od2
= «(}d, ),,'
«(}d2)"2 ... «(}dJ an · (8.1 6
§8.1 Irreducible Representations of SU(N)
Being a tensor, the basis tensor transforms in
=
(Ouf)d, .. dn)a, ... a n
L = L
Ou
.35
and in R as follows:
Uo,b,·· .Uanb" (f)d, ... dnh, .. b"
b, .. . b"
= UQ.,d, .. ,Uandn
(f)b, ... bn)a, ... a n Ub,d j
•• •
1l b n d n ,
b, ... bn
(Rf)d, ... dn)a, ... a" = (f)d, ... d")or, ... ar" = Sd,ar,Sd2ar2 .. "Sd"a r"
= Sdr , a, Sdr2 02
...
namely, Ouf)d, .. dn
Sdr n an = (f)dr , ... dr " ) a, ... a" '
=
L
(8.17
f)b, .. b" V'b,d, ... Ub"d",
b, ... b"
(8.18
transforms a basis tensor f)r1, ... r1" to another basis tensor f)d r , .. .dr " ' wher the jth index d j moves to the rjth position, NOT to the Tjth position Equivalently, the Tjth index drj , NOT the rjth index d rj , moves to the jt position. In a simple example R = (1 2 3) = (1 2)(23), R
Any tensor T can be expanded with respect to the basis tensors To, ... o"
=
L
Td, ... d" (f)r1, ... d,.)o, .. o"
= To, ... a,,·
(8.19
d, ... d"
In the SU(N) transformation
is a tensor, but
Ou, To, .. a"
Ta, ... a"
is a scala
f)r1, ... d" is the basis tensor of the tensor space T, and YlAl f)d,.d" is th basis tensor of the subspace ?jAl. Since the dimension of ?jAl is less tha that of T, some yl,AI f)dj ... d" may be vanishing or linearly dependent. Ou task is to find the complete set of the basis tensors of ?jAl, which are linearl independent. After the application of a Young operator, yf)rh.d" is a linear comb
nation of
f)d, ... d n .
yf)11233
For example, if the Young tableau
= [E + (1
2)
+ (1
4)
+ (2
4)
. [E  (1 3)] [E  (2 5)] =
+ (1
24)
y
+ (2
is
~,
1 4)] [E
+ (3
5)]
f)ll233
[E + (3 5)1 [E + (1 2) + (1 4) + (2 4) + (1 24) + (2 1 4)] .
[f)1l233 
f)13231 
f)21133
+ f)2313d
358
Chap. 8 Unitary Groups
+ 031213 + 013213] + 2 [01l332 + 031312 + 013312]  2 [013231 + 031231 + 0 33211 ]  2 [013132 + 031132 + 0 33112 ]  [021133 + 012133 + 013123 + 0 31123 + 0 23113 + 0 32113 ]  [(}21331 + 0 12331 + 013321 + 031321 + 02331l + 0 3231 d + 4 [023131 + 032131 + 0 33121 ],
= 2 [0 11233
(8 .20)
Usually, the expansion of YOabcde is quite long. In order to simplify till notations we introduce a graphic method to denote the tensor expansion. YOa,.a" is denoted by a tensor Young tableau where the box filled with j in the Young tableau Y is now filled with the subscript aj. For exampk, the tensor in Eq. (8.20) is denoted by a tensor Young tableau I ; I ~ I 3 I Another example is that if the Young tableau Y is
~
~' the tensor YOa,.a7 is denoted by a tensor Young tableau
A tensor Young tableau is a linear combination of the basis tensor Oa, ... an which describes a tensor in a given tensor subspace. In differen tensor subspaces, the same tensor Young tableau describes different ten sors. For example, in two tensor subspaces y[2, li T and y~2,I]T for SU(3), [2 I]
Young tableau Y 1
'
= []]I] ~ ,
the tensor Young tableau [2,1]
Y1
[2,11
Y2
0 123
[J]IJ
= ~ ,
B::fD describes two different tensors,
[]]I]
[2,1]
T,
[2,1]
T
0 123 
0 321
+ 0 213

0 231 E
Y1
[]]I] ~ = 0 132 
0 312
+ 0 231

0 213 E
Y2
= ~=
_ 0 132 
[21] '
Young tableau Y2
They are related by a permutation (2 3), (2 3)y 1[2,I]ll U123

y[2,1](2 2
3)ll
_
u123 
y 2i2 ,1]llu132·
(8. 2 1
§8.1 Irreducible Representations of S U (N )
35
Please do not get confused between a tensor Young tableau and a Youn t.ableau. Study the symmetry of the tensor Young tableaux based on the sym metry (6.28) of a Young operator and the Fock condition (6.31)
(8.22
Denote by Qo a vertical transposition of the Young tableau yL"j· Qo in terchanges two digits in the same column of the Young tableau yL"j, sa i and j. Rightmultiplying Qo on yL"j, one obtains a minus sign. Bu leftmultiplying Qo on the basis tensor Oa, ... a" interchanges two subscript ai and aj in Oa, .. a". Two digits ai and aj are located in the same co umn of the tensor Young tableau yL"j QoOa, ... a" , namely, two digits in th same column of a tensor Young tableau are antisymmetric. A tensor Youn tableau with the repetitive digits in the same column must be O. The row number of a nonvanishing tensor Young tableau for SU (N) is not large than N. The Focle condition also gives some relations between the tenso Young tableaux in a tensor subspace. For example,
Y1[2,ljo abc
[ill] ~


_y[2,1]
= _
1
(1 3)0 abc,
GTbl , ~
[ill] ~
~+~. ~ ~
(8.23 Remind that although leftmultiplying with a horizontal permutation P o a Young operator y does not change the Young operator y as well as th tensor Young tableau YOa, ... a" , P does not make a symmetry of digits i the tensor Young tableau. For example, the tensor Young tableau yI2,1lOab is invariant by leftmultiplying it with (1 2). The expansion of yI2,ljoabc symmetric between the first two indices, but not between a and b, (1 2)y1[2,I]O abc 
(1 2) {O abc  0 eba + 0 bac  O} bea Obac  0 bca. + 0 abc  0 cba  y\[2,11 0 abc·
Now we return to the problem how to find the complete set of th linearly independent basis tensors in the subspace 'lj"l = y!:,IT, wher yL"] is a standard Young operator. Discuss a useful example first. Th general form of a tensor Young tableau in the tensor subspace 7;[2,1] of th tensor space T of rank 3 of SU(3) is
360
Chap. 8 Unitary Groups
[H1J= Y
[2,11
1
Oabe
= {E 
(1 3)
+ (1
2)  (2 1)(1 3)} Oabe
The tensor Young tableau is vanishing when a tableaux with a pair of same digits are
(8.2
c. The tensor YOU
Y112 ,1 1 0113 _ Y 1[2,1 10122 _
Y 1[2,1]O 133
_

and those with three different digits are
Y I[2,1]O 123

_
Y 1[2.1]O 132

Y 1[2,1]O 213

_
_
A tensor Young tableau is said to be standard if the digit in each column the tableau increases downward and the digit in each row does not decre rightward. The dimension of the tensor subspace ~[2, 1] is 8. The ei linearly independent tensor Young tableaux are all standard.
Theorem 8.2 The standard tensor Young tableaux constitute a compl set of basis tensors in the tensor subspace 'Jj"].
§8.1 Irreducible Representations oj S U (N )
36
Proof For any basis tensor (h,.b n , there is a permutation S to arrang it.s subscripts in the increasing order:
On the other hand, yL.\1 S belongs to the rightideal nL.\I, generated by th standard Youn operator yL.\1 in the group algebra of Sn . The basis vector in nL.\1 are yL.\ R itv · Thus, any tensor in the tensor subspace TYI can b expressed as a linear combination of the following tensor Young tableaux:
J
(8.25
Namely, the tensor Young tableaux given in Eq. (8.25) constitute a com plete set of basis tensors in the tensor subspace 01.\1. Our task is to show that Y1.\1 R,..,AJ a ] an are standard or vanishing. R,.." is a permutation transforming the standard Young tableau y~\1 to the standard Young tableau yL.\1 so that
yfIR,.."Oal .. an = R,.." {yL.\IO a1 ... a "
} .
(8.26
The tensor in the curve bracket belongs to the tensor subspace T,j.\I, no
7;YI. Equation (8.26) shows that two tensors belonging to different tenso subspaces are related by R,.." . It can be checked as follows that those tw te nsors are described by the same tensor Young tableau. Let R
,..,., 
(7'11 T22 ..... Tn) , n
Arbitrarily choose a box in the Young pattern [AJ. The box in the Youn tableau yL.\1 is filled with a digit, say j. The same box in the Young tableau yL.\1 is filled with Tj, that in the tensor Young tableau yL.\IOa] .. an is filled with aT j , and that in the tensor Young tableau yL.\1 R,..vO"] .. a,, is filled with bj = o.Tj. Therefore, two tensor Young tableaux are the same.
Since 0.1 .'S 0.2 .'S ... .'S an and yL.\1 is a standard Young tableau, th tensor Young tableau yL.\IOIl.] .. an satisfies that the digit in each column o yL.\I Oo1 ... a" does not decrease downward and the di~it in each row does no decrease rightward. Thus, yL.\j Oat ... a n as well as yl.\ R,.."Oa] ... a" is standard or vanishing depending on whether there are repetitive digits in a column of the tableau. The different standard tensor Young tableaux are linearl independent although yL.\j R,..,.,Oa].a" with different R,..,., may be denoted
362
Chap. 8 Unitary Groups
by the same tensor Young tableau. For example, from Eq. (8.24)
where RII
8.1.3
=E
and RI2
= (23).
Chevalley Bases of Generators in SU(N)
Calculate the Chevalley bases of generators in the selfrepresentatio SU(N). From Eqs. (7.79) and (7.81) one has (rv)) =
v'2[V./ 
Vv+tlj =
v'2
[(T~3~HI)v  (T~~HI)v+J
JV: 1 6,/(N_j)  Jv ~ 1 8v
(N_Hl) '
In the selfrepresentation of SU(N), the Chevally bases (see Eq. (7. where dv = 1) are H
v
=~( ).H.=~(V+l)T(3) _~(Vl)T(3) L rv J J v+1 v V
)=1
V
(8
_ T(I) T(l) vv (v+l)(v+I)'
E v  E "'v ( v+l)  T(J) v(v+l)
+ I'T(2) v(v+l)'
F
v =
T(l) 'T(2) v(v+I)  't v(v+l)"
Namely, each of the matrices of H v , and Fv contains only a non ishing twodimensional submatrix in the vth and (v + l}th columns rows, which is the same as that in the selfrepresentation of SU(2) (see (7 .144)). Their actions on the basis vectors are E//I
H v 8,/ = 8 v ,
H v 8 v + 1 = 8 v + 1 ,
Ev8v+1 = 8 v ,
Fv 8 v =
(8
8 v+ 1 ·
The remaining actions are vanishing. In the SU(N) transformation, e index in a tensor plays the role of a vector index. The action of a gener on a basis tensor are the sum of the basis tensors, each of which is obta by the action of the generator on one tensor index according to Eq. (8. Thus, the standard tensor Young tableau is the common eigenstate of with the eigenvalue to be the number of the digit v fill ed in the tabl subtracting the number of the digit (v + 1). The eigenvalues constit the weight m of the standard tensor Young tableau . Two tensor Yo
§8.1 irreducible Representations oj SU(N)
363
tableaux have the same weight if their sets of the filled digits are the same. The action of Fv on the standard tensor Young tableau is equal to the sum of all possible tensor Young tableaux, each of which is obtained from the original one by replacing one filled digit 1/ with the digit (1/ + 1), and the action of Ev is the sum of those by replacing (1/ + 1) with 1/. The obtained Young tableaux may not be standard, but they can be transformed to the sum of the standard tensor Young tableaux by the symmetries (8.22) and (8.23). For example,
ITIIJ = _ ITIIJ ITIIJ = [TTll c:rrc:rr ' c:rrc:rr ' 1[TTll = [TTll E[TTll=ITIIJ c:rr c:rrc:rr c:rr ' [TTll = [lT2l I2Til = [lT2l_ [ll3l . c:rr c:rr + c:rr c:rr ~ H
H
2
2
!
E
F
2
!
2
F
0,
2
J
8.1.4
Inequivalent and Irreducible Representations
Theorem 8.3 The tensor subspace 7;).\J = yl.\J T corresponds to an irreducible representation of SU(N) with the highest weight M, N!
M
=
L
lvlvw v ,
(8.29)
v=!
where Au is the box number in the I/th row of the Young pattern [AJ and w" are the fundamental dominant weights of 8U(N). The tensor subspaces are equivalent if and only if their Young patterns are the same. The basis tensors with the same tensor Young tableau in all tensor subspaces 7;).\J with the Young pattern [AJ constitute a complete set of basis tensors of the irreducible representation [AJ of the permutation group 8 11 •
Proof The key for proving the representation corresponding to the tensor subspace 7;).\] to be irreducible is whether there is one and only one highest weight state in 7;).\1 which satisfies the condition (7.115). One defines a standard tensor Young tableau to be smaller than another by the concept of "the dictionary order" (see §6.2.2). Compare the filled digits in two standard tensor Young tableaux in TPJ from left to right of the first row, and then those of the second row, and so on. For t.he first different filled digits, a smaller filled digit corresponds to a smaller standard tensor
364
Chap. 8 Unitary Groups
Young tableau. Obviously, the standard tensor Young tableaux, den by ¢o, where each box located in the j th row of the tableau is filled the digit j, is the smallest one in 7jA]. ¢o satisfies Eq. (7.115) because raising operator E,/ replaces a digit (1/ + 1) filled in the tableau with digit 1/ such that the resultant tableau contains two 1/ in the same colu The highest weight M of ¢o is calculated by Eq. (8.29). Compare the filled digits in any other standard tensor Young tab ¢ in 7;:] A] with ¢o, from left to right and row by row. If the first diffe filled digits occurs at the ith column of the jth row, where the filled in ¢ is k > j, ¢ is now denoted by ¢(j,i,k,o:). The ordinal number ¢(j, i, k, 0:) stands for the case when ¢ with this property is not uniqu 7jA]. ¢(j,i,k,o:) is smaller than ¢(j',i',k',o:') if j > j', or i > i' w j = j', or k < k' when j = j' and i = i', or 0: < 0:' for the same j, i, k. E k  1 ¢(j, i, k, 0:) =I 0 because it is a combination of the standard te Young tableaux in 7jA] where the smallest one is ¢(j, i, k  1, /3) wi positive coefficient n ;:::: 1. n is the number of k filled in the jth ro ¢(j, i, k, 0:). For any linear combination of the standard tensor Y tableaux with a weight m in 7jA], E k  1 I 0 if the smallest stan tensor Young tableau in the combination is ¢(j, i, k, 0:). Thus, the sor subspace 7jA] = Y1LAlT corresponds to an irreducible representatio SU(N) with the highest weight M given in Eq. (8.29). Due to Eqs. (6 and (8.26), the remaining conclusions in the Theorem is obvious.
8.1.5
Dimensions of Representations of SU(N)
The dimension d[A](SU(N)) of the representation [Aj of SU(N) is equ the number of the standard tensor Young tableaux in the tensor subs 7jA]. There is a simpler way, called the hook rule, to calculate the dim sions. Please first review the hook rule for calculating the dimension representations of Sn (see §6.2.2). For a box at the jth column of the ith row in a Young pattern define its content mij = j  i and its hook number h ij to be the num of the boxes on its right in the ith row of the Young pattern, plus number of the boxes below it in the jth column, and plus l. The dimen d[A](SU(N)) of the representation [Aj of SU(N) is expressed by a quot
d]A](SU(N))
= II ij
(8
§8.1 Irreducible Representations of 5 U (N )
.36
Y1'~] is a tableau obtained from the Young pattern [A] by filling (N + mij into the box located in its ith row and jth column, and y~A] is a tablea obtained from [A] by filling h ij into that box. The symbol y1'~] means th product of the filled digits in it, so does the symbol y~A]. When [A] = [n] is a onerow Young pattern, T[n] is the set of the totall symmetric tensors, and its dimension is n II d[n](SU(N)) =
N+jl n  j +1
j=1
=
(N+nl)! n!(N  I)!
=
(n+Nl) N _ 1 . (8.31
This formula can be understood that the standard tensor Young tableau with a onerow Young pattern [n] are characterized by the positions of th (N  1) dividing points between each two neighbored digits. When [A] = [Ji, v] is a tworow Young pattern, N
d[IL v]
,
N
1
...
N+v
2
(SU (N)) = '==::::;:=~~~~:::::::;:==~:::::::::;:::==;==:;:=:;
IJL:llv~11:::IJL~+2IJLvl
(N
III
+ Ji 1)!(N + v  2)!(Ji  v + 1) (N  1)!(N  2)!(Ji + 1)!v l
(8.32
d[n](SU(2)) d[IL,v]
= d[lI+v,v] (SU(2)) = n + 1,
(SU(3))
= (Ji + 2)(v + I)(Ji v + 1)/2.
In fact, the representation Dj of SU(2) given in Chap. 4 is equivalen to [n] with n = 2j. For SU(3) one has d[l](SU(3)) = d[1 ,1](SU(3)) = 3 d[2,t](SU(3)) = 8, dI 3](SU(3)) = d[3,3](SU(3)) = 10, and d[4,2](SU(3)) = 27 For a onecolumn Young pattern [In], n ~ N, a standard tensor Youn tableau is a tableau filled with n digits downward in the increasing orde so that its number is the combinatorics of n among N,
= II n
d[lnl(SU(N))
Nj+l n  j +1
=
N! n!(N  n)!
=
(N) n .
(8.33
J=l
vVhen n
=
N, there is only one standard tensor Young tableau.
LR b(R)R is an antisymmetrized operator, = cal ... aN' y[l"'IOal .. aN = Cal ( y[1 N]812 .. N)
Sinc
y[l"'] =
.. (LN (y[1 N]8 12 N)
aI···af\.'
(8.34
Define E =
y[lN]8 12N .
Due to the Weyl reciprocity,
366
Chap. 8 Unitary Groups
L
(y[lNJOal' .aN)
i . a~:l
... aNUall" .UaNN
Ou E = y[lNJOu012 ... N = (y[lNI 012 ..N)
Ua l l · ·
.UaNN
al.··aN
(y[l NI 012 ..N) detu = y[lNI 012 .N = E.
E is an invariant tensor in SU(N), and [IN] describes the identical 1'0 ]11 sentation of SU(N). The numbers of the standard tensor Young tableaux with the followil two Young patterns are evidently equal to each other [>'l,A2, ... ,ANl,AN] and
[(AI  AN), (A2  AN), ... , (ANI  AN),O],
because there is only one way to filling the digits in the first AN columns a standard tensor Young tableau with the first. Young pattern, namely, ti digits in those columns are filled from 1 to N in the increasing order. T W representations with the Young patterns given in Eq. (8.35) will be show to be equivalent later.
8.1.6
Subduced Representations with Respect to Subgroups
SU(N  1) is a subgroup of SU(N) where the Nth component preserw invariant. In this way one obtains a subgroup chain of SU(N) SU(N) :J SU(N  1) :J ... :J SU(3) :J SU(2) .
(8. 3
An irreducible representation [A ] of SU(N) can be reduced with respect the subgroups one by one in the subgroup chain. In fact, the basis tensor the representation [A] of SU(N) is the standard tensor Young tableau wh c' N has to be filled only in the lowest boxes of some columns. Removing ti boxes filled with N, one obtains a standard tensor Young tableau of ti subduced representa tions [IL] of SU(N  1). For example,
[A] = [5,2,2,1]' [IL] = [3 , 2,1].
Removing the boxes filled with N from all standard tensor YOU) tableaux in the representation [A] of SU (N), one obtains all standard t O sol' Young tableaux in a few representations [IL] of SU(N  1). Th is is L
§8.2 Orthonormal Irreducible Basis Tansors
367
"lethod of reducing the subduced representation [A] of SU(N) with respect 1.0 its subgroup S(N  1):
E9
[A] t
d[AJ(SU(N))
[11],
=L
d[IlJ(SU(N 1)),
[IlJ
(8.37)
AN .::; I1NI .::; ANI'::; I1N2 .::; ... .::; 112 .::; A2 .::; III .::; AI.
n.emind that in the reduction the multiplicity of each representation [11] is Ilot larger than one. By the successive applications of this method one is able to reduce an irreducible representation of SU(N) with respect to the subgroup chain (8.36). The subduced representations of SU(N + M) and SU(N M) with respect to the subgroup SU(N)xSU(M) are discussed in §8.4 of [Ma and Gu (2004)]. The Casimir invariants of orders 2 and 3 of SU(N) can be calculated by the method of the subduced representations of SU(N + M) (see §8.5 in [Ma and Gu (2004)]).
8.2
Orthonormal Irreducible Basis Tensors
Define the inner product in the tensor space T such that the basis tensors are orthonormal to each other. The tensor representation of SU(N) wi th the basis tensors ()al" .a n is the direct product of selfrepresentations, which is unitary. Through the projection of a standard Young operator y},"'J, the tensor space reduces to its subspace 'JjAJ = yLAJT, whose basis tensors are the standard tensor Young tableaux. The standard tensor Young tableaux are the integral combinations of ()al ... a n , but they are generally not orthonormal. For example, in t.he tensor subspace ~ [2 , IJ of SU (3), where the general form of the basis tensor is given in Eq. (8.24), y12,li()123 .IS not or th ogonaIt 0 y[2,li() I 132· y[2,l I i() 112 an d y[2,li() 1 123 are normaI'Ize d t,0 ()a l .. a n
6 and 4, respectively. Furthermore, the highest weight states both in ~[2,li and in Tzi", IJ of SU(3) are denoted by the standard tensor Young tableau but they are not orthogonal,
EEfTI,
YI[2,11() 1I2 [2,IJ
Y2
()12l
= 2() 112 =
()
23
()

211 
[2,11
YI
()112
()
121,
(8.38)
= 28 121  ()211  ()JJ2.
Namely, the standard tensor Young tableaux both in one representation of SU (N) and in one representation of Sn are generally not orthonormal.
368
Chap. 8 Unitary Croups
8.2.1
Orthonormal Basis Tensors in
7jAJ
Usually, it is by a similarity transformation that a nonunitary represe tion of a compact Lie group changes to be unitary and the basis state combined to be orthonormal. As far as the problem of finding the ortho mal basis tensors in an irreducible subspace 1jA] of SU(JV) is concer one prefers to use the method by applying the lowering operators FJl cessively to the highest weight state because the highest weight is single the highest weight state is orthogonal to any other state. This is not but the essence of the method of the block weight diagram. But now multiplicity of a weight is easy to count because the standard tensor Y tableaux with the same set of the filled digits have the same weight. standard tensor Young tableaux with different weights are orthogon each other. The modules of the calculated basis states by this method normalized, instead to 1, to the module of the highest weight state. method is explained by some examples in SU(3) as follows. The block weight diagrams of two fundamental representation and (0,1) are given in Figs. 7.3 (a) and 7.3 (b). The standard te You~ableaux with the highest weights in two representations are and Qj, respectively. The standard tensor Young tableaux are calcu
from the highest weight states by the lowering operators as given in 8.1 where the block weight diagrams are also listed for comparison. S two representations (1,0) and (0,1) are conjugate to each other, their tensors can be related through Eq. (7.153).
(a) (1,0)
Fig. 8.1
(b) (0,1)
Block weight diagrams and basis tensors of fundamental representations of SU (3).
The Young pattern of the representation of symmetric tensors of 3 of SU(3) is [A] = [3,0]. Its highest weight is M = (3,0). There are t typical standard tensor Young tableaux which are normalized to 36 and 6, respectively,
OrUwnormal Irreducible Basis
3
lalala I a I bib
Ia Ib Ic
{J!2
(()abb
+ (hrrb + ()bba)),
1= V6{6 1/ 2 (()abc + ()rrcb+ lhrrc + ()bca+Ocab + Ocba)} ,
where a, b, and c are three different digits. For each set of .filled digits the is only one standard tensor Young tableau, so that there is no multip weight in the representation.
v'31
1
I
2
I
2
v'3121213
v'31
1
13 13
!
v'3121313
Fig. 8.2
dIagrams and basis [3,0] of SU(J)
weight standard tensor Young are listed in Fig. 8.2 and some calculations are as follows:
12 1+ I 112111+121 1 I 1 I = 31 1 11 12 I, 11 11 131=1112131+121 13 I = 21 1 12 1 3
FJ .
Fl
the app tablea The blo of SU(
1 1 I 1 1= 11
I 1
,
370
Chap. 8 Unitary Groups
1( 3, 0), (1, 1)) =
If
FJ 1( 3, 0), ( 3, 0)) =
=J31 1 1 1 12I, 1(3,0), (0,0)) =
If
Fi 1(3,0), (2, I)) =
If
VI
FJ 1 1 1 1 [ I
FJ 1 1
I1
13
I
I
=J61112131·
The conjugate representation of [3 , 0] is [3,3]. Its highest weigh is 1 ; 1 ; 1 ~ I , where M = (0,3) . The block weight diagrams a standard tensor Young tableaux of [3,3] of SU(3) are listed in Fig. 1~
0,3
v'3
v'3 1
1
I ~ 1~
1
1~ 1~ 1
v'31
1v'31 ~ 1 ~ 1 ~ 1v'3 1 3 1 3 1
1
f,;+i+:ill
1 ; 1 ; 1 ~ 1 /61
3
1 3 1 2 I'
3
I' v'3 1 3 1
I
2
1 2 1 1 I'  1 2 1 2 2
v'3
1~ 1~ 1
 v'31
2
1
I
1 1 I'
1~ 1~ 1~ 1 Fig. 8.3
Block weight diagrams and basis tensors of the representation [3,3] of SU(3).
The Young pattern for the mixed symmetric tensors of rank 3 of S
[A) = [2 , 1, which is the adjoint representation of SU(3). Its highest state is ; i , where M = (1,1). The general form of the exp of the tensor Young tableau in ~[2,il is given in Eq. (8.24). Two tensor Young tableaux are
They are normalized to 6 and 4, respectively. The block weight dia and the standard tensor Young tableaux of [2,1] of SU(3) are listed 8.4. Some calculations related to the multiple weight (0,0) are as fo
§8.2 Orthonormal Irreducible Basis Tansors
37
f{ [IfD f{ {[IpJ + lIfO} = J2 [IpJ f{ GPJ ' 1(1, Oh) = I[ {F2 1(1, Cl, f{ 1(1,1), Oh) }
Itt, I), (0,0)1) =
f{
Fl 1(1, I), (2,1))
I), (0,
=
I),
=
Fl
2)) 
(0,
I[ {F2 GfD [IpJ + ~ GPJ}
=
j[GPJ·
(8.39 Note that the orthogonal basis tensors are combined by
(~ ~~~)
X[;\J =
which is the similarity transformation matrix for the orthogo
nal bases in the representation [2,1] of the permutation group 8 3 (see Eq (6.85) and Prob. 24 of Chap. 6 in [Ma and Gu (2004)]),
= J'i
1(1, I), (0, O)a)
(X1;\J)
al
[IpJ + J'i (X1;\I) W'
IIW
3
I
al
GPJ,
(8.40
I
v0W
2
1_
nw:=u
l,l
Fig. 8.4
Block weight diagrams and basis tensors of the adjoint representation [2,1] of 8U(3).
The orthonormal basis state with a weight m = "£~=/ wJ.1.mJ.1. in an ir reducible representation [A] of 8U(N) is denoted by a symbol, usually calle the Gelfand bases [Gel'fand et al. (1963)], where N(N + 1)/2 parameter Wab, 1 S a S b S N, are arranged as a regular triangle upside down: WIN
W2N ... W(Nl)N WI(NI)
WNN W(Nl)(NI)) ,
WI2
W22
:.\72
Chap. 8 Unitary Groups
t>+l mt>
= 
2:
I'
Wd(t>+I)
+ 22:
d=1
(H II
),1
2:
Wdt> 
d=l
wd(t>I),
WaD
= O.
d=1
The representation matrix entries of the Chevalley bases of generators :'1' t>
Et>lwab)
=
2:
A"t>(Wa b) IW ab
+ (jav(jbl') ,
v=l
(S A:.!) The Gelfand bases for the representation [2,1] of SU(3) are listed as all example, 11) =ffiTI
1(1,1)) = [2 2
~ 10)
G::fTI
1(1,2)) = [2 2
~ 10)
12) =
13)=[E=O = 14) =
J2
1(2,I))=[22~00),
tI:fD  Jl72 ffiTI
15) =
[IfTI
16) =
J372ffiTI
17) = 18)
=
tI:fD
[IfTI
= 1(2, 1)) = [2 2
=
= 1(0 , 0)1) = [2 2
~ 0 0)
= I(O,Oh) = [21
1(1,2))
= [2 1 ~ 00)
1(1,1)) = [21
~ 0 0)
~ 00)
,
(S.43) ,
~ 10)
,
§8.3 Direct Product of Tensor Representations
t{ .2. 2
Orthonormal Basis Tensors in Sn
The highest weight states in '0»'] of SU(N) with different f.l constitute a ('omplete set of the basis tensors for the representation DI>'I (S,,), which is 1I0t unitary. Denote by bL~ = yl>'1 Ob,b" the highest weight state in ?j>'I. Then, due to Eq. (8.26), bl>'l = yt>. ] RVI'0b1b" is the highest weight state ill TJ>'], and  R b[>'] b[>'] VI'_ //1' 1'1"
(8.44)
Letting X[>.] be the similarity transformation which changes D [>'] (Sn) to the real orthogonl representation D[>'](Sn) (see Eq . (6.84)), one obtains the orthonormal basis tensors ¢L~ for Sn from Eq. (6.85), (8.45) p
p
For example, the orthonormal highest weight states for the mixed tensors of SU(3) can be calculated from
X[2,1]
=
(~ ~~~)
(sec Prob. 24 of Chap.
6 in [Ma and Gu (2004)]) [2,1] _ cPll 
cP~21,IJ
8.3
8.3.1
y[2,I]Ll 1
_
17112 
2Ll
17112 
Ll
17211 
Ll
17 121 ,
= Jl13 { y[2,1]0112 + 2y12 ,1] 8 121 } = J3 {8 121

8211 }
(8.46) .
Direct Product of Tensor Representations
Outer Product of Tensors
Let T~: ),a and T:,2) .b no be two tensors of rank n and of rank m of SU (N), n
res pectively. Merge them to be one tensor T~,I)an T~,2).. bno of rank (n + m), called the outer product of two tensors. Its product space is denoted by T. After the projections of two Young operators acting on two tensors, respectively, T reduces to a subspace T[>'lIl'l :
where the Young patterns [A] and [f.l], whose row numbers are not larger than N l contain nand m boxes, respectively. Here the ordinal index v for the standard Young operator yt>.] is omitted for simplicity. The tensor subspace T[>'lIILI c T is invariant in the SU(N) transformation and corresponds
374
Chap. 8 Unitary Groups
to the representation [A] x [It] with the dimension d[>,] (SU(N))d[fL] (SU(N where we denote the representation directly by its Young pattern for CO venience. Generally, the direct product representation is reducible. It c be reduced as follows. Applying a Young operator y[w] to T, where [ contains (n + m) boxes and its row number is not larger than N, one ha
is invariant in the SU(N) transformation and corresponds to the re resentation [w]. If
T[w]
where ta is a vector in the group algebra of the permutation group Sn+1 there is a subspace corresponding to the representation [w] in T[AllfL],
(8.4
Remind that the leftmost operator Y[A]Y [fL] determines that the tensor su space after the projection belongs to T[AllfL] because the tensor in the cur bracket belongs to T , and the rightmost operator y[w] determines the pro erty of the t ensor subspace in the SU(N) transformations owing to th e We reciprocity. In comparison with Eq. (6.105) for the outer product of two repr sentations of the permutation group, one can borrow the technique of t Littlewood  Richardson rule to calculate the reduction of the direct produ representation of SU(N)
[A] x [It] ~
EB a);'fL[w].
(8.4
[wi
However, Eq. (8.48) is different from Eq. (6.100) because the represent tions in Eq. (8.48) are that of the SU(N) group, not that of the perm tation group. If a Young pattern [w] in Eq. (8.48), which is calculated the Littlewood  Richardson rule, contains the row number larger than N [w] should be removed from the ClebschGordan series for SU(N). T dimension formula of the reduction (8.48) becomes d[AJ(SU(N))d[fL](SU(N))
= La);'fLd[W](SU(N)).
(8.4
[wi
For example, the direct product of two adjoint representations [2,1] x [2, of SU(3) and their dimension formula are
§8. 3 Direct Product of Tensor Representations
8
X
8 = 27 + 10 + 10'
+ 2 X 8 + 1,
37
(8.50
where 10* denotes the representation [3,3] which is conjugate with [3,0 Compare the reduction with Example 1 in §6 .5.2. An important example for the reduction is the direct product of th totally antisymmetric tensor representation [IN] of rank N and an arb trary representation [AJ of SU(N). In the CG series calculated by th Littlewood  Richardson rule, there is only one representation with the row number not larger than N so that two Young patterns are directly adhib ited,
(8.51
Since [1 N] is the identical representation, [A'] is equivalent to [AJ, whic was mentioned in Eq. (8.35). Therefore, the irreducible representations o SU(N) can be characterized by a Young pattern with the row number les than N, namely by (N 1) parameters where (N 1) is the rank to SU(N) In order to calculate the ClebschGordan coefficients, one needs to writ the expansions of the standard tensor Young tableaux with the highes weights appearing in the ClebschGordan series. In writing an expansio for the highest weight M one first finds out all possible products of tw standard tensor Young tableaux in two tensor subspaces where the sum o two weights is M. The coefficient in front of each term can be determine by the condition (7.115) that the expansion is annihilated by each raisin operators EJi.. In the following the expansions of the products of standar tensor Young tableaux for the reduction of [2,1] x [2,1] of SU(3) are liste as examples. The expansions of the products of the basis states are als listed for comparison. I
~ I~
11
11
11(2,2), (2,2))
I
~
8j=O 8j=O, x
= 1(1,1))1(1,1)),
Chap. 8 Unitary Groups
376
~ ~ IlTll ITJIJ _ ITJIJ ITJIJ tij ~ ~ ~ ~' X
11(3 , 0), (3,0))
=
/lfi {1(1, 1))1 (2, 1)) 
X
1(2,1)) I(1,1)) } ,
1illTil~[JJllxliT2l_[Tl]]xITJIJ ITIIITI ~ ~ ~ ~' 11(0,3), (0,3)) = /lfi {1(1, 1))1(1,2)) 1(1,2))I(l,l))},
~~ITJTIxITJJJ+ITJJJxITJIJ ~S
~ ~ ~ ~
[TT2lxllTllllTll xITJJJ
~ ~ ~ ~' 11(1,1), (1, 1))s = J1/20 {J3[ 1(1,1))1(0, 0h) + 1(0, O)l)l(l, 1)) 1 + 1(1,1))1(0, 0h) + I(O,Oh)l(l, 1))  y'6 [ 1(1,2)) 1(2,1)) + 1(2,1))1(1,2)) l},
~~llTllxITJJJ_ITJJJxITJIJ ~A
~~~~  2
ITJIJ ITJIl 2 ITJIl ITJIJ X
X
~ ~+ ~ ~
ITDlxITTIlllTll xllf2J ~~+~~' 11(1,1), (1, l))A
= J17i2 { 1(1 , 1)) I(O,O)d 1(0,0)1)1(1,1))  J3 [ 1(1,1)) I(O,Oh) 1(0 , 0h)11, 1) 1
 12 [ 1(1,2))1(2,1)) 
1(2,1))1(1,2))
l} ,
~ ~ ITJIJ X ITITI +ITITI X 1lTll_ [iJJJ X [II ti:tij ~~~ ~~~
~X8fDBfDXSPJSPJXBf 2
+
ITJJJ [TT2l 2 [lf3l ITJIl ~ ~+ ~ ~ X
X
II
0),
378
Chap. 8
Ou
(..f!...
~
Unitary Groups
Cb ' ... ) T cal ···
U a , a' ... U * b 1 b'1 1
C=!
cdd'
L
. ••
d ' b', ... T da't ".
(a')(b')
· (L
1t a , a'I . .. 1tb I b't '"
(a')(b')
( T ddba 'I; ,.... )
.
d
The trace tensor of a mixed tensor of rank (n, m) is a tensor of rani 1, m  1). There is a special mixed tensor D~ of rank (1, 1) of SU (N) w component is the Kronecker (j function when a = b, when a =1= b,
(OlLD)~ =
L
• 1taa' Ubb'
a'b'
Db'a'
=L
U aa' 1t
• J:b' J:b Dba ' bb' va' = va =
a'b'
The onedimensional tensor subspace composed of D is invariant in S and corresponds to the identical representation of SU(N). A mixed t can be decomposed into the sum of a series of traceless tensors with dif ranks in terms of the invariant tensor D~. For example,
Ti
= {
Ti 
D~ ( j~ ~ T~) } + D~ C~ ~ T~) .
The first term is a traceless mixed tensor of rank (1,1) and the trace t in the bracket of the second term is a scalar. The decomposition of a m tensor into the sum of traceless tensors is straightforward, but tedious. has to write all possible terms and calculate the coefficients by the trac conditions. For example,
N
=
T:;b + D~ L p=]
N
{CITtp+ C2 T:b} + Dg L {C3 Ttp + C4 T;a} p=l
Solving the traceless conditions La
§8 . .'7 Direct Product oj Tensor Representations
H.:L3
3
Traceless Mixed Tensors
I ' J's t , we study the condition whether the traceless mixed tensor spa oI"lloted by a pair of Young patterns [A]\[tt]* is a null space or not. It is IIldl space if the number of constraints from the traceless conditions is n It'ss than the number of independent tensors. Denote by nand m the ro 1IIImbers of two Young patterns [A] and [tt]' respectively. For an arbitra pair of tensor Young tableaux in the space, assume that there are £ pai , ,I' repetitive digits in the first columns of the two Young tableaux. Fixi 1,lte different digits in the two first columns and the digits in the remaini ('o lumns, one changes the pairs of repetitive digits from 1 to N where som 1 . 1~llsors are vanishing owing to antisymmetry of indices. The number illd ependent tensors is
( N  (n ~ m 
2£)) .
The traceless condition is written as a sum of tensors to be 0 where t (f  1) pairs of digits are fixed and only one pair of digits runs over from (.0 N. The number of the traceless conditions is the number of the possib values of the (£  1) pairs of digits, that is, (
N  (n + m £1
2£))
The condition for the traceless mixed tensor space not to be a null space that the number of traceless conditions is less than the number of indepe dent tensors, namely, [N  (n + m  2£)]/2 ;::: £. The solution is
n
+m
::; N.
(8.5
Second, discuss a totally antisymmetric contravariant tensor subspa T[1'nr of rank m, whose basis tensors are standard tensor Young tablea y{l ~IOb" .b m . Multiplying the basis tensors with a totally antisymmetr tensor of rank N, one obtains
(8.5
From the viewpoint of replacement of basis tensors in one tensor subspac the correspondence of two sets of basis tensors is onetoone. In fact, on one basis tensor appears in the righthand side of Eq. (8.58). The numb of the basis tensors of two sets are the same
380
Chap. 8 Unitary Groups
The difference of two sets of basis tensors is only in the arranging ord!'I The representations with respect to two sets of basis tensors are equivalelll From the transformation of the basis tensors in SU(N), Ou<)?a! ... aN_ TrI
1
L
= m!
L
L
bl ... bm. dl ... d N
l'7l
cd, ... dN_mbl ... bm
(U~,dIUc la,)
...
C\ ... CN_m.
X (u * CNm d Nm U c Nm a Nm )
y[l m I 0 1 , .. tm u ', L bI
'"""' ~
. ..
ut'm b_
l1 .. .tYn
1
m'
C\ ... CNrn
L
X {
t\ .. tm
L
U;,d, ... U;N_mdN_mu;,b, . . . U;mbmcdl .dN _ m b1 ... bm
d\ ... dNm vI·· ·b
TTl
L
~! CI
L
cC, ...CN_ mL, ... Lmy[lm I OL,
. Imuc,a, ... UCN_maN_m
· .CN_m ll .·. l m
L
(fJ C I .. eN  1ll UCla] . . . UCNtn(tNnt'
C \ ... CN  m
(8.59 where the formula for
Ca l.aN
is used,
Uald l ·· .UaNdNCd, .. d N
= (det
=
CIl, ... aN
L d, ... dN
d, .. dN u)ca, ... aN
=
Cal .. aN·
(8.60 Equation (8.59) shows that <Pa, .aN_m is proportional to the basis tenso ypNmIOCq .. n N_m in the tensor subspace T[IN'"I. Namely, the represen tations of two tensor subspaces T[Iml' and T[lN"'1 are equivalent,
(8.61
[Im r ml' Third, generalize T [I to the traceless tensor subspace 7[>1 ' corre sponding to a pair of Young patterns [A]\[lm]* where the row number o [A] is not larger than N  m. Denote the basis tensors in 7[~lmr by n~lb which are traceless between c and each bj
§8.:1 Direct Product of Tensor Representations
N
381
N
LL
(8.62)
c=1 bj =1
Multiplying the basis tensors with a totally antisymmetric tensor of rank N, one obtains (8.63) a, ... aN _~ c ... belongs to a covariant tensor subspace corresponding to a direct product representation, [IN mj x [AJ, which is calculated by the Littlewood  Richardson rule. From the traceless condition (8.62)
a,
L
fa,
.. aNrrtC
1
m!
a, ... aN_mcb, .. bm
because each term in the sum contains m factors of 15 functions including a factor related with c, say I5g. Thus, in the reduction of [1 N mj X [A], there is only one nonvanishing term whose Young pattern is obtained by adhibiting [I N  m ] and [A] directly, )
1
m.
(8.64)
[A'] is the Youn g pattern obtained by adhibiting [INm] and [A] directly. Similarly, one has [lm]\[A]* ~ [X]*, or equivalently,
1::::: k::::: m,
(8.65)
where the row number of [T] is m, and [T'] is obtained from [T] by removin g its first column. The replacement of basis tensors for the equivalent tensor subspaces in Eq. (8.65) is (8.66) where ~,: .. aNm is traceless. At last, discuss the general case where n~,: .. b~C. is the traceless basis t.ensor in the representation [A]\[T]*. The row number of [T] is m and that of [A 1 is not larger than (N  m). The tensor is traceless between each pair of one covariant and one contravariant indices, say d and bj or d and c. Let
382
Chap. 8 Unitary Groups
(8
The covariant part of cJ>~;:.aN_~d.. corresponds to the representa [I N  m ] X [A]. In its reduction by the LittlewoodRichardson rule, cept for the Young pattern by adhibiting [I N  m ] and [A] directly, Young pattern in the ClebschGordan series contains an operation of a symmetrizing one covariant index, say d, and all new covariant indices However, fa, .. aN mdb; .. b;n annihilates the antisymmetrized term bec the product of two f. makes a factor sg which annihilates the trace tensor n~, . . . b~c. Thus, the following two representations are equivale )
rj
[A]\[r]* :::' [A']\[r']*,
= rj
A~ =

Ak
1,
1
~ j ~
+ 1, 1 ~ k
~
m,
N  m,
(8
wh ere [A'] is the Young pattern obtained by adhibiting [1 Nm] and directly, and [r'] is obtained from [r] by removing its first column. Suc sively applying the replacement (8.68), one is able to transform any trace mixed tensor subspace denoted by [I1J\[r]* into a covariant tensor subs denoted by a Young pattern [A] so that [11]\[r]* is irreducible. Furtherm a contravariant tensor subspace denoted by a Young pattern [r]* is equ lent to a covariant tensor subspace denoted by a Young pattern [A], w
[r]* :::' [AJ,
l~j~N.
(8
Adhibiting the Young pattern [r] upside down to the Young pattern one obtains an N x rl rectangle. The relation (8.69) was shown in F 8.1 and 8.3 as examples.
8.3.4
Adjoint Representation of SU(N)
As shown in §7.6.I, the highest weight of the adjoint representatio SU(N) is M = WI +WNI, corresponding to the Young pattern [2 , I N [1]\[1]*. In this subsection we are going to discuss the adjoint representa of SU(N) by replacement of tensors. From the definition (7.8), the adjoint representation of SU(N) satis N 2 _1
UTAU
I
=
L
TBDsdA(u),
(8
B=1
where TA is the generator in the selfrepresentation of SU(N). TA i N x N traceless Hermitian matrix. A traceless mixed tensor Ti of r
§8·4 SU(3) Symmetry and Wave Functions of Hadrons
(1,1) has a similar transformation,
(OuT)~ =
L
uaa,T:: (u1)b'b·
(8.7
a'b'
T~ can be looked as the matrix entry of an N x N traceless Hermitia matrix at the ath row and the bth column so that it can be expanded wi respect to the generators (TA)ab where .j2FA are the coefficients, N 2 _1
Tab
= v£.'2
'" ~
(T) A ab F A,
FA
= J22:
(TA)ba T~.
(8 .7
ab
A=l
From the vi ewpoint of replacement of tensors, T~ and FA are two tensors the traceless tensor subspace of rank (1,1) so that they correspond to th sa.me representation [1]\ [1] * Calculate the transformation of FA in SU(N N 2 _1
OuT = uTu = ,j2 1
L
uTAu 1 FA
N 2 _1
=.j2
2:
B= 1 N 2 _1
OuT = .j2
L
TB (OuF)B·
B=l
Namely, N 2 1
(OuFh =
2:
D'BdA(u)FA.
(8 .7
A= l
FA corresponds to the adjoint representation. It shows that the adjoint re resentation ofSU(N) is equivalent to the representation [1]\[1]* ~ [2 , 1 N Both T: and FA are the tensors transformed according to the adjoint re resentation of SU(N). Two forms of T~ and FA are commonly used particle physics.
8.4
SU(3) Symmeh'y and Wave Functions of Hadrons
As an example of physical applications of group theory, we are going study the flavor SU (3) symmetry in particle physics in this section . Th rank of SU(3) is 2 so that the planar weight diagram is more convenient demonstrate the basis states in an irreducible representation of SU(3) tha
384
Chap. 8 Unitary Groups
the block weight diagram. We will derive the mass relations of hadrons a calculate the wave functions of hadrons in the multiplets of SU(3).
8.4.1
Quantum Numbers of Quarks
In the theory of modern particle physics, the "elementary" particles divided into four classes. The particles participating in the strong inter tion are called the hadrons. Those without the strong interaction are cal the leptons. The particles mediating the interactions are called the ga particles. The particles in the fourth class, called t.he Higgs bosons , are troduced in the theory for providing the static masses of other particles, a are not observed yet in experiments. There are three generations of lept as well as their antiparticles. They are the charged leptons e, fl, and T a their neutrinos V e , vIJ.' and V T • The gauge particles include the photon m diating the electromagnetic interaction, the neutral boson ZO and char bosons W ± mediating the weak interaction, and the gluons mediating strong interaction. The graviton mediating the gravitational force is be studied in theory and in experiment. Some other particles presented in supersymmetric theory are still not observed in experiments. We focus our attention on the hadrons. Among hadrolls, the fermi are called the baryons and the bosons are called the mesons. All hadr are constructed by the more elementary particles called the quarks a antiquarks. In the modern theory there are 18 quarks and 18 antiquar Usually, the quarks are described by a visual language, "color" and "flav which are not in the common sense on color and flavor. There are th color quantum numbers, say red, yellow, and blue. The quarks with th colors are the bases of the color SU(3)c group in the theory of quant chromodynamics. They participate in the SU(3)c gauge interaction me ated by the gluons. The theoretical and experimental researches expect socalled color confinement that a state with color cannot be observed the recent experimental energy level. Thus, the quarks in the low ene have to appear in the colorless states, or called the color singlet of SU(3 Namely, three quarks construct the basis states of totally antisymme tensors of rank three, and a quark and an anti quark construct the tr state of the mixed tensor of rank (1,1), and
(8.
abc
Both states correspond to the identical representation of SU (3)0. The st
§8·4 SUr;]) Symmetry and Wave Functions of Hadrons
38
with three quarks is a baryon state with baryon number 1. The pair of quark and antiquark is a meson state without the baryon number. Therefore, quark brings the baryon number 1/3 and an anti quark brings the baryo number 1/3. Some composite colorless states composed of the state (8.74) are being studied recently. The quark and antiquark states appearin in the following are understood to be the colorless states. There are six flavor quantum numbers for quarks. They are divided into three generations in weak interaction. Each generation contains tw quarks: the up quark u and the down quark d; the charm quark c and th strange quark s; and the top quark t and the bottom quark b. The firs quark brings 2/3 electric charge unit and the second 1/3 unit. In eac gen eration, their lefthand states constitut.e a doublet with respect to th weak isospin, and the righthand states are singlets. The u quark and the quark are very light, and the s quark is a little bit heavier. They are calle the light quarks. The c, b, and t quarks are heavier one by one, and ar called the heavy quarks. The up quark u and the down quark d constitute a doublet in the isospi SU(2) group. Although they have different electric charges, the isospin i conserved approximately in the strong interaction and plays an importan role in particle physics. Generalizing the isospin symmetry, one assume that three light quaTks constitute a triplet of the flavor SU(3) group, whic is a broken symmetry because the s qu ark is heavier than u and d quarks However , the flavor SU (3) symmetry made some historical contribution in discovering new hadrons and predicting their properties. Even recently the flavor SU(3) symmetry helps the research of hadron physics in som respects. In this section we pay attention to the application of SU (3) t the hadron physics only from the viewpoint of group theory. Exce()t for color and flavor there a re a few inner quantum numbers fo quarks such as the baryon number B, the electric charge Q, the isospin T and its third component T 3 , the strange number 5, and the supercharge Y They are related by
Y
= B + 5,
Q = T3 + Y/2.
(8.75
The isospin SU(2) is a subgroup of the flavor SU(3) group. T3 and Y spa the Ca.rtan subalgebra of the flavor SU(3) group,
o 1
o
(8.76
Chap. 8 Unitary Gmups
The quantum numbers of quarks are listed in Table 8.1. The quantu numbers of antiquarks are changed in sign except for T. In addition, t quarks have some spatial quantum numbers such as the spin and the pari Table 8.1
8.4.2
Quantum numbers of light quarks
Quark
B
T
T3
S
Y
Q
u
1/3
1/2
1/2
0
1/3
2/3
d
1/3
1/2
 1/2
0
1/3
 1/3
s
1/3
0
0
1
 2/3
 1/3
Planar Weight Diagrams
Corresponding to the Chevalley bases of SU(3), there are two SU(2) su groups. One is called the Tspin where the generators are HI = 2 EI = TI = TI + iT2 , and Fl = T_ = Tl  iT2 . Y is constant in multiplet of Tspin. The other is called Vspin where the generators H2 = 3Y/2  T 3, E2 = V+ = T6 + iT7 , and F2 = V_ = T6  iT7 . Q constant in a multiplet of V spin. Since three quarks are the basis sta of the flavor SU(3), one has
TId = u,
T+u = T+s = 0,
T_d = T_s = 0,
V+s = d,
V+u = VId = 0,
V_u
TIu = d,
TId
= T+s = 0,
T_d = u,
T_u = T_s = 0,
s,
V+u
= V+s = 0,
V_s = d,
V_u=V_d=O.
V+d =
= V_s = O.
(8.7
In the planar weight diagram for the flavor SU(3), the abscissa axis a the ordinate axis are the eigenvalues of T3 and Ts, respectively. The sim roots and the fundamental dominant weights are
where the unit vectors along the abscissa axis and the ordinate axis denoted by e2 and el, respectively, such that r J.I are positive roots. move the basis states along the abscissa axis so that both T3 and Q chan by ±1, but Y does not change. V± move the basis states along the directi with an angle 27f/3 to the abscissa axis so that T3 changes by =f1/2, changes by ±1, but Q does not change.
§8·4 SU(3) Symmetry and Wave Functions of Hadrons
387
The hadrons are composed of three quarks or a pair of quark and antiquark, denoted by a suitable standard tensor Young tableaux. In the covariant tensor Young tableaux, 1, 2, and 3 are replaced with u, d, and s, respectively. In the contravariant tensor Young tableaux, 3, 2, and I are replaced with s, cr, and il, respectively. In drawing the planar weigh diagram of a given representation [A], one first determines the position o the highest weight state, where the box in the first row is filled with u and that in the second row with d. Then, applying the lowering operators T_ and U_, one calculates the positions of the other basis states. For a multiple weight, each basis state has definite quantum numbers T, T 3 , and Y. All basis states in the representation have the same spin and parity. There are three types of the planar weight diagrams for the flavor SU(3).
(a) For a onerow Young pattern [A, 0], the planar weight diagram is a regular triangle upside down. The length of the edge of the triangle is A. Al weights are single. The conjugate representation of [..\, OJ is [A, A] ~ [A,O] whose planar weight diagram is a regular triangle. The highest weight state in [..\, 0] is described by a standard tensor Young tableau where each box is filled by u so that Y Y
d
= A/3,
(8.79)
u
•
•
Y
T3 S
N*
N'o
N*+
N'++
•
•
•
•
~.
~ 0
~.+
a) [1,0]
T3
Y
• ::::_0
:=* •
S
T3
•
u
•
d
b) [1, OJ' Fig. 8.5
fl
c) [3,0]
The planar weight diagrams of [A, 0] of SU(3).
:IKH
Chap. 8 Unitary Groups
From the highest weight state one constructs a Tmultiplet by applicat of the lowering operator T _, where u quark is replaced with d quark by one. In the planar weight diagram, the states in the Tmultiplet located in a horizontal line with Y = >../3. Recall that the onerow ten Young tableau describes a totally symmetric tensor so that the quark the tableau can be interchanged symmetrically. From each state in the multiplet one constructs the Umultiplets by applications of the lowe operator U _, where d quark is replaced with s quark one by one. In planar weight diagram, the states in the Umultiplet are along a line w the angle 21f /3 to the abscissa axis where Q is fixed. All weights in representation are single. The planar weight diagrams of the represe tions [1,0], [1, 1] ~ [1,0]', and [3,0] are listed in Fig 8.5. [1,0] and [1 describe the quarks and antiquarks, respectively. [3 , 0] describe the bar decuplet, observed in experiment with spin 3/2 and the positive parity. baryons in the decuplet are denoted by N*, I;', :::;* , and 0, J'/*++ N*o 2":*+ I;*
=I u
Id
d
Iu
s
Id
s
= v'3 1u I u Id I , N* = I d I d Id I , I;'o = J6 1u I d I s I , :::;,0 = v'3 1u I s I s I,
Is
s
D
lu lu I ,
= v'3 1u = v'3 1u = v'3 1d
:::;' = v'3 1d
N*+
= Is
(8
I sis I·
(b) For a Young pattern [2\>..] ~ [>..,0]\[>..,0]*, the planar weight agram is a regular hexagon with the edge length >... The weights on edge are single. The multiplicities of the weights increase one by on their positions go from the edge toward the origin. The representatio [2>", >..] is selfconjugate and its planar weight diagram is symmetric in inversion with respect to the origin. The highest weight state in [2>", >..] is described by a standard te Young tableau where each box in the first row is filled by u and each in the second row is filled by d so that
Y
= >..,
(8
From the highest weight state, one constructs a Trnultiplet by applicat of the lowering operator T _, where u quark is replaced with d quark by one. The u quark on the column with two rows cannot be replaced d quark otherwise two d's are filled in the same column. The state the Tmultiplet are located in a horizontal line with Y = >.. in the pl
§8.4 SUr:;) Symmetry and Wave Functions of Hadrons
389
weight diagram. From each state in the Tmultiplet one constructs the
Umultiplets by applications of the lowering operator U_, where d quark
is replaced with s quark one by one. Since the s quarks can be filled in two rows, the multiple weight appears. The basis states in one Tmultiple have the same number of s quarks as each other . The planar weight diagrams of the representation [2, 1] for the baryon and the representation [1]\[1]* for the mesons are listed in Fig. 8.6. In experiments, the observed baryons in the octet [2,1], called P, N, I;, A and ::::, have spin 1/2 and positive parity. The observed mesons in the octet [1]\[1]* have negative parity. When the spin is 0, they are the scala mesons, called K, 7f, T), and K. When the spin is 1, they are the vecto mesons, called K*, p, ¢, and K*. The tensor Young tableaux of the (1/2)+ baryon octet and the 0 meson octet are listed as follows. They can be calculated from the highest weigh state by the lowering operators (see Eqs. (8.39) and (8.83)). Scalar meson
Baryon P _ uu d N
K+ = us
= ci d
I;+ = UU S
(8.82 2:;
= dd s
J3!2 cis
A =
T)
::0 _ us   s _ ds   s
7f o = T)
=
J176 {Ull + dd 
Ko = 
AFJ
/l
= 
(7f+)
{F2
KO 
2ss}
sd
A{GJ\[~J'  00 *}, A
=
7f O }
Vi {2F2 00 *  ~~. + []J\0 *} =Vi{20\[~r  ~~* 0\[~r}·
=

(8.83
Chap. 8 Unitary Groups
l lil
Two sets of the tensor Young tableaux are related by Eqs.
(8.23)
(1) ..':)8). In fact,
3~=2~ + [lfJ + []¥J = 2
c!]\[~J • + ~\~. +
00' ,
3[]¥J=2[]¥J + ~ + ~
(8
= 2 0\[~J ·  [~J{~'  [TIC!]'
v'2~

v1[]¥J =v1{0~*  C!]\C!]*} ,
~[]¥J =VI{2~0*  ~\~*  [TI[IJ'}. y
y
P
N
•
2;0
•
::0
~
a) Baryons [2,1] Fig. 8.6
2;+
•
1f+
1f0
1f
T3
A
•
•
•
2;
K+
KO
T
'T}
•
K
•
It
b) Mesons [1]\[1]*
The planar weight diagrams of adjoint representation of SU
(c) For a Young pattern [A1,A2], Al > 2A2 > 0, the planar we diagram is a hexagon where the lengths of two unneighbored edges are same. The length of the top edge is Al  A2, and the length of the bot edge is A2. Th e weights on the edge are single. The multiplicities of weights increase one by one as their positions go inside until the hexa becomes a triangle upside down. The weights inside the triangle are A2 The highest weight state has
(8
§8.4 SU(3) Symmetry a.nd Wa.ve Functions of Hadrons
391
The conjugate representation of [A1' A2] is [A1' Al  A2]' Their planar weigh diagrams are inverted with respect to the origin.
8.4.3
Mass Formulas
Present a simple model to study the mass formula of the hadrons in a multiplet of the flavor SU(3), which are made by the quarks and the antiquarks Assume that the binding energy V for the hadrons in one multiplet o SU (3) are the same, and the difference of the masses of hadrons comes from the different quarks. The masses of u and d quarks and their antiquarks are ml and the masses of s quark and its anti quark are m2. The masses in the baryon decuplet are M(N*) = 3m1  V, M(='*)
= m1 + 2m2 
M('E*) = 2ml
V,
+ m2
 V,
M(n) = 3m2  V.
Then, one obtains the mass formula M(n)  MC='*)
= M(3*)
 M('E*) = M('E*)  M(N*).
(8.86)
From the experiments, the observed average masses for the isospin multiplets are MN" = 1232 MeV, ME' = 1384.6 MeV, Ms. = 1531.8 MeV, Mo. = 1672.5 MeV. M(n)  M(='*) = 140.7 MeV, MC='*)  M('E*) = 147.2 MeV, M('E*)  MCN*) = 152.6 MeV.
In the beginning of sixties of the last Century the simple model predicted that a baryon n with the spin 3/2, positive parity, supercharge Y = 2 and electric charge Q = 1 should exist at the mass near 1680 MeV. It was found in 1962 as expected. This model is too simple to explain the masses of the baryon octet because both the baryons 'E and A are composed of one s quark and two quarks of u and d, but have different masses in experiment. Further analysis shows that the different masses of s quark and the quark of u or d can be demonstrated by a broken mass matrix M. In addition to the symmetric mass Ma, M  NIa has the transformation property like the supercharge Y called the "33" symmetry broken. Namely, the Hamiltonian contains a mass term '?jjM'lj; which is invariant in SU(3). How many parameters appear in the mass term '?jjM'lj;? Since NI  Ma belongs to the adjoint representat.ion
Chap. 8 Unitary Groups
392
the parameters come from the reduction of [2,1] x [A] to [A]. When [A] a onerow Young pattern, [2,1] x [A] contains one [A] so that there is tw mass parameters as shown in Eq. (8.86). When [A] is a tworow YOUI pattern, [2,1] x [A] contains two [A] so that a new mass parameter appea GellMann, Nishijima, and Okubo expressed the mass operator !VI the sum of generators and their products,
where T(T + 1) is the eigenvalue of the operator T2. Because there a only three mass parameters, the terms of higher order are not needed a a redundant parameter c should be determined. The antisymmetric co bination of generators in the polynomial of order 2 is proportional to t linear term of generators (sec Eq. (7.5)), and the symmetric combinatio as shown in the last formula of Fig. 7.7, contains the representations [4, [2,1]8, and [0,0], whose dimensions are 27, 8, and 1, respectively. The p rameter c should be determined to exclude the representation [4,2]. T mass formula holds for the baryon decuplet such that
y2 + cT(T + 1)
= a + bY.
For the baryon D, Y = 2 and T = 0, one has 4 = a  2b. For the bary :=:*, y = 1 and T = 1/2, one has 1 + 3c/4 = ab. For the baryon 2 y = 0 and T = 1, one has 2c = a. The solution is a = 8, b = 6, a
c = 4. The solution meets the condition from the baryon N*, where Y = and T = 3/2. Thus, the GellMannNishijimaOkubo mass formula is
M(T, Y) = Mo + M1Y + M2 {y2  4T(T +
I)} .
(8.8
For the baryon octet, M(N) = Mo + M,  2M2, lv£(2:,) = Mo  8M lvf(A) = M o, and M(:=:) = Mo  AIl  2lvh Then,
M(N) + MC~.) 2
lvI(2:,)
+ 3M(A)
(8.8
4
The prediction fits the experiment data,
= 938.9 MeV, M(A) = 1115.7 MeV, M(N)
M(2:,)
= 1193.1
MeV,
M(:=:) = 1318.1 MeV.
The lefthand side of Eq. (8.88) is 1128.5 MeV, and the righthand side 1135.1 MeV. The formula (8.88) holds approximately for the mass squa
§8·4 SU(:J) Symmetry and Wave Functions of HadTOns
393
of the scalar meson octet. The experiment data are m(7r)
= 138.0 MeV,
= 495.7 IvleV,
m(K)
m(T})
= 547.5 MeV.
The lefthand side of Eq. (8.88) for the mass square is 0.2457 Ge V2, and the righthand side is 0.2296 Ge V 2 . The formula is not in good agreement with the vector meson octet because there is a mixture between the meson octet and the meson singlet w.
8.4.4
Wave Functions of Mesons
A meson in low energy is composed of a quark and an antiquark with zero orbit angular momentum. The wave function of a meson is a product of the color, the flavor, and the spin or wave functions. It is not needed to consider the permutation symmetry because the quark and the antiquark are not the identical particles. The mixed tensor of rank (1,1) is decomposed into a traceless tensor and a trace tensor (scalar),
Ox D
*
= 00 . ffi 1.
Due to color confinement, the color wave function of a meson has to be in the colorless state, namely in the singlet of SU(3k The flavor wave function of a meson can be in the octet (traceless tensor) or singlet (trace tensor). For the spinor wave functions, the traceless tensor describes the vector mesons and the trace tensor the scalar mesons. The vector meson octet and singlet (w) with the negative parity and the scalar meson octet and singlet (T}I) with the negative parity have been observed in experiments Denote by (1/J+, 1/J) and (1/J, 1/J+) the spinor wave functions for a quark and an antiquark, respectively. The spinor wave functions for the scalar meson and for the vector mesons are
5 5
= 0, = 1,
53
= 0: =1: = 0:
53
= 1 :
53 53
V172 (1/J+ 1/J+ + 1/J1/J) , 1/J_ 1P+ , V172 (1/J+ 1/J+  1/J1/J) , 1/J+ 1/J .
(8.89)
The flavor wave function of the singlet meson is the trace tensor,
~ {~\[~J'
+
[TIm * + 0\[~J *} .
(8.90)
Chap. 8 Unitary Groups
The flavor wave functions of the octet mesons are given in Eq. (8.82). B in the particle physics, the wave functions are preferred to be expressed a matrix of three dimensions, where the row index denotes the covari one and the column index denotes the contravariant one. The basis tens of the scalar mesons are, for example,
The traceless tensor is expanded with respect to the basis tensors wh the coefficients are written by the names of the mesons, wO
rJ +
J2
v'6
w+
K+
wO
w
rJ +
K
KO
M=
V2
v'6
KO
(8.
2rJ
v'6
M transforms in the flavor SU(3) as follows: M
~
nMv. J •
(8.
Through Eq. (8.72), the flavor wave functions can be expressed in those the real orthogonal representation of eight dimensions,
/V2 = w, + iM5 ) /V2 = K, + iM7 ) /V2 = KG,
/V2 = w+, iM5 ) /J2 = K+, iM7 ) /V2 = KG,
(MJ + iM2 )
(MJ  iM2 )
(M4
(M4 
(M6 M3
(M6 
= wG ,
Ms
(8.
= rJ·
Similarly, the flavor wave functions of the baryon octet are also expres in a matrix of three dimensions,
A
~o

J6
V2
B=
~
p
(8.
where the minus sign comes from the definition of the particles as sho in Eq. (8.82).
§8·4 SU(3) Symmetry and Wave Functions of Hadrons
8.4.5
395
Wave Functions of Baryons
A baryon in low energy is composed of three quarks which are identical particles satisfying the Fermi statistics. Its total wave function has to be antisymmetric in the transposition between the quarks. Assume that the orbital angular momentum of the low energy baryon is vanishing such that its total wave function is a product of the color, the flavor, and the spinor wave functions. Three quarks are described by a tensor of rank 3, which is decomposed by the Young operators,
DxDxDol
I I IffiEPffiEPffi§,
(8.95)
[1] x [1] x [1]:::: [3] E9 [2,:1.] E9 [2,1] E9 [1 3 ].
Those wave functions belong to the representations of the permutation group denoted by the same Young patterns. Due to the color confinement, the color wave function is in the color singlet [1 3 ] which is totally antisymmetric in the quark transposition. The product of the flavor and the spinoI' wave functions has to be totally symmetric. There are three choices for the flavor wave functions. The representation [3] of the flavor SU (3) describes the decuplet which is the totally symmetric states in the permutations. The representation [2,1] describes the octet which is the mixed symmetric states. The representation [1 3 ] describes the singlet which is the total antisymmetric states. However, there are only two choices for the spinor wave functions because the representation [1 3 ] of SU (2) corresponds to the null space. The representation [3] of the spinor SU(2) describes the quadruplet (5 = 3/2) which is the totally symmetric states in the permutations. The representation [2,1] of the spinor SU(2) describes the doublet (5 = 1/2) which is the mixed symmetric states. Since the product of the flavor and the spinor wave functions are totally symmetric, the wave function of flavor decuplet has to multiply that of spinor quadruplet, and the wave function of flavor octet has to multiply that of spinor doublet where a suitable combination is needed such that the multiplied wave functions are combined to be totally symmetric with respect to the permutations. This coincides with the experimental data that the observed lowenergy baryons are the baryon decuplet with spinparity (3/2)+ and the baryon octet with spinparity (1/2)+.
(a) The (3/2)+ baryon decuplet. The wave function is a product of the flavor and the spinor wave functions. Two examples are given in the following.
Chap. 8 Unitary Groups
396
N;/;, T
= 3/2, T3 = 1/2, Y = 1,
and S3
= 1/2.
Nth = ~ {u+u+d_ +u+d+u_ +d+u+u_ +u+u_d+ +u+d_u+ + d+ u_ u+
E:'?1/2' T
".0 _ L.,_l/2 
+ u_ u+d+ + u_d+u+ + d_'u+u+}
= 1, T3 = 0, Y = 0,
and S3
.
= 1/2.
1
M {u+d_s + u+sd_ + d+u_s + d+su_ 3v 2 + s+u_d_ + s+d_u_ + u_d+s_ + u_s+d_ + d_U+,L + d_s+'u_ + s_u+d_ + sd+u_ + lLd_s+ + u_sd+ + d_u_s+ + d_su+ + Lu_d+ + Ld_u+}.
(b) The (1/2)+ baryon octet. Both the flavor and the spinor wave functions are in the mixed symm of the permutations, Their product has to be combined as the wave fun with total symmetry, The representation matrices of generators o permutation group S3 are calculated in Table 6.4. Their direct prod and the eigenvectors to the eigenvalue 1 are
1111) D[2,lj
D[2,lj
[(12)] x
D[2,lj
[(1 23)] x
[(1 2)]
D[2,lj
= ( 0 1 0 1
[(1 23)]
o o
=
0 11 0 0 1
'
1111) ( 1 0 10 1 1 0 0 ' 1 0 0 0
The combination corresponding to the identical representation is the mon eigenfunction v to the eigenvalue 1, v T = (2,1,1,2). Write the wave function of a proton with S3 = 1/2 as exampl proton has the quantum numbers T = T3 = 1/2 and Y = 1. Take Young tableau Y = The basis tensors of the flavor wave func are
Eff2:]
GTul m = 2uud 
dUll  udu,
(23)
~= m
2udu  duu  'uud
Exercises
39
Similarly, the basis tensors of the spinor wave functions are
C±I:J ~ = (23)
(+  ) + ( + )  2(  +),
~ ~ =
(+  ) + (  +)  2( + ).
Thus, the wave function of a proton with S3 = 1/2 is
Pl/2=[]¥J{2~+[(23)~l +
[(2 3) [ ] ¥ J
1 {~ + 2
}
[ (2 3)
~ 1}
{2uud  duu  udu} . 3 {( +  )  (  +)} {2udvd11.11.11.11.d} ·3{(+)(+)} 3 {11.+u_d_  2d+11._1L + ll+d_ll_  211._u_d+ + d_u_u+ + 11._d_11.+ + u_u+d_ + d_11.+1l_  2u_d+u_}. (8.96 The normalization factor should be changed to
+
Jl7l8.
8.5
 Exercises
1. Calculate the dimensions of the irreducible representations denoted b
th e following Young patterns for the SU(3) group and for the SU(6 group, respectively: [3],
[2,1]'
[3,3],
[4,2]'
[5,1].
2. Calculate the ClebschGordan series for the following direct produc representations, and compare their dimensions by Eq. (8.30) for th SU(3) group and for the SU(6) group, respectively: (a) [2,1] 0 [3,0], (b) [3,0]0 [3,0], (c) [3,0]0 [3,3], (d) [4,2]0 [2, 1].
3. Try to express each nonzero tensor Young tableau for the irreducibl representation [3,1] of SU(3) as the linear combination of the standar tensor Young tableaux.
Chap. 8 Unitary Groups
4. Write the explicit expansion of each standard tensor Young tabl e' ~ the tensor subspace y~3,1Ir, where r is the tensor space of rank ' the SU(3) group and the standard Young tableau of the Young op 1" Y2I3,lj.IS 3 2 I 4 I.
W
5. Transform the following traceless mixed tensor representations of SU(6) group into the covariant tensor representations, respectively, calculate their dimensions: (1) [3,2,1]*,
(2) [3,2,1]\[3,3]*,
(3) [4,3,1]\[3,2]*.
6. Prove the identity:
7 . Expand the Gelfand bases in the irreducible representation [3,0] of SU(3) group with respect to the standard tensor Young tableaux making use of its block weight diagram given in Fig. 7.3. 8. Expand the Gelfand bases in the irreducible representation [3,3] of SU(3) group with respect to the standard tensor Young tableaux making use of its block weight diagram given in Fig. 7.3.
9. Express each Gelfand basis in the irreducible representation [4,0] the SU (3) group by the standard tensor Young tableau and calcul the nonvanishing matrix entries for the lowering operators FI'" Dr the block weight diagram and the planar weight diagram for the re resentation [4,0] of SU(3).
10. Express each Gelfand basis in the irreducible representation [3,1 the SU(3) group by the standard tensor Young tableau and calcu the nonvanishing matrix entries for the lowering operators Fil . D the block weight diagram and the planar weight diagram for the resentation [3,1] of SU(3).
11. Calculate the ClebschGordan series for the direct product repre
tation [2,1] x [2,1] of the SU(3) group, and expand the highest we state of each irreducible representation in the ClebschGordan se with respect to the standard tensor Young tableaux.
12. A neutron is composed of one u quark and two d quarks. Const the wave function of a neutron with spin 53 = 1/2, satisfying correct permutation symmetry among the identical particles.
Chapter 9
REAL ORTHOGONAL GROUPS
In this chapter we will study the tensor representations and the spinor representations of the SO(JV) groups, and then, the irreducible representations of the proper Lorentz group .
9.1
Tensor Representations of SO(N)
The tensor representations are the singlevalued representations of SO(JV). In this section, the reduction of a tensor space of SO(JV) is studied and the orthonormal irreducible basis tensors are calculated.
9.1.1
Tensors of SO(N)
Similar to the tensors of SU(JV), a tensor of rank n of SO(JV) has JVn components and transforms in R E SO(JV),
Ta,u".!!:....,,(ORT)u, ...a"= ~
Ru,b, . .. Ranbnn, .. b".
(9.1)
b, ... b"
A basis tensor equal to 1,
()d, . .d"
contains only one nonvanishing component which is
(9.2)
oR()d, ... d" =
~ ()b, .. b" R b , d, ... Rbnd n '
(9.3)
b, ... bn
Any tensor can be expanded with respect to the basis tensors,
Tu, ... u" =
~ Td, ... d" (()d, ... dJ", .. .u" =Tu, .. u" . d, ... d"
(9.4)
400
Chap. 9 Real Ortho90nal Groups
In a permutation of Sn the tensors and the basis tensors are transforme Eqs. (8.6) and (8.18), respectively. The tensor space is an invariant l space both in SO(N) and in Sn. The SO(N) transformation is commu with the permutation (the Weyl reciprocity), so that the tensor space be reduced by the projection of the Young operators. The main difference between the tensors of SU(N) and SO(N) is the transformation matrix R E SO(N) C SU(N) is real. As a result tensors of SO(N) have the following new characteristics. First, the real and the imaginary part of a tensor of SO(N) transform separately in (9.1) so that only the real tensors are needed to be studied. Second, t is no difference between a covariant tensor and a contravariant tenso the SO(N) transformations. The contraction of a tensor are accompli between any two indices so that before the projection of a Young ope the tensor space has to be decomposed into a series of traceless te subspaces, which are invariant in SO(N). Third, denote by T the trac tensor space of rank n. After the projection of a Young operator, 7j; yLAi T is a traceless tensor subspace with a given permutation symm A similar proof to that in §8.3.3 shows that 7jAi is a null space if the of the numbers of boxes in the first two columns of the Young patter is larger than N. Fourth, when the row number m. of the Young pa [A] is larger than N /2, the basis tensor y1"1 edl .. d m c .. can be changed a dual basis tensor by a totally antisymmetric tensor Eal .aN' *
[yIAJeLI
··"Nm
e ..
1
Its inverse transformation is EVI .. bmQm+1
.. h," ~.,
•
[y[ AIll] U
aN···am+lc",
In fact, the correspondence between two sets of basis tensors are oneto and their difference is only in the arranging order. Thus, a traceless te subspace 7jAI, where the row number m. of the Young pattern [A] is la
than N/2, is equivalent to a traceless tensor subspace T,Vi, where the number of the Young pattern [A'] is N  m. < N/2,
9.1
{
}
I)
1
+
{ {
,J
(() 1
402
Chap. 9 Real Orthogonal Groups
In summary, the traceless tensor subspace 1j>'] corresponds to resentation [A] of SO(N), where the row number of [A] is less than When the row number £ of fA] is equal to N /2, yj>.] is decomposed in selfdual tensor subspace '0t(+)>'] and antiselfdual tensor subspace T corresponding to the representation [(±)A], respectively. Through a s but a little bit complicated, proof as that for Theorem 8.3, the repre tions [A] and [(±)A] are irreducible. In fact , there is no further con to construct a nontrivial invariant subspace in their representation s The highest weight.s of the representations [A] and [(±)A] are calc later. All the irreducible representations are real except for [(±)A] N = 4m + 2. yj>.] is a null space if the sum of the numbers of bo the first two columns of the Young pattern [A] is larger than N.
equivalent to 1:J>.'] where [A] and [.\'] are mutually the dual Young pa (see Eq. (9.7)). As far as the orthonormal irreducible basis tensors of SO(N) cerned, there are two problems. One is how to decompose the sta tensor Young tableaux int.o a sum of the traceless basis tensors. T composition is straightforward, but tedious. The second is how to co the basis tensors such that they are the common eigenfunctions of H orthonormal to each other. For SU(N), the standard tensor Young ta are the irreducible tensor bases, but not orthonormal. Because the h weight is simple and the standard tensor Young tableau with the h weight is orthogonal to any other standard tensor Young tableau irreducible representation , the orthonormal basis tensors for SU(N) obtained from the high est. weight state by the lowering operators terms of the method of the block weight diagram. The merit of the m based on the standard tensor Young tableaux is that the basis tenso known explicitly and the multiplicity of any weight is equal to the n of the standard tensor Young tableaux with the weight. For SO(N), the key in finding the orthonormal irreducible basis t is to find the common eigenstates of H j and the highest weight s an irreducible representation . For the groups SO(2£ + 1) and SO(2 generators Tab of the selfrepresentation satisfy
(Tab)cd
= i {Oae Obd 
OadObc} ,
[Tab, Ted] = i {obeTad + oadT&c  obdTac  oacT&d} . The bases H j in the Cartan subalgebra are Hj
= T(2jl)(2j),
1 ~ j ~ N/2.
§9.1
9.1.2
Tensor Representations of SO(N)
Irreducible Basis Tensors of 80(2£
403
+ 1)
The Lie algebra of SO(2t' + 1) is Be. The simple roots of SO(2t' + 1) are Te = ee·
(9.14)
= 1 and rt is the shorter root with de = 1/2 From the definition (7.141), the Chevalley bases of SO(2t' + 1) in the selfrepresentation are
Til are the longer roots with dJ1.
HJ1. = T(2J1.1)(2J1.) T(21'+1)(2'k+2), EI'
= ~
{T(2J1.)(2J1.+l)  iT(21'1 )(21'+!)  iT(2J1.)(21'+2)  T(2Il1 )(21'+2)} ,
~ {T(21')(21'+1)
FJ1. =
+ iT(21'1)(21l+1) + iT(2J1.)(2J1.+2)
 T(21'1)(21'+2)} ,
He = 2T(2tl)(U) , Ee = T(2l)(2l+1)  iT(2l1)(2e+l» Fe
=
T(2e)(2t+l)
+ iT(2t1)(U+l)·
(9 .15) (}Q is not the common eigenvector of HI'" Generalizing the spherical har monic basis vectors (4.180) for SO(3), one defines the spherical harmonic basis vectors in the selfrepresentation of SO(2t' + 1) (_1)ta+1 <Pa =
{
Jlfi ({}2al + i{}2a)
)
a=t'+1)
(hl+!)
J1fi ({}4f2a+3
1 ~ a ~ t',
t' + 2
~
a
~
2t' + 1. (9.16) The spherical harmonic basis vectors <Pa are orthonormal and complete. In the spherical harmonic basis vectors <Pa, the nonvanishing matrix entries of the Chevalley bases are  it94e2a +4) ,
HJ1.
HI'
H/1.
HI'
He
H t
EJ1.
=
E Il
Ee
E e
FIl
F Il
Ft
F e
(9.17)
=
= V2
where 1 ~ f.I ~ t'  l. Namely, the diagonal matrices of Hil and He in the spherical harmonic basis vectors <Pa are
404
Chap. 9 Real Ortho9onal Groups
HJ1. = diag{O, ... ,O,I,I,O, ... ,O,l,I,O, ... ,O}, 'v'
'v'
J1.1 U2J1.1 He = diag{O, ... , 0, 2,0, 2,0, ... , O}. 'v' el
'v'
J.LI
'v' iI
The spherical harmonic basis tensor CPC'l .. D: n of rank n for SO(2t' + the direct product of n spherical harmonic basis vectors CPD: I ... CPD:n standard tensor Young tableaux yl,l,lcpD:I ... D: n are the common eigen of HJ1., but generally not orthonormal and traceless. The eigenvalue in the standard tensor Young tableaux y];'l CPD:I . .D: n is equal to the nu of the digits J1 and (2t'  J1 + 1) in the tableau, minus the number of (J and (2t'  tL + 2). The eigenvalue of He in the standard tensor Young ta is equal to the number of t' in the tableau, minus the number of £ + then multiplied with 2. The action of FJ1. on the standard tensor Y tableau is equal to the sum of all possible tensor Young tableaux, e which is obtained from the original one by replacing one filled digit J the digit (tL + 1), or by replacing one filled digit (2t'  tL + 1) with the (2£  J1 + 2). The action of Fe on the standard tensor Young tabl equal to the sum, multiplied with a factor /2, of all possible tensor Y tableaux, each of which is obtained from the original one by replacin filled digit e with the digit (t' + 1) or by replacing one filled digit ( with (t' + 2). The actions of EJ.L and Ee are opposite. The obtained Young tableaux may be not standard, but they can be transformed sum of the standard tensor Young tableaux by the symmetry (8.22). Two standard tensor Young tableaux with different sets of the digits are orthogonal to each other. For an irreducible representation SO(2t' + 1), where the row number of [AJ is not larger than £, the h weight state corresponds to the standard tensor Young tableau where box in the ath row is filled with the digit a because every raising op Ell annihilates it. The highest weight M = LJ1. wJ1.MJ1. is calculated Eq. (9.17), 1 :S J1
< £,
Me = 2A£.
The t.ensor representation [AJ of SO(2£ + I), where Me is even is a s valued representation. It will be known later that the representation odd l"\1[e is a doublevalued representation, called the spinor one. Although the standard tensor Young tableaux is generally not trac the standard tensor Young tableau with the highest weight is tra because it only contains CPD: with a < t' + 1 (see Eq. (9.16)). For exa
§9.1
40
Tensor Representations of SO(N)
the tensor basis (h ()j is not traceless, but 4>14>1 is traceless. Since th highest weight is simple, the highest weight state is orthogonal to any othe standard tensor Young tableau in the irreducible representation. Simila to the method of finding the orthonormal basis tensors in SU(N), one able to find the remaining orthonormal and traceless basis tensors in [A] o SO(2e + 1) from the highest weight state by the lowering operators FIJ i terms of the method of the block weight diagram. As an example, calculate the traceless symmetric tensor of rank 2 i SO(7) whose Lie algebra is B 3 . Its representation is denoted by the Youn pattern [2,0,0]' where the highest weight is M = 2w], or denoted b (2,0,0). The simple roots TIJ are expressed as
Two typical standard tensor Young tabl eaux are 0:
I
f3 1= y[2,O,ol4>a/J = 4>a/J + 4>a/J,
0:
I
0:
1= y[2,O,oi4>aa = 24>aa·
The trace tensor is
+ ... + ()7()7 + 4>26 + 4>62
= 4>17  4>71 =
 I 1 I 7 I+ I 2 I 6 II
Using the brief symbols, 1M, m)
(9.19  4>35  4>53 3
+ 4>44
I 5 I+ ~ 1;4'14 'I .
= 1m), one has
IM)=1(2,0,0))=1 1 1 1 1=24>1l'
In terms of the method of the block weight diagram, the remaining basi states can be calculated from the highest weight state \(2,0,0)) by th lowering operators FJ.L . The calculated results are listed in Fig. 9.l. There is only one standard tensor Young tableau I 1 I 2 Iwith the dom nant weight (0, 1,0) so that it is a single weight. The representation contain a dominant weight (0,0,0) with the multiplicity 3 because there are fou standard tensor Young tableaux with the weight (0,0,0) , I 1 I 7 I, I 2 I 6
I 3 I 5 I, and I 4 I 4 I, where one combination is the trace tensor. Through some steps one has
Chap. 9 Real Orthogonal Groups
405
2
v'2 @] v'2 ~ v'2
EJ
2
~ v'2E] v'2~ v'2~ v'2@]
~ v'2~ A
Fig. 9.1
=E}l~ v'2E]
B =
JT{ B+~2~}
c=
ji;{8~~3B}
El
The block weight diagram and the tensor Young tableaux in the representation [2,0,0] of SO (7).
I1
5 I = J2 1 1
6
I'
1(1,2,2)) = Fl 1(1,1,2)) = J2FJ 1 1
5 1= J21 2
5
I,
1(2, I, 0)) = F2 1(1, 1, 2)) = J2 F2
1(0, 1,2)) = (1/2)F3 1(0,2,4)) = (1/2)F3
1
3
I3
1= J21 3
I 4 I·
1(0, 1,2)) belongs to an A 3 quintet. From 1(2, 1,0)) and 1(1,2,2)) one construct an A)triplet and an A2triplet, respectively. Since the multip ity of the dominant weight (0,0,0) is 3. Define that 1(0,0,0))) belongs the A,triplet, a suitable combination of 1(0,0, Oh) and 1(0,0, Oh) belo to the A 2 double, and a suitable combination of I(O,O,Oh), I(O,O,Oh) a
§9.1
Tensor Representations of SO(N)
1(0,0,0)3) belongs to the A 3 quintet. 1(0,0, 0h) is the AIsinglet. 1(0,0, Oh is both the Atsinglet and the A 2 singlet.
1(0,0,0)1) = jlfiFl 1(2,1,0)) =F1
1
1 161= 11 171 + 1 2 161,
F2 1(1,2,2)) = al 1(0,0,0)1) +a2 I(O,O,Oh), F31(0,1,2))=b J 1(0,0,0)l)+b2 1(0,o,O)z)+b3 1(O,O,Oh), El 1(0,0, O)z) = EJ 1(0,0, 0h) = E2 1(0,0, Oh) = 0,
' 1 where a 21 + az2 = 2 and bi~ + b22 + b2 3 = 6. Applymg El F2 = F2El to (1,2,2)) one has
E 1 F2 1(1,2,2)) = V2al 1(2,1,0)) = F2El 1(1,2 , 2)) = F2 1(1, 1,2)) = 1(2, 1,0)).
The solution is al = jlfi. Choosing the phase of 1(0,0, Oh) such that a2 a positive number, one has a2 = )2  ai = Applying ElF,] = F3E and E2F3 = F3E2 to 1(0, I, 2)), one has
fi72.
EIF3 1(0, 1,2)) = V2b 1 1(2, 1,0)) = F3EI 1(0, 1,2)) = 0,
= (jlfib 1 + fi72b2) 1(1,2,2)) = F,3E2 1(0, I, 2)) = F3 1(1, 1,0)) = V2 1(1,2,2)).
E2 F3 1(0,1,2))
J4i3.
The solutions are b1 = 0 and b2 = Choosing the phase of 1(0,0, Oh such that b3 is a positive number, one has b3 = )6  b~ = )14/3. Thus the remaining two states with the weight (0,0,0) are
I(O,O,Oh) =
J273 {Fz
= J4i3F2
1(1,2,2))  jlfi 1(0, 0, Oh) }
I 2 I 5 I 
Vf73 { I 1
I 7 I
+
I 2 I 6 I}
=Ji73{ 1 1171+ 1 216 1 +21315 1} , 1(0,0, Oh) = =
J37i4 {F3
J377 F3 I 3
I 4 I 
1(0, 1,2)) 
J2fii { 
= )2/21 { I 1 I 7 I  I 2 I 6 I
J4i3 1(0,0, Oh)} I 1 I 7 I + I 2 I 6 I + 21 3 I 5 I }
+ I3
I 5 I
+ 31
4 I 4 I}
In comparison with Eq. (9.19), one knows that three states with the weigh (0,0,0) are all traceless.
408
Chap. 9 Real Orthogonal Groups
9.1.3
Irreducible Basis Tensors of SO(U)
The Lie algebra of 80(2£) is De. The simple roots of 80(2£) are l~IL~£l,
re
= ee  I + e£. = 1. From the
(9.2
The lengths of all simple roots are the same, d p definiti (7.141), the Chevalley bases of 80(2£) in the selfrepresentation are t same as those of 80(2£ + 1) except for IL = £,
= T(2£3)(U2) + T(2fl)(2e), Ee = ~ {T(2f2)(2tl)  iT(2e3)(UI) + iT(2e2)(2£) + T(U3)(2t)} Fe = ~ {T(2e2)(2tl) + iT(u3)(2tl)  iT(2l  2)(U) + T(U3)(2e)}
He
,
.
(9.2 Oa is not the common eigenvector of Hp. One defines the spherical harmon basis vectors in the selfrepresentation of 80(2£), which are the generali tion of t.hose for 80(4) (see Eq. (9.144)) ,
¢O' = {
(l)iO'Jl72 (02al
Jl72 (04f2o+1 
+ i0 20') ,
1 ~ a ~ £,
£+ 1
i0 4 l 2a +2) ,
~
a ~ 2£.
(9.2
The spherical harmonic basis vectors ¢CY, are orthonormal and complete. the spherical harmonic basis vectors ¢CY" the nonvanishing matrix entr of the Chevalley bases are
H,,¢p
= ¢J1.'
HJ1.¢J1.+1 =
¢i11I,
H,,¢Upll = ¢2tp+l,
H p¢2tJ1. = ¢2lJ1.' He¢eI = ¢el,
He¢e = ¢e,
He¢e+! = ¢e+l,
H e¢e+2 = ¢f+2,
EjL¢p+1 = ¢J1.'
EJ1.¢up+1 = ¢UJ1.' Ee¢f+2 = ¢e,
Ee¢tH = F,.¢J1.
¢tI,
= ¢J1.+1 ,
Fe¢el = ¢f+!,
(9.2
Fp¢uJ1. = ¢2lp+l, Fe¢e
= ¢e+2,
where 1 ~ IL ~ e 1. Namely, the diagonal matrices of HJ1. and He in spherical harmonic basis vectors ¢a are
H,. = He
=
diag{O, ... ,O,l,l,O, ... ,O,l,l,O, ... ,O}, ~
~
~
J1.1 U2J1.2 diag{O,,,. , O,l,l,l,l,O, . . . ,O}. ~
£2
~
t 2
pI
§9.1
Tensor Representations of SO(N)
~
The spherical harmonic basis tensor CP(Y.I .. On of rank n for SO(2£) is th direct product of n spherical harmonic basis vectors CP(Y.I ... CP(Y.n' The sta dard tensor Young tableaux y1'xlCP(Y.'' (Y.n are the common eigenstates of H but generally not orthonormal and traceless. The eigenvalue of HI" in th standard tensor Young tableaux yJ}1 CPo I ".On is equal to the number of th digits p, and (2f.  p,) in the tableau, minus the number of (p, + 1) an (2f.  p, + 1). The eigenvalue of Hi in the standard tensor Young tablea is equal to the number of the digits (f.  1) and f. in the t.ableau, minus th number of (f + 1) and (f + 2). The eigenvalues const.it.utes the weight. Tn the standard tensor Young tableau. The action of FI" on the standard te sor Young tableau is equal to the sum of all possible tensor Young tableau each of which is obtained from the original one by replacing one filled dig p, with the digit (p, + 1), or by replacing one filled digit (2f.  p,) with t digit (2f.  P, + 1). The action of Fe on the standard tensor Young tablea is equal to the sum of all possible tensor Young tableaux, each of which obtained from the original one by replacing one filled digit (e  1) with t digit (f. + 1) or by replacing one filled digit f. with the digit (f. + 2). The a tions of EI" and Ee are opposite. The obtained tensor Young tableaux m be not standard, but they can be transformed to the sum of the standa tensor Young tableaux by the symmetry (8.22). The standard tensor Young tableaux with different weights are orthog nal to each other. For an irreducible representation [A] or [( + )A] of SO(2f \',;h ere th e row number of [A] is not larger than f, the highest weight sta corresponds to the standard tensor Young tableau where each box in t o:th row is filled with the digit 0: because every raising operator E" ann hilates it. In the standard tensor Young tableau with the highest weig of the representation [(  )AJ, the box in the o:th row is filled with the dig 0:, but the box in th e fth row is filled with the digit (e + 1) . The highe weight M = Lf.L w"Mf.L is calculated from Eq . (9.23),
= A"  Af.L+1, Me l = Me = A( I, Me 1 = Ae 1  At, Mf.l = Ael + At,
l :S p,
Mf.L
Me = A( l + Ae, Me
= At  l
 At,
Ae = 0, for [(+)A],
(9.2
for [(  )A] .
Th e tensor representation [AJ of SO(2£) where (Me  l + Me) is even is singlevalued representation . The representation with odd (Mel + Me), shown later, is a doubl evalued representation, called t.he spinor one. Although the standard tensor Young tableaux is generally not traceles
410
Chap. 9 Real Orthogonal Croups
the standard tensor Young tableau with the highest weight is traceless cause it only contains
BJ
=
y[l,J)
=
= (1/2) { (0 1 + iOz) (0 3 + i( 4) + = (1/2) {_y[1,l)013  iy[1,l)023 = Wt3 
(0 3 + i(4) (OJ + i( 2 )} iy[J,l)014
iwt,
If BJ If ([E  BJ) If F2
=
=
Jl78 { (OJ + iOz) (0  (0 3 + i(4) (0 3
= iV1f2 {y[l,l)012

1 
i(2)
i(4) + (0 3
+ y[1,l)034}
= (1/2) {(03  i(4) (OJ  i(2) = (1/2) {y[1,l)031  iy[J,l)04J =
+ y[l,1)OZ4}
=
{
+ (0 1 
i(2) (OJ
+ i(2)
+
i( 4 ) (0 3 + i( 4 )}

=
iV2wt2'
(OJ  i( 2 ) (03  i( 4 )} iyll,1)032  y[l,l)042}
Wt3 + iWt4'
(9
§9.1
Tensor Representations of SO(N)
41
For [(  ),\], one has
tE
= yP,l]¢13 = ¢13  ¢31
= (1/2)
+ i(2) (0 3  i(4) + (0 3  i(4) (0\ + i(2)} {_y[I,I]013  iy[I,1]023 + iy[1,1]014  y[1,1]024}
= W13
+ iW14'
= (1/2) { (0\
I[ =
Fl
tE I[ (tB
J1!8 { (lh + i(
tE) I[ =
{¢14  ¢41
+ ¢23
 ¢32}
+ (OJ  i(2) (0 1 + i(2 ) + (0 3 + i(4) (0 3  i( 4)  (0 3  i( 4) (0 3 + i(4)} 2)
=
iJl72 {yl l,l]012 
=
(1/2) {(03
=
+
=
(0 1

i( 2 )
y[1,l]034}
+ i( 4 ) (Ol  i( 2 ) (1/2) {y[l,l]031 + iy[1,l]04l 
=
i.j2w12'
(Ol 
i( 2 )
iy[1,l]032
(0 3 + i( 4 )}
+ y[l,1]042}
= W13  iW14'
(9.26 9.1.4
Dimensions of Irreducible Tensor Representations
The dimension d[>,] (SO(N)) of the representation [,\] of SO(N) can be ca culated by the hook rule [Ma and Dai (1982)]. In this rule the dimensio is expressed as a quotient, where the numerator and the denominator ar denoted by the symbols YJ>'] and Y~),], respectively: yl),]
d[(±),] (SO(2£))
=
l>:r' 2Y
yl),] d[),] (SO(N)) =
T),],
where '\e oj::. 0,
h
(9.27 the remaining cases.
Yh
The meaning of two symbols Y~),] and yJ),] are as follows. The hook pat (i, j) in the Young pattern [,\J is defined to be a path which enters th Young pattern at the rightmost of the ith row, goes leftward in the i row turns downward at the j column, goes downward in the j column, an
412
Chap . .9 Real Orthogonal Groups
leaves from the Young pattern at the bottom of the j column. The inver hook path (i, j) is the same path as the hook path (i, j) but with t opposite direction. The number of boxes contained in the path (i, j), well as in its inverse, is the hook number h ij . Yf~AI is a tableau of the Youn pattern [,.\] where the box in the jth column of the ith row is filled with t hook number h ij . Define a series of the tableaux y;.gl recursively by t
rule given below. yfl is a tableau of the Young pattern [,.\] where each b is filled with the sum of the digits which are respectively filled in the sam l in the series. The symbol y;.j,>l means the produ box of each tableau
yt of the filled digits in it, so does the symbol yFI. The tableaux yt are defined by the following rule: l
(a) yfol is a tableau of the Young pattern [,.\] where the box in the jth column of the ith row is filled with the digit (N+ji). (b) Let [,.\(1)] = ["\]. Beginning with [,.\(1)], one defines recursively the Young pattern [,.\(a)] by removing the first row and the first column of the Young pat.tern [/\(a1)] until [,.\(a)] contains less than two columns.
yt
l to (c) If [,.\(a)] contains more than one column, define be a tableau of the Young pattern [,.\] where the boxes in the first. (a  1) row and in the first (a  1) column are filled with 0, and the remaining part of the Young pattern is nothing but [,.\(a)]. Let [,.\(a)] have r rows . Fill the first r boxes along the hook path (1, 1) of the Young pattern [,.\(a)], beginning with the box on the rightmost, with the digits (,.\i a )  1), (,.\~a)  1), "') (,.\~a)  1), box by box, and fill the first (,.\;a)  1) boxes in each inverse hook path (i, 1) of the Young pattern [,.\(a)], 1 ::; i ::; r, with 1. The remaining boxes are filled with O. If a few 1 are filled in the same box, the digits are summed. The sum of all filled l with a > 0 is O. digits in the pattern
yr
The calculation method (9.27) is explained through some examples. Ex. 1
The dimension of the representation [3 , 3,3] of 80(7).
Tensor Representations oj SO(N)
§9.1
dI 3 ,3,3](SO(7)) = ~~~~
Ex. 2
= 11
X
9
X
7
X
= 1386.
2
The representation of onerow Young pattern [n] of SO(N). I
N
=
I
N+
I
N  1
(N
N
I [ I
I
1
I
I
N
+n
 3
N
+ 2n 
2
n  1
/nl
+ n  3)!(N + 2n  2) (N  2)!n!
2(N + n  3)
+ 2n 
N
N 2
n
(9.28
+1 , = (n + 1)2 , = (n + l)(n + 2)(2n + 3)/6
dl n ](SO(3)) = 2n dl n ](SO(4)) dl n ](SO(5))
Ex. 3
.
The representation of tworow Young pattern [n, m] of SO(N)
N tV + I yTIn, m] = r,;,':':''+";;i't'''+';i;;==:Tt';;';Nf+=~:;'rl'''...L....:.N,+,",,=2 ,"Ne.+c...:",'J ' N 1 N+m 2 N N+m N+~
1
I N N
[ 1
I
 2
1
I
I
N+~
N+m
Y ln,m] h
dln,m](SO(N))
=
I
 1 N
N
J
[ )
I
I
I
m  2
+ Tn + 2m
2 4
I N
+
n
n 7"'+21
_ 
1
N
nm
+n +
I
TTl.
3
N
+
2n
2
I 1
(n  m + l)(N + n  4)!(N + m  5)! (n + l)!m!(N  2)I(N  4)! (9.29 X (N+n+m3)(N+2n2)(N+2m4).
d[n,m](SO(4)) = (n  m d[n,m](SO(5))
= (n 
m
+ l)(n + m + I), + l)(n + m + 2)(2n + 3)(2m + 1)/6 .
The factor 2 in the denominator of Eq. (9 .27) for SO( 4) was considered.
414
Chap. 9 Real Orthogonal Groups
For the representation with onecolumn Young pattern [In], there is traceless condition so that when n < N /2, N! n!(N  n)!·
9.1.5
(9.
Adjoint Representation of SO(N)
As shown in §7.6.2, the highest weight of the adjoint representation W2, corresponding to the Young pattern [I, 1,0, ... , OJ. this subsection we are going to discuss the adjoint representation of SO( by replacement of tensors. The N(N  1)/2 generators Tab in the selfrepresentation of SO( construct the complete bases of Ndimensional antisymmetric matric Denote Tab by TA for convenience, 1 ::::; A ::::; N(N  1)/2. Tr (TATB) 26AB . For R ESO(N), one has from Eq. (7.8)
SO(N) is M =
N(Nl)/2
RTAR J
=
z=
TBDjlA(R).
(9.
B+l
The antisymmetric tensor Tab of rank 2 of SO(N) satisfies a similar relat in the SO(N) transformation R
(ORT)ab =
z=cd RacTcd (R1)db
= (RTR1)ab·
Tab can be looked like an antisymmetric matrix and expanded with resp to (TA)ab N(Nl)/2
Tab =
z=
(9.
(TA)ab FA,
A=!
where the coefficient FA is a tensor which transforms in the SO(N) tra formation R as
(ORT)ab =
(RTR1)ab
=
z=
(RTAR1)ab FA
A
; ; (TB)ab { (ORT)ab =
~
D'BdA(R)FA} ,
z= (TB)abORFB . B
I,
§9.1
Tensor Representations of SO(N)
'll 5
Thus, FA transforms according to the adjoint representation of 80(N)
(ORF)B =
L
DElA(R)FA'
(9.33)
A
The adjoint representation of 80(N) is equivalent to the antisymmetric tensor representation [1,1] of rank 2. [1,1]:::: [1] for 80(3). The adjoint representation of SO(N) where N = 3 or N > 4 is irreducible so that SO(N), except for N = 2 and 4, is a simple Lie group. The 80(2) group is Abelian. The adjoint representation of 80(4) is reducible, and the direct product of two 8U(2) is homomorphic onto 80(4) through a twotoone correspondence (see §9.5). 9.1.6
Tensor Representations of O(N)
The group O(N) is a mixed Lie group, whose group space falls into two disjoint regions corresponding to det R = 1 and det R = l. Its invariant subgroup 80(N) has a connected group space corresponding to det R = 1. The set of elements related to the other connected piece where det R = 1 is the coset of 80(N). The property of O(N) can be characterized completely by 80(N) and a representative element in the coset. For an odd N, the representative element in the coset is usually chosen to be a = 1. a is selfinverse and commutable with every element in O(N) so that the representation matrix D(a) in an irreducible representation of O(N) is a constant matrix
D(a) = el,
D(a)2
= 1,
c = ±1.
(9.34)
Denote by R the element in SO(N) and by R' = aR the element in the coset. From each irreducible representation DP,j (SO(N)) one obtains two induced irreducible representations D[A]± (O(N)), (9.35)
Two representations DIAj±(O(N)) are inequivalent because the characters of a in two representations are different. For an even N = 2£, a = 1 belongs to SO(N), and the representative element in the coset is usually chosen to be T. T is a diagonal matrix where the diagonal entries are 1 except for TNN = 1. T2 = 1, but T is not commutable with some elements in O(N). Any tensor Young tableau yg,jOa, "an is an eigentensor of T with the eigenvalue lor 1 depending on whether there are even or odd number of filled digits N in the tableau. In
416
Chap. 9 Real Ortho90nal Groups
the spherical harmonic basis tensors, T interchanges the filled digits g a (e + 1) in the tensor Young tableau yL"']¢al .. an· Thus, the representat matrix D[>'] (T) is known. Denote by R the element in SO(2£) and by R' = TR the element in coset. From each irreducible representation DI\] (SO(2£)) where the number of [AJ is less than £, one obtains two induced irreducible repres tations DI\]±(0(2£)),
(9.
Two representations DI\]±(O(N)) are inequivalent because the charact of T in two representations are different. When the row number of [AJ is £ = N/2, there are two inequival irreducible representations DI(±)\1 of SO(2£) whose basis tensors are gi in Eq. (9.9). Two terms in Eq. (9.9) contain different numbers of subscripts N such that T changes the tensor Young tableau in [(±),\J that in [(=r=)A], namely, t he representation spaces of both DI (±)\](SO(2 are not invariant in 0(2£). Only their direct sum is an invariant sp corresponding to an irreducible representation D I\1 of 0(2£),
DI\](R) = DI(+)\](R) EEl Di()\](R),
DI\](TR) = DI\I(T)DI\](R),
(9. where the representation matrix DI\1 (T) can be calculated by interchang the filled digits £ and (e + 1) in the tensor Young tableau yL\I¢a, .an ( Eqs. (9.25) and (9.26)). Two representations with different signs of DI\I are equivalent because they can be related by a similarity transformat
X (~ ~1). =
9.2
r
Matrix Groups
Dirac introduced four " matrices, which are the generalization of the Pa matrices. Similar to the Pauli matrices, four 'Y matrices also satisfy anticommutative relations. In terms of the tool of 'Y matrices, Dirac tablished the equation of motion for the relativistic particle with spin 1 called the Dirac equation. In the language of group theory, Dirac fou the spinor representation of the Lorentz group. The tool of'Y matrice generalized for finding the spinor representations of SO(N) in this secti The set of products of the 'Y matrices forms the matrix group f. Its gro a lgebra is called the Clifford algebra in mat.hemat ics .
r
§9.2
9.2.1
Matrix Groups
417
Property of r Matrix Groups
Define N matrices la satisfying the anticommutative relations { la , Ib } = lalb
+ Ibla
1:::; a, b:::; N,
= 20ab1,
. (9.38)
namely, I~ = 1 and In.lb = Ibla when a ::f: b. The inverse of a product of la matrices is the same product but in the opposite order . The set of all products of the la matrices, in the multiplication rule of matrices, satisfies four axioms of a group and forms a group, denoted by r N. In a product of la matrices two la with the same subscript can be moved together and eliminated by Eq . (9 .38) so that f N is a finite matrix group. Choose a faithful irreducible unitary representation of f N to be its selfrepresentation. As shown in Eq . (9.49), the representation does exist. Thus, from Eq. (9.38) la is unitary and Hermitian, ",t la
= 'VI la
= ~I/(1 '
(9.39)
The eigenvalue of I'a is 1 or l. Let (9.40)
When N is odd, liN) is commutable with every la matrix so that it is a constant matrix owing to the Schur theorem, (N) _
Ix.

{±1
when N when N
±i 1
= 4m + 1, = 4m  1.
(9.41)
Namely, 1~4m+l) is equal to either 1 or 1 , which is not a new element f 4m+l. T wo groups WI'th d'ff I erent IX.(4m+l) are isomorp h ic t h rough a onetoone correspondence, say
. In
1:::;
a:::; 4m,
can b e expresse d as a pro d uct of other la matrices. Thus, all elements both in f 4m and in f 4m+ I can be expressed as the products of matrices I'a, I :::; a :::; 4m so that they are isomorphic. On the other hand, since 4m  I ) is equal to either i1 or iI, f 4m  1 is isomorphic onto a group composed of f 4m  2 and 'if 4m  2 , 'D.
!'Lift
hermore, for a given . IX.(4m+l) ,
(9.42) (4m+l)
14m+l
11
f 4m 
1 ~
{f 4m 
2,
if 4m 
Z}.
(9.43)
418
9.2.2
Chap. 9 Real Orthogonal Groups
The Case N
= 2.e
First, calculate the order g(2£) of ru. Obviously, if R E r2{, R belon too. Choosing one element in each pair of elements ±R, one obtains a r;e containing g(2t) /2 elements. Denote by Sn a product of n different Since the number of different Sn contained in the set r;e is equal to combinatorics of n among 2£, then
(9.4
Second, for any element Sn E r zt except for ±1, one is able to f a matrix la which is anticommutable with Sn. In fact, if n is even a la appears in the product Sn, one has A(aSn = Snla' If n is odd, th must exist at least one la which does not appear in the product Sn so th laSn = Snla' Thus, the trace of Sn is 0,
Namely, the character of the element S in the selfrepresentation of X(S)
={
±d(U)
0
where d(2e) is the dimension of irreducible,
2 (d(2t)
la'
f=L
when S = ±1, when S oj:. ±1,
r 2t
(9.4
Since the selfrepresentation of rzt
Ix(SW = g(2t) = 22l+1 ,
SEf 2l
(9.4
Due to Eqs. (9.39) and (9.45), det I'a = 1 when £ > l. Third, since is anticommutable with every la, one is able to def l,jU) by multiplying 1,~2e) with a factor such that lye) satisfies Eq. (9.3
,re)
( IJ(2f))2
lYe)
= 1.
(9.4
In fact, may be defined to be the matrix 12t+l in r 2 (+I' Fourth, the matrices in the set r~t are linearly independent. Otherwi there is a linear relation Ls C(S)S = 0, where S E r;e' Multiplying it w R 1 / d(2l) and taking the trace, one obtains that any coefficient C(R) = Thus, the set r;e contains 22t linearly independent matrices of dimensi
§9.2
r
Matrix Groups
419
= 2f so that they constitute a complete set of basis matrices. Any matrix M of dimension d(2i) can be expanded with respect to S E f~e,
d(2f)
M
=
L
C(S)S,
C(S)
=
1 Tr (SM. I) dIU)
(9.48)
sEr~l
Fifth, due to Eq. (9.38), ±S construct a class. 1 and 1 construct two classes, respectively. The f2l group contains 2u + 1 classes. It is a onedimensional representation that arbitrarily chosen n matrices la correspond to 1 and the remaining matrices Tb correspond to 1. The number of the onedimensional inequivalent representations is
The remaining irreducible representation of fu has to be d( 2tL dimensional which is faithful. The fa matrices in the representation is called the irreducible la matrices. The irreducible la matrices may be chosen as follows [Georgi (1982)J. Expressed la as a direct product of e twodimensiona matrices, which are the Pauli matrices (5a and the unit matrix 1: T2nl
=
T2n
=
(2l)
If
1 x ... x 1
X (51 X (53 X .. . X (53,
'v' n]
'v" in
1 x ... x 1
X(52 X (53 X . .. X (53,
'v' nl
'v" in
(9.49)
(53 X ... X (53 .
'v"
e
Since I?l) is diagonal, the forms in Eq. (9.49) are called the reduced spinor representation. Remind that the eigenvalues ±1 are arranged mixed in the diagonal line of T}21) At last, there is an equivalent theorem for the fa matrices .
Theorem 9.1 (The equivalent theorem) Two sets of d(2l)dimensional matrices la and ;Ya' both of which satisfy the anticommutative relation (9.38) where N = 2f1., are equivalent 1 Sa
S 2£.
(9.50)
The similarity transformation matrix X is determined up to a constant factor. If the determinant of X is restricted to be 1, there are d(2l) choices for the fact.or: exp (i2mr /d(U)), 0 S n < d(2l).
Chap. 9 Real Ortho90nal Groups
420
Proof The irreducible representations of ru constructed by two sets of matrices la are equivalent because the characters of any element 5 E r 2l are equal to each other (see Eq. (9.45)). Assume that there are two similarity transformation matrices X and Y,
Thus, Y XI can commute with every ~(a so that Y = cX. 0 Being an important application of the equivalent theorem, the charge conjugation matrix C(U) used in particle physics is defined based on the theorem. From irreducible unitary matrices la satisfying the anticommutative relation (9.38), define "fa =  ("fa) T, where T denotes the transpose of the matrix . "fa also satisfy Eq. (9.38) so that "fa are equivalent to la, ( C(2l)) t C(U) = 1,
(C (2f))
I
detC(U) = l. (9.51)
If(2f)C(2f) _ (_ ~.)f (II )T ()T 12 ... (12i )T _ (_1)£ ( If(2l))
T
(9.52) Taking the transpose of Eq. (9.51), one has la =  (C(2f))T ,
= [(C(2e))T
!
[(C(2t))lf
(C(2t)r 1 ]
la [(C(2f))T
(C(2f))lr l
Thus, (C(2£))T (C(2t))  l = A(2£)1, (C(2f))T = A(2e)C(2t) , and C(2t)
= A(2t)
(
C(2t))
T
= (A (20)
2
C(U),
A(2t)
= ±l.
The constant A(2f) can be determined as follows. Remind that r;e is a complete set of d(2t)dimensional matrices and is composed of Sn, 0 :S n :S 2£, where Sn is a product of n different la matrices. Since
is either symmetric or antisymmetric. The number of Sn as well as SnC(2f) is the combinatorics of n among 2£. Because the number of the symmetric matrices of dimension d(2C) is larger than that of the antisymmetric ones, SeC(2t) has to be symmetric, SnC(2t)
r
§9.2
T
( C(2e))
Matrix Groups
421
= A(2e) C(2f) ,
(9.54
The charge conjugation matrix C(21!) satisfies Eqs. (9.51) and (9.54). In the reduced spinor representation (9.49), one has C (4m) = (at x
a2 )
x (al x
a2 )
x ... x (at x
a2 ) ,
'
c(4m+2) _ a2
x
c(4m).
~
m
In the particle physics, the strong spacetime reflection matrix used,
B( 2e l
= IjU)C(2t) =
(B(U)) 1/a B (2e) 9.2.3
The Case N
(_1)i/\(2f)
= (fa)T ,
(B(2e))T,
detB(2f)
(B(2£)) t B( 2e l
B(2£)
(9.55 is also
=1
(9.56
= 1.
= 2£ + 1
Since Ij2£) and (2e) matrices la in f 2e , 1 ::; a ::; 2£, satisfy the antisym metric relations (9.38), they can be defined to be the (2£ + 1) matrices la in f 2 £+1. In this definition, 1~2e+ll in f U+ l has been chosen,
'x
(2f)
121.+1
Obviously, the dimension in f 2e ,

(2 f + l) _
='/ '
d(2 f+ l)
_
II .. ·/2£+1 
(·){1 l
.
(9.57
of the matrices in f 2 f+ 1 is the same a
d(2f)
(9.58
When N is odd, the equivalent theorem has to be modified because the multiplication rule of elements in f2t+l includes Eq. (9.41). A similarity transformation cannot change the sign of 1~2t+l). Namely, the equivalen condition for two sets of la and 'Ya has to include a new condition IX = "Ix in addition to those given in Theorem 9.l. T Letting "Ia =  (fa) ,one has
'x
_(Z f + l)
_
_
= 11·· ·/2£+1 = 
{
12' + 1··· A/I
(_1) £+ 1 {/~2e+I)} T =
}T
(_l)e+I/~2e+I),
(9.59
namely, C(4ml) satisfying Eq. (9.51) exists, but C(4m+l) does not. In the same reason, B(4m+l) satisfying Eq. (9.56) exists, but B(4ml) doe not. In fact, due to Eq. (9.52), 12 £+ 1 = satisfies Eq. (9.51) when N 'c 4m  1, but does not when N = 4m + 1. Thus,
I?£J
422
Chap. 9 Real Orthogonal Groups
c C4ml)
B(4m+l)
= C(4m2), = B(4m) = Ij4m) C(4m),
(CC 4m 
1) {
=
(_1)mcC4 mI),
(BC4m+I))T = ( _ 1)mB C4m+1).
(9.6 Note that cC4mI), BC4m+l), cC4m2) and C C4m ) have the same symme in transpose.
9.3
Spinor Representations of SO(N)
Covering Groups of SO(N)
9.3.1
From a set of N irreducible unitary matrices la satisfying the anticomm tative relation (9.38), define N
'Ya =
L
R E SO(N).
Rab/b,
(9.6
b=l
Since R is a real orthogonal matrix, 'Ya satisfy 'Ya'Yb
+ 'Yb'Ya =
L
RacRbd hCld
+ Idlc} =
2
L
Rac R bc 1
= 2Jab l.
cd
Due to Eq. (9.38) and
I:a RlaR2a = 0, 1
L
Rial R2a2Ia"a2 =
2" ~
a,a2
RIal R 2a2 (ralla2  la2lal) ,
alia2
'YI'Y2· · ·'YN=
L L
Rlal · · ·RNaNlalla2'·"aN
al .. ·aN
Rial'·' RNaN ca, .. aNI112 .. 'IN
From the equivalent theorem, la and 'Ya can be related through a unita similarity transformation D(R) with determinant 1, N
D(R)I,o,D(R) =
L
det D(R)
Radld ,
= 1,
(9.6
d=l
where D(R) is determined up to a constant exp (i2mr/d CN )) ,
o ::; n < d(N).
(9.6
The set of D(R) defined in Eq. (9.62), in the multiplication rule of mat ces, satisfies four axioms of a group and forms a Lie group G~v. There is
Spinor Representations of so (N)
§.9.3
d(NLtoone correspondence between the elements in G~ and the element
in SO(N), and the correspondence is invariant in the multiplication of ele ments. Therefore, G~, is homomorphic onto SO(N). Since the group spac of SO(N) is doublyconnected, its covering group is homomorphic onto i by a twotoone correspondence. Thus, the group space of G~ must fal into several disjoint pieces, where the piece containing the identical elemen E forms an invariant subgroup G N of G~. GN is a connected Lie group and is the covering group of SO(N). Since the group space of GN is con nected, based on the property of the infinitesimal elements, a discontinuou condition will be found to pick up G N from G~. Let R be an infinitesimal element. Expand Rand D(R) with respec to the infinitesimal parameters Wab,
L Wab (Tab)ed = Oed a
Red = Oed  i D(R)
=1
Wed,
a
where Tab are the generators in the selfrepresentation of SO(N) given in Eq. (7.85) and Sab are the generators in G N . From Eq. (9.62) one has
be,
Sab] =
L
(Tab) ed I'd = i {Oael'b  Obel'a} .
(9.64
d
The solution is
Sab
=
1 4i bal'b  I'na) .
(9.65
Sab is Hermitian because D(R) is unitary. For simplifying the notations, define C= Thus, C
I
_
{
when N = (4£ + 1), when N 1 (4£ + 1).
E(N) C(N)
SabC   (Sab)
T _
(9.66
•
 Sab'
C I D(R)C = {D(Rl)} T = D(R)*.
(9.67
The discontinuous condition (9.67) restricts the factor in D(R) (see Eq (9.63)) such that there is a twotoone correspondence between ±D(R) in GN and R in SO(N) through the relations (9.62) and (9.67). Namely, G N is the covering group of SO(N),
(9.68
Chap . .9 Real Orthogonal Groups
424
When N = 3, G 3 ~ SU (2). G N is called the fundamental spinor represe tation, or briefly, the spinor representation, denoted by Dis](SO(N)). S is called the spinor angular momentum operators. The irreducible tens representation [A] is a singlevalued representation of SO(N), but a no faithful one of G N. The faithful representation of G N is a doublevalu onc of SO(N). Since the products Sn span a complete sct of the d(Nldimension matrices, it can be decided by checking the commutative relations of with the generators Sab whether there is a nonconstant matrix commuta with all Sab. The result is that only 'Y~N) is commutable with all S . . . "Ix(2[+1) IS a constant matnx so t h at t h e f un d amenta]spmor representatI D Is](SO(2J!+ 1)) is irreducible. Due to Eqs. (9.67) and (9.60), Disl(SO(N is selfconjugate when N = 2£ + 1 and is real when N = 8k ± 1. 'Y~2t) is not a constant matrix so that the fundamental spinor represe t a tion Disl (SO(2£)) is reducible. Through a similarity transformation is changed to 0"3 X and s](SO(2£)) is reduced into the direct su of two irreducible representations,
'Y?l)
1
X
~l lsi D
DI
. _ (DI+S](R) 0 ) (R)X 0 DI~s](R)·
(9.6
Two representations Di±s](SO(2£)) can be proved to be inequivalent reduction to absurdity. In fact, if Z~IDI~sl(R)Z = Di+sl(R) and Y 1 ffi Z, then all generators (Xy)~ISabXY are commutable with 0"1 x 1, b t.heir product is not commut.able with it,
2e(Xy)~1 (S12S3~ ... S(2 f~ I)(2t)) XY = y ~ 1 [X~I'Yj2f)X] Y = 0"3
X
l
It leads to cont.radiction. Introduce two projective operators P±,
P±
=~
(l±'Y?fl),
P±DISI(R) =DIsI(R)P±,
X~l P+X =
(6 ~) ,
X~l P+Disl (R)X = ( DI+~ (R) ~),
X~I P~X =
0 ~) ,
X~l P~Dlsl(R)X = (~ DI~~I(R))
(9.7
.
From Eq. (9.52) one has
C~ I Dis] (R)P C = {DIS] (R)* P± ±
DIs](R)* P:r::
when N when N
= 4m, = 4m + 2.
(9.7
§9 . .'1
Spinor Representations oj SO(N)
425
Two inequivalent representations DI±s] (R) are conjugate to each other when N = 4m + 2, are selfconjugate when N = 4m, and are real when N = 8k owing to Eq. (9.54). The dimension of the irreducible spinor representations of SO (N) is
(9.72)
9.3.2
Fundamental SpinoTs of SO(N)
In a SO(N) transformation R, I}i is called the fundamental spinor of SO(N) if it transforms by the fundamental spinor representation Dls](R):
(9.73) 1/
where qi is a column matrix with drs] components. The Chevalley bases HJ1.(S) EJ1.(S) and FJ1.(S) with respect to the spinor angular momentum can be obtained from Eqs. (9.15) and (9.21) by replacing Tab with Sab' In the chosen forms of fa given in Eq. (9.49), the Chevalley bases for the SO(2f + 1) group (Be Lie algebra) are 1 HI" (S) = 1 x ... x 1 x  {0'3 '.v'
J1.1
He(S)
=1
2
X
1 1
X
0'3} x 1 x ... xl, '.v'
[J1.1
x ... x 1 X0'3,
'.v'
e1 EJ1.(S) = 1 x ... x 1 X {O'+ x O'_} x 1 x ... x 1 = FJ1.(S)T, '.v' '.v' J1.1 eJ1.1 E t (S), = 0'3 X ... X 0'3 XO'+ = Ft(S) T ,
(9.74)
~
e I
where 1 =:; JL < e. The Chevalley bases for the SO(2£) group (De Lie algebra) are the same as those for SO(2£ + 1) except for JL =
e,
1
He(S) = 1 x .. . xl x {o3 X 1 + 1 x 0'3} , '.v' 2 e2 E e(S) =  1 x ... x 1 x {o+ x O'+} = Fe(S)T.
(9.75)
'.v' [2
The basis spinor x[rn] of SO(N) is also expressed as a direct product of twodimensional basis spinors x( a),
e
(9.76)
426
Chap. 9 Real Orthogonal Groups
(~) ,
x( +) =
x()
=
(~) .
(9.
When N is even, the fundamental spinor space is decomposed into subspaces by the project operators P±, 'l1± = P±'l1, corresponding to i ducible spinor representations D[±s]. The basis spinor in the representat space of DI+s] contains even number of factors x( ), and that of D contains odd number of x( ). The highest weight states X[M] and th highest weights Mare
[s] of SO(2€ + 1),
M=(O, ... ,O,I),
x( +) ... X( +) X( +),
'v"
'v"'
£1
£1
M = (0, ... ,0,0, 1),
X(+)···X(+)X(+),
[+s] of SO(2€),
'v"
'v"' iJ
(9.
f2
M = (0, ... ,0, 1,0),
X(+)·· ·X(+)X(),
[s] of SO(2€).
'v"
'v"' l J
£2
The remaining basis states can be calculated by the applications of lower operators F," (5).
9.3.3
Direct Products of Spinor Representations
The spinor representation is unitary so that
(9.
.
'l1t'l1 =
L
f.L OR ('l1t'l1)
'l1:'l1f.L =
=
L
'l1:0f.L v'l1 v, f.LV 'l1 t D ls](R)1 Dis] (R)'l1 = 'l1t'l1.
(9.
'l1*'l1 is invariant in the SO(N) transformations. It is a scalar of SO(N). the language of group theory, the products of 'l1: and 'l1 v span an invari space, corresponding to the direct product representation Dls]* x Dis] SO(N). In the reduction of Dis], x Dis] there is an identical representat where the ClebschGordan coefficients are O/tv. Generally,
L bl
R albl
···
Ra"b n 'l1 Lfbl
.. ·"b"
'l1.
(9.
.. . b"
'l1 Lfn, .. . /a" 'l1 is an antisymmetric tensor of rank n of SO(N) correspo ing to the Young pattern [In]. n is obviously not larger than N, otherw
§9.3
Spinor Representations of SO(N)
1.27
the repetitive la can be moved together and eliminated. When N = 2e + 1, I;U+l) is a constant matrix so that the product of (N  n) matrices la can be changed to a product of n matrices la' Thus, the rank n of the tensor (9.81) is less than N/2, and the ClebschGordan series is
[s]* x [s]::: [s] x [s]::: [0] EB [1] EB [12] EB . . . EB [It],
for SO(2e+ 1). (9.82)
The matrix entries of the product of la are the ClebschGordan coefficients . The highest weight in the product space is M = (0, ... ,0,2), corresponding to the representation [If]. When N = 2e, due to the property of the projective operators P±,
P+P_
= P_P+ = 0,
(9.83)
P~lal .. ·la.2m P± = 0,
the product of (N  n) matrices la can still be changed to a product. of n . matnces la' If n  ,one h as /112·· 'Il  (  Ii)l 12e12tl'" ~11+11/(2t)
e
1112··
"e P ± =
~
bI12"
'Ie ± (i)l'2e1U_I" ' ,HI) P±.
(9.84)
If N = 4m,
[±s]* x [±s]::: [±s] x [±s]::: [0] EB [12] EB [14] EB . .. EB [(±)12m],
['fs]* x [±s]::: ['fs] x [±s] ::: [1] EB [1 3] EB [1 5] EB ... EB [1 2m If N = 4m
(9.85) I ].
+ 2,
[±s]* X [±s]::: ['fs] X [±s]::: [0] EB [12] EB [14] EI7 . .. EI7 [12m] ['fs]* X [±s]::: [±s] X [±s]::: [1] EI7 [1 3] EB [1 5] EI7 ... EI7 [(±)12m+J].
(9.86)
The selfdual and antiselfdual representations occur in the reduction of the direct product [±s] X [±s], but not in the reduction of [+s] X [s]. The highest weights are M = (0, ... ,0,0,2) in the product space [+s] x [+s], M = (0, ... ,0,2,0) in the product space [s] x [8], and M = (0, ... ,0, I, 1) in the product space [+s] X [s]. 9.3.4
Spinor Representations of Higher Ranks
In the 80(3) group, D 1/ 2 is the fundamental spinor representation, and the spinor representations Dj of higher ranks can be obtained from the reduction of the direct product of the fundamental spinor representation and a tensor representation,
428
Chap. 9 Real Orthogonal Groups
(9
The spinor representations of higher ranks of SO(N) can be obtaine the same way. A spinor \]I alan with the tensor indices is called a spintensor transforms in R E SO(N) as follows
(9 b I·· .b n
The tensor part of the spintensor can be decomposed like a ten Namely, the tensor is decomposed into a direct sum of the traceless ten with different ranks and each traceless tensor subspace can be reduce the projection of the Young operators. Thus , the reduced subspace o traceless tensor part of the spintensor is denoted by a Young patter or [(±)A] where the row number of [AI is not larger than N/2. How this subspace of the spin  tensor corresponds to the direct product o fundamental spin or representation [s] and the irreducible tensor repre tation [AJ or [(±)AJ, and it is still reducible. This is the generalizatio Eq. (9.87). It is required to find a new restriction to pick up the irredu subspace like the subspace of D t +!/2 in Eq. (9.87) for SO(3). The res tion comes from the socalled trace of the second kind of the spinte which is invariant in the SO(N) transformations: N
~al . .. ai_lai+I ... an ==
L
')'bwal.
Uil
ba i +I
... a
n'
b=!
RalDI ... Ranbn bl
. . . bn
b'
[~"bRW]
Di s)(R) \]I bI . b, _ I b' b, + I ... b
n
L
Ralb l ... R anonD[s)(R) [L'Yb'\]Ib l .. b'_1 0'O'+1 .. b
b, ... bn
=
n ]
~
L
R alb1 ··· Ranb n D s)(R)<[>b 1 .b'_l b'+lbn· 1
b 1 ·.·b n
The irreducible subspace of SO(N) contained in the sp intensor spac addition to the projection of a Young operator , satisfies the usual trac conditions of tensors and the traceless conditions of the second kind:
L b
'l/Ja· b.b ... c = 0,
L
'Yb'l/Ja·b··c
= O.
(9
b
The highest weight M of the irreducible representation is the highest w
§9.3
Spinol' Representations of SO(N)
429
in the direct product space. The irreducible representation is denoted by [s, A] for 80(2£ + 1)
[8] x [A] ::: [s, A] EB .. . M = [(AI  A2)"'" (AtI  Ae), (2Ae + 1)],
(9.90)
and [±8, A] for 80(2£),
[+8] M
X
[A] or [+8]
= [(AI
X
[(+)A]::: [+8,A]EB ...
 A2),··., (Ael  Ae), (AeI
+ Ae + 1)],
[8] X [A] or [8] x [(  )A] ::: [8, A] EB ... M = [(AI  A2), . .. , (AeI + Ae + 1), (AeI  At)],
(9.91)
[+8] x [(  )A] ::: [8, AI , A2,"" AtI, (At  1)] EB .. . M
= [(AI 
A2), ... , (AtI
+ Ae), (AeI  At + 1)],
[8] x [( +) A] ::: [+8, AI, A2 , ... , AeI' (Ae  1)] EB . .. M = [(AI  A2), " " (AtI  At + 1), (AtI + At)].
These irreducible representations [8, A] of 80(2£ + 1) and [±8, A] of 80(2£) are called the spinor representations of higher ranks. Remind that the row number of the Young pattern [A] in the spinor representation of higher rank is not larger than otherwise the space is null. For example, in the space o the spinor representation [8,1 n] of 80(2£ + 1), the number of spin  tensors before taking the traceless conditions is d(U+I) times the combinatorics of n among (2£+ 1). There is no traceless condition of the first kind. The number of traceless conditions of the second kind is d(2t+ 1) times the combinatorics of (n  1) among (2£ + 1). Thus , the space is null when n > £ because the number of the traceless conditions is not less than the number of tensors . The remaining representations in the Clebsch Gordan series (9.90) and (9.91) are calculated by the method of dominant weight diagram. Fo example, when [A] is a onerow Young diagram, one has
e,
50(2£
+ 1):
50(2£):
[8] X [A, 0, ... ,0] ::: [8, A, 0, ... ,0] EB [8, A  1,0, ... ,0]' [±s] x [A, 0, ... ,0] ::: [±8, A, 0, ... ,0] EB [=f8, A  1,0, ... ,0].
(9.92) [=fs, AI , 0, ... ,0] ap pears in the second reduction because the factor i'b in Eq. (9.89) is anticommutable with "If in P±.
430
Chap. 9 Real Orthogonal Groups
9.3.5
Dimensions of the Spinor Representations
The dimension of a spinor representation [s, A] of SO(2£ + 1) or [±s of SO(2£) can be calculated by the hook rule [Dai (1983)]. In this the dimension is expressed as a quotient multiplied with the dimens of the fundamental spinor representation, where the numerator and denominator of the quotient are denoted by the symbols yyl and y respecti vely:
(9
The concepts of a hook path (i, j) and an inverse hook path (i, j) discussed in §9.1.4. The number of boxes contained in the hook path (i is the hook number h ij of the box in the jth column of the ith row. yr~ a tableau of the Young pattern [A] where the box in the jth column of ith row is filled with the hook number h ij . Define a series of the table yJ:I recursively by the rule given below. yJAI is a tableau of the Yo pattern [A] where each box is filled with the sum of the digits which respectively filled in the same box of each tableau yJ:l in the series.
symbol yyl means the product of the filled digits in it, so does the sym y:[Al h
.
The tableaux yJ:1 are defined by the following rule: (a) YJ~1 is a tableau of the Young pattern [A] where the box in the jth column of the ith row is filled with the digit (N  1 + j  i). (b) vet [A(1)] = [A]. Beginning with [A(1)], we define recursively the Young pattern p, (a)] by removing the first row and the first column of the Young pattern [A(al)] until [A(a)] contains less than two rows.
ltl
to be (c) If [A(a)] contains more than one row, define a tableau of the Young pattern [A] where the boxes in the first (aI) row and in the first (aI) column are filled with 0, and the remaining part of the Young pattern is nothing but [A(o)]. Let [A(a)] have r rows. Fill the first (r 1) boxes along the hook path (1, 1) of the Young pattern [A(a)], beginning with the box on the rightmost, with the digits A~a) , A~a), ... , A~a), box by box, and fill the first A;a) boxes in
§9.:J
Spinor Representations of SO(N)
;1:3
each inverse hook path (i, 1) of the Young pattern [>,(a)], 2 ::; i ::; r, with 1. The remaining boxes are filled with O. If a few 1 are filled in the same box, the digits are summed. The sum of all filled digits in the pattern Yl~] with a > 0 is O. Ex. 1 Dimension of the representation [+s, 3, 3, 3] of SO(8).
Y5[3,3,3]
_ 
Y h[3,3,3]
_ 
7 6
8 7
5
6
5 4 3
4 3
7 5 3
9 8 7
11 6 4
12 10 5
3 2 1
2
d[+s(3,3,3))(SO(8)) = 23 x 11 x 7
X
52 = 15400.
Ex. 2 Dimension of the representations [±s,n] of SO(2£) and [s,n] o SO(2£ + 1). [n] _
Ys
I
y~n]
=I n I n  1 I .. . I 1 1= nl,
d
N  1 [ N [ ... [ N
+n
_
 2
1
(N
+n 
2)1
(N _ 2)!
'
(SO(2£)) = 2[1 (2£ + n  2)! = 2£1 (2£ + n  2) [±s,n] n!(2£  2)! n' (SO(2£
d [s,n]
+
1)) = 2£ (2£ + n  I)! = 2l (2£ + n n!(2£  I)! n
(9.94
1) .
Ex. 3 Dimension of the representations [±s, In] of SO(2£) and [s, n of SO(2£ + 1). N 1 N2 y[I"] _ 5 
N N 1
1 n+ 1
N n+2 N  2n + 1
+ N n+l Nn
d[±s,l"](50(2£))
d
1 1
=2
, (50(2£ )) [±s,I'] +1
£1
(2£)1(2£  2n + 1) I( £ )1 ' n. 2  n + 1 .
= 2£
(2£ + 1)!(2£  2n
+ 2).
n!(2£n+2)!
(9.95
432
Chap. 9 Real Orthogonal Groups
Rotational Symmetry in NDimensional Space
9.4
9.4.1
Orbital Angular Momentum Operators
The relations between the rectangular coordinates coordinates rand (}b in an Ndimensional space are
= rcos(}l sin(}2 ... sin(}N_I, X2 = r sin () 1 sin (}2 ... sin eN  I , Xb = r cos eb  I sin eb ... sin () N  I , x N = r cos eN  I ,
and the spher
Xa
XI
3 :::; b :::; N  I,
(9.
N
L
x~
= r2
a= ]
The unit vector along x is usually denoted by of the configuration space is
x/r. The volume elem
NI
N
II
x=
dx"
= rNIdrdO,
a=1
0:::; r <
00,

Jr :::;
dD.
e1 :::; Jr,
=
II
(sin e"t 1 d(}n.,
(9
a=1
0 :::;
eb :::; Jr,
2
< b< N 
1.
The orbital angular momentum operators Ln.b are the generators of transformation operators PR for the scalar function, R ESO(N), N
L
L2 =
L~b'
(9
a
where the natural units are used, Ii = c = 1. From the second Lie theor Lab satisfy the same commutation relations as Eq. (9.12) of the genera • Tab in the selfrepresentation of SO(N).
9.4.2
Spherical Harmonic Functions
The Chevalley bases HI,(L), EJ1.(L), and FJ1.(L) can be obtained from E (9.15) and (9.21) by replacing Tab with Lab. Because N
L"bXd =
L c=1
N
Xe (Tab)ed'
OROd
=
L
OcRcd,
(9
c= 1
the common eigenfunctions X", of HJ1.(L) can be obtained from ¢'" gi in Eq . (9.16) for SO(2£ + 1) and in Eq. (9.22) for SO(2£) by replacing
§.9.4
Rotational Symmetry in IV Dimensional Space
4
basis vector ()a with the rectangular coordinate Xa· The runction y~l (x) is the bital angular operators HJ.L(L) for a
of the o
(9.100
HJ.L where C2 ( ) represental.ion of from Eq. (7.137)
invariant of order 2. ([A]) or C2 ([(±)A])
ble tenso direct
e
C2 ([/\]) =
L
AJ.L(AJ.L + N  2f.J),
= 2e + 1
N
or 2£.
(9.10
J.L=1
Since there is only one coordinate vector x, the representation [A] f the spherical harmonic function y~l (x) has to be the totally symmetr representation the onerow Young ,\,0, ... ,0 C 2 (["\,0,..  2) for SO(N). highe weight state proportional to Xl, ),
t=
+ ,
{ ££, _ 1,
(9.10
Generally, the highest weight state y,tl (xl with [,,\] = [A, 0, ... ,0] and M (A, 0, ... ,0) is (see Prob. 6 of Chap. 9 in [Ma and Gu (2004)])
yl"l(x)=C M
{
N,"
(Xl)" r
(2,\
+ 2£ 
7i
i
[A I
=C
N,)..
{(I)I(Xl+ iX 2)}"
1)1
("+£1)!' f  I)!
r,J2
,
(9.10
N
N
The remailting barmonic function y~l m ca be calculated ng operators FJ.L(L) I I"y~l(x is called the hannonic polynomial which is a homogeneous polynomial degree A with respect to the rectangular coordinates Xa and satisfies th Laplace equation
\7;'
=
N
82
a=l
8x~'
L
(9.10
Chap. 9 Real Orthogonal Croups
434
9.4.3
Schrodinger Equation for a Twobody System
An isolated nbody quantum system is invariant in the translatio spacetime and the spatial rotation. After separating the motion o centerofmass (see §4.9), there is only one Jacobi coordinate vector a twobody system in Ndimensions, denoted by Rl == x for simpli Remind that a factor of the square root of mass has been included in (see Eq. (4.217))., The eigenfunction of angular momentum of the sy has to be proportional to the spherical harmonic function y~l (x), w [A] = [A, 0, ... ,0] is a onerow Young pattern
(9.
where ¢1.\J(r) is the radial function. Since the system is spherically s metric, one can discuss the wave function with the highest weight. calculation is simplified by making use of the harmonic polynomials, \7; [¢l.\l(r)y);l(x)]
= \7;
[r.\¢l.\l(r)y);l(x)]
+ 2 \7", {r.\¢l.\l(r)} . \7
= y);l(x)
\7; {r.\¢l.\l(r)}
= yl.\l(x)
{r 1  N ~rNl ~ [r.\¢I.\l(r)]} dr dr
M
+2
:r
{r.\¢l.\l(r)}
{~. \7,y);l(x)}
= yl.\l(x) {~¢I.\l (r) + dr2
M
+2
aYt I(x)
N  2A  1 ~qPl(r) _ A(N .A  2) qPl r dr r2
{Ar ¢I.\l(r) + ~¢l.\l(r)} ~yl.\l(x) dr r M
= yl.\l(x) {~¢I.\l(r) + M
•
dr2
N 1 r
~¢l.\l(r) dr
_ A(N +.A  2) ¢I.\l(r)}. r2
Substituting 1jJ\C) (x) into the Schrodinger equation in the coordinate sy of the centerofmass
(9.
one obtains the radial equation
_~ {~¢I.\l(r) + N 1 ~¢[.\l(r) _ A(N + A  2) ¢[J,] (r)} 2
dr 2
= {E
r dr  V(r)} ¢I.\l(r).
r2
(9.
in NDimensional Space
a Threebody System
9.4.4
After separating centerofmass (see §4.9), Jacobi coordinate vectors in a threebody system. Rl := x describes t massweighted separation from the first particle to the centerofmass the last two particles, and R2 := y describes the massweighted separati of last two particles. There are 3N degrees of freedom for a threebo system, where N degrees of freedom describe the motion of centerofma N(N  1)/2  (N  2)(N  3)/2 = 2N  3 degrees of freedom describe t global rotation of the system, and the remaining three degrees of freedo describe the intrrnal motion Thr internal variables are denoted by 6 x . x, 6 = y which are invariant in the Since there vectors, the angular momentum of the systern tworow Young pattern [ eigenfunction of the [>'1, A2, 0, ... ,01, of the products of yJ:'1 (x) a tum is expressed where [w] and Young patterns and [AJ, contained in [T]. From the calculation preceding subsection, it is convenient to express the eigenfunction of the a l instead of the spheric gular momentum by the harmonic polynomial harmonic function y~l. The ClebschGordan (CG) series of [w] x [T] of SO(N) consists of tw parts. One is calculated by the LittlewoodRichardson rule, just like t reduction of [w] x [T] of SU(N). The other comes from the trace ope ation between two indices belonging to two representations, respective Without loss of asslirnes w 2: T,
Yk
([w] x
T1,1]EEl ... EEl[W,T]
,s], [wJ
X
[T
 1] T
CD([w
[T  1JhR EEl ...
Tjx[ODLREBEB [w+Ts2t,sj. s=O
°
X
TS
t=o
For SO(3), s = or 1 and [A,l] ~ [A,O] For SO(4), the representati with s i 0 is reduced into the direct sum of a selfdual representation a an antiselfdual representation. As far as the independent basis eigenfunctions of the angular mome tum is concerned, one has to exclude the functions in the form (x ·y)F(x,
436
Chap. 9 Real Orthogonal Groups
because the factor (x· y) is an internal variable and can be incorporated in the radial functions. The function, which belongs to a representation wi t of 0 in the reduction (9.108), is nothing but that in the form (x ·y)F(x, y Thus, the independent eigenfunction with the angular momentum [AI, A2] expressed as the combination of y.k'J(x)Y,t) (y), where ([w] x [T])LR co tains the representation [AI, A2], namely, Al + A2 = w + T , A2 ~ wan A2 :::; T. Denoting w by q, one has T = Al + A2  q and A2 ~ q ~ AI. Th Young pattern [AI, A2] appears in the reduction, min{q,(AI +A2 q)}
EB
([q]X[Al+A2q])LR~
[Al+ A2 8,8].
(9.10
8=0
The number of the independent basis eigenfunctions of the angular mome tum [AI, A2] is equal to (AI  A2 + 1). Each independent basis eigenfunctio is related to a tableau of [AI, A2] where q boxes in the left of the first ro are filled with x and the remaining boxes are filled with y. For exampl for [AI, A2] = [8,3] and q = 5, the tableau is
(9.11
Fortunately, due to the spherical symmetry of the system, only the wa function with the highest weight M = (AI  A2, A2, 0, . .. ,0) is needed in th derivation of the radial equation and th e normalization does not matte The wave function with the highest weight is denoted by Q~Al ,A 2J (x, y satisfying
HdL)Q~A"A2J(X,y)
= (A1
H2(L)Q~A' ,A2J(X, y)
= A2Q~AI ,A2J(X, y),
HLI(L)Q~AI'A2 J(X,y)
= Efl(L)Q~AI'A2J(X,y) = 0,
3~
II
~
€,
 A2)Q~A'hl(x,y),
(9.11
1 ~ fJ ~ €.
The partners of Q~I hJ (x, y) can be calculated, if necessary, by the low ering operators Ffl(L) wh ere L ab = Lab(X) + Lab(Y) . Being the highe weight state, Q[,AI ,A 2J(x, y) is proportional to a product where each bloc in the first row of the tableau (9.110) corresponds to a factor Y([11,~,'.. ..,~} an
each block in the second row corresponds to a factor Y(~"10,~, : :~~) . Two box in the same column have to correspond to an antisymmetric combinatio
Rotational Symmetry in NDimensional Space
§9.4
namely, Q[)q
,A2] (x, y) ex:
q
[y[J
,0, ... ,0] (x)] q A2 [y[1 ,0, ... ,0] (y)] A,q (1,0, ... ,0) (1,0, ... ,0)
. {y(l,a, .. ,oJ(x)y[l,a, ... ,oJ (y) _ y[l'O"'O] (X)yll,0,.,0](y)}A2, (l,a, ... ,O)
y[l,o, ... ,a]() (1,0, ... ,0)
X
ex:
(1,I,a, ... ,O)
X

1 
,0,.,0] () y _ (l,a, .. ,a) y ex: 1 
. + ~X2,
Xl
y[1
YI
(1,1,0, ... ,0)
(1,0, ... ,0)
yil,a,.,a] () X . (1,1,0, ... ,0) x ex: 2  X3 + ~X4, yll,a, ,OJ () y _ . (1,1,0, ... ,0) y ex: 2  Y3 + 2Y4·
.
+ 1Y2,
Introducing a factor [(q  A2)!('>Q  q)!]l only for simplification in th derivation of the radial equation, one obtains
(9.11
Q~Al ,A2J (x, y) is a homogeneous polynomial of degree q and degree (AI A2 q) with respect to the components of x and y, respectively, and satisfi Eq. (9.111) and the Laplace equations v~Q~A' ,A2] (x, y)
= v;Q~A"A2] (x, y) = Va: . VyQ~A' ,A2] (x, y) = O.
(9.11
For SO(3), A2 in Eq. (9.112) has to be equal to 0 or 1, and X 2 = X3, Y2 = Y For SO( 4), the representation [AI, A2] is reduced into the direct sum a selfdual representation r(+ )Al ,A2] and an antiselfdual representatio and [()A1,A2]' QU+)A"A21(X,y) has the same form as Eq. (9.112), b QU)A"A2J(X,y) is obtained from Eq. (9.112) by replacing X 2 with X3 X3  iX4 and Y2 with Y, = Y3  iY4' The Schrodinger equation for a threebody system in the coordina system of the centerofmass is
~ {v; + v;} 1P~"A2](X,y)
=
[E
1I(~1,6'~3)11P~I,A21(x,y).
(9.11
Let A,
1P~,A2](X,y) =
2:
¢~A"A2J(6,6,6)Q~A"A2J(x,y).
(9.11
q=A2
The a.ction of the Laplace operator on 1P~"A2](X,y) is divided into thr parts. The first part is its action on the radial functions ¢~Al ,A2J, which ca be calcuJated by replacement of variables (see Eq. (9.117)). The second pa is its action on the basis eigenfunctions Q~A' ,A2] (x, y) which is vanishin
Chap. 9 Real Orthogonal Groups
438
The third part is its mixed action
2{(8~1 ¢~1'),2])
2x
+
+ 2 {(8~2 ¢~),1'),2])
(8~3 ¢~),1'),2]) 2y
+
y}.
(8~3 ¢~),1 '),2] )
V'",Q~),1'),2](X,y)
x} . V'yQ~),1'),2](X,y).
From Eq. (9.112) one has X. V'"Q~),I'),2] = qQ~),I'),2 ] , Y . V' yQ~),1 ,),2] = (AI
+ A2 _
q)Q~"\1 ,),2],
y. V'"Q~),I'),2] = (A I  q + 1)Q~),':i),2], x· V'yQ~),1 ' ),2] = (q  A2
(9
+ 1)Q~~;),2].
Thus, the general radial equation for the radial function ¢~),l ,),2] is (see et a\. (2001c)]) {V';,
+
V'~} ¢~),l h] + 4q 8~1 ¢~),l ,),21 + 4(Al + A2 
+ 2(A 1

q)~I,[),1'),2] + 2(q 86 'l'q+l
_ A'
2
q) 8~2 ¢~),l ,),21
)~I,[),I ' ), 2 ] 86 'l'ql
= 2 (E  V) ¢~),l ,"\2],
(9
This method can be generalized to a quantum multiplebody system [Gu et a\. (2003b)]) .
9.4.5
Dirac Equation in (N
+ I)dimensional
Spacetime
The transformation between two inertial systems in fourdimens spacetime is the Lorentz transformation. There are two common used of coordinates and metric tensors for the fourdimensional spacetime. is (XO,Xl ,X2,X3) with the Minkowski metric tensor 1] =diag(l, 1, 1 where Xo = ct (see [Bjorken and Drell (1964)]). The other is (Xl,X2,X3 with the Euclidian metric tensor oilV where X4 = ict (see [Schiff (1 Marshak et a\. (1969)]). The Lorentz transformation matrices for two
4
Rotational Symmetry in N Dimensional Space
§9.4
are related by a similarity transformation,
XlAX
X=
(~ ~ ~
~l)
010 001
XI
0 0
=
= A,
(Hn)
(9.11
000
l
In this textbook we adopt the second set and generalize it to the (N + 1 dimensional spacetime. The formula for the first set can be found in [G et al. (2002)]. The Dirac equation in (D + I)dimensional spacetime can be expresse as (see [Schiff (1968)]) N+l
~ I I"
(a
ieAI"
aXil 
)
w(x,t)
+ Mw(x , t)
where M is the mass of the particle, and (N anticommutative relation:
+ 1)
= 0,
(9.11
matrices I I" satisfy th
(9 .12
For simplicity, the natural units Ii. = c = 1 are employed in this subse tion. Discuss the special case where only the time component of A N + 1 nonvanishing and spherically symmetric: Aa = 0
eAN+l = iV(r),
when 1:S: a
:s:
(9.12
N.
The Hamiltonian H(x) of the system is expressed as i
:t
w(x, t) = H(x)w(x, t),
a
N
H(x) =
L
IN+lla
a=l
a
(9 .12
+ V(r)
+'N+1M.
Xa
The orbital angular momentum operator Lab is given in Eq . (9.98). Th spinor operator S ab is given in Eq. (9.65). The total angular momentu operator is ) ab = Lab + Sab' There are three Casimir operators of order for the total, orbital, and spinor wave functions, respectively, N
N
)2
=
L a < b= 2
)~b'
L2
=
L a
N
L;b,
52 =
L a < b=2
5~b '
(9.12
440
Chap. 9 Real Orthogonal Groups
Due to t.he spherical symmetry, Jab is commutable with the Hamilton H(x). There is another conservative operator K which is commutable w both Jab and H(x) (see [Schiff (1968)]) 1
N
K
= i'yN+l
L
la/bLab
a

S2
+ 21'N+l (N  1)
+ (N
 1) /2}
(9.1
.
The set of mutually commutable operators consists of H(x), J2, K, and the Chevalley bases HJ1(J). Their common eigenfunctions are ca lated from the products of the spherical harmonic functions y~l (x) the fundamental basis spinor x( ms) by the Clebsch Gordan coefficie where [>.] = [>. , 0, ... ,0]. Due to the spherical symmetry, the eigenfunc with the highest weight is only interesting for deriving the radial equat Introduce (2f!. + 1) unitary matrices f3a, which are 2£dimensional and sat
(9.1
The concrete forms of f3a are the same as those 1'0. given in Eq. (9.49). (a) The case of N = 2f!.+ 1
Let 1 :::: a :::: 2f!.
The spinor operator
Sab
and the
K
L
/4, N1
N
= i
(9.1
operator become the block matrices
Sab = i (/3a/3b  f3b/3a) K,
+ 1.
/3"/3b L ,,b
+ 2 '
(9.1
a
They are commutable with 1'N+l' The relation between 5 ab and Sa similar to that between the spin or operators for the Dirac spinors and the Pauli spinors. At the level of the Pauli spinors, the fundamental sp x[m] belongs to the fundamental spinor representation [s] with the hig weight (0, ... ,0, 1). Due to Eq. (9.92), there are two sets of the w functions belonging to the representation lJ] == [s, >., 0, ... ,0] of SO(2f!. whose highest weight is M = (>',0, ... ,0,1). Calculating from Eq. (7.1 one has C 2 ([j]) = >.(>. + 2f!.) + f!.(2f!. + 1)/4, C 2 ([>']) = >.(>. + 2f!. 1), C 2 ([s
§9·4
Rotational Symmetry in NDimen sional 8p(,ec
'II
£(2£ + 1)/4, and
+ £ = .\ + £ = IKI, C 2 ([j])  Cz([.\ + 1])  C 2 ([s]) + £ = .\  £ = WI.
C 2 ([j])  C 2 ([.\])

C 2 ([s])
(0.128
Thus, two eigenfunctions of the total angular momentum with the highes weight M of U] and the different eigenvalues of K are calculated by th ClebschGordan coefficients (see Prob . 14 of Chap. 9 in [Ma and Gu (2004)], also see Eq. (4.189)) ¢IKWj(X)
= yg,Jo"O)(x)X[(O, . . . , 0,1)]
= C(2f+I),A ¢I{(J,[jJ(x)
{
(_1)l(xJ + iX2)} A r"f'2 X[(O, .. . , 0,1)]'
= L y~+lj(x)X[M  m]Cl:E~~l,,[jJ,M "Tn
C(2f+I ),( Hl)
(ljlAJ>:+l (
= (r)A+12 A/ 2 J2e + 2,\ + 1 x {X2 f+ 1X[(O, ... ,0,1)]
')A Xj+1.X2
+ (X2f J + 'iX2c)X[(0, ... , 0,1, I)]
+ (X2e3 + iX2f2)x[(0, . .. ,0,1, 1,1)] + ... + (X3 + iX4 )x[(I, I, 0, ... ,0,1)] + (Xl + iX2)x[(I, 0, ... ,0, I)]}
,
(9.129
Both ¢ ±IKJ, [J j(X) are annihilated by every raising operator EIJ.(J). Remind that two coefficients in ¢±JKI,[jl(x) are the same except for a factor r,
(9.130 Introducing two operators N
(3 . X
= r I
L
f3a x a,
(9.131
a=l
one has ((3 . x) 2
= 1,
((3. x) ¢1(,U j(x)
= ¢_I(,[jl(x) ,
442
Chap. 9 Real Orthogonal Groups
((3 . \7) r(Nl) / 2!(r)¢I<,[j] (X)
= ((3. X)2 ((3 . \7) r(Nl)/2f(r)¢I<.lj] (X) = ((3. X)
[:r
+~
L
f3af3b L ab] r  (N  l)/2 f(r)¢K,IJ](X)
(9.1
a
= r(Nl) / 2 [df(r) dr
_ Kf(r)] ¢K [.'](x). r ,J
At the level of the Dirac spinors, the wave function q, I< ,U] (x) with highest weight in the irreducible representation [j] and with the differ eigenvalues K of K can be expressed as
(9.1
Its partners can be calculated from it by the lowering operators Fp. (J) . T factor i in the radial function iGI«r) can be removed by replacing (f2 Eq. (9.126) with (fl. Substituting q, I<,Uj (x, t) into the Dirac equation (9.122) one obtains radial equation
(9 .1
Equation (9 .134) holds for SO(3) although the algebra of SO(3) is AI'
(b) The case of N = 2e with e > 2 The spinor representation of SO(U) is reduced into two fundam tal spinor representations [±,s] with the highest weights (0 , . . . , 0, 1) a (0, . .. ,0 , 1,0), respectively. Let 1
<
a
< 2e.
(9 .1
'YN+l is a diagonal matrix where half of the diagonal elements are eq to + 1 and the remaining to 1. Because the spinor operator Sab and
operator K are commutable with 'YN+l, each of them becomes a direct s of two matrices, acting on the fundamental spinors x±(m), respectively
(9.1
§9. 4
Rotational Symmetry in NDimensional Space
44
For a given total angular momentum there are two kinds of representation of SO(21:') at the Pauli spinor level, [j±] = [±s,.\, 0, ... ,0] . Because o Eq. (9.92), two sets of the wave functions for each of representations [j± are calculated to have different eigenvalues of K.. From Eq. (7 .137) on calculates that C2 ([±s]) = £(2£  1)/4, C 2 ([.\]) = .\(.\ + 2£  2), C 2 (U±]) .\(.\ + 2e  1) + £(2e  1)/4, and
+ e 1/2 = .\ + e 1/2 = IKI, C2([±s]) + £  1/2 = .\  e+ 1/2 = IKI.
C 2 ([j±])  C2(['\])  C2 ([±s]) C 2 ([J±])  C2([.\
+ 1]) 
(9.137 Two wave functions with the highest weight M+ = (.\,0 ... ,0,1) of U+ are
'"
'I'IKI.[j+1
( A)
x
=
"y[A+ I) ( A)
L
17l
X
X [M+ 
Tn
]C[8).["+1)
M + 17l.17l.lj+).M+
17l
(I)(el)"JI+l C(2t).(.>'+I)
.
(1'j2)"+1~ (XI + tX2)"
(9.138
+ iX2e)x[(O, ... , 0, 1,0)] + iX2i2)X  [(0, ... ,0,1, 1,0)] + iX2i4)X  [(0, ... ,0,1, I, 0,1)] + ...
x {(X2i1
+ (X2C3 + (X2f5
+ (X3 + iX4)x_[(I, 1,0, .. . ,0, 1)] + (Xl + iX2)x[(I, o, ... ,O,I)]}.
Two wave functions with the highest weight M_ = (.\,0 ... ,0,1,0) of [j_ are
444
Chap_ 9 Real Orthogonal Croups
+ (X2e3 + iX2e2)X+ [(0, ... ,0,1,0, I)] + (X2e5 + iX2e4)x+ [(0, ... ,0,1, 1,1,0)] + ... + (X3 + iX4)x+[(1, 1,0, ... ,0,1,0)]
(9.13
+ (Xl + iX2)x+ [(I, 0, ... ,0,1,0)]}.
The ClebschGordan coefficients in Eqs. (9.138) and (9.139) are calcula by the condition that the highest weight state is annihilated by every rais operators EJ1. (J). The coefficients can be found in Prob. 14 of Chap. 9 [Ma and Gu (2004)]. Due to Eq. (9.130), Eq. (9.132) still holds for case with N = 2£. At the level of the Dirac spinors, the wave functio 'l11<,fjl (x) with the highest weights in the irreducible representation [ and with the different eigenvalues K of K can be expressed as 'l1 IJ
= rf+ 1/2 e iE t {FlJ
'l1 II< I.Ij _I (x)
= rf+l/2e i Et {F_ \I<\(r)¢_II<j,(ij_I(x) + G_\I
= ±I Klw±\J<j,[J±J (x),
IKI
=
A+
e 1/ 2.
(9.1 Their partners can be calculated from them by the lowering operat
FJ1.(J).
Substituting 'l1 K[j±J(X) into the Dirac equation (9.122) one obtains radial equations, which are the same as Eq. (9.134) . \Nhen D = 4, SO ~ SU(2)xSU(2), and the representations [j±J belong to two differe nt SU groups, respectively. When D = 2, the SO(2) group is an Abelian gro and the radial equation (9.134) holds for SO(2) but K = ±1/2, ±3/2, [Dong et a1. (2003); Dong et. aJ. (1998c)J. The radial equations (9.134) the ordinary differential equations for the variable r, and can be solv easily. Their solutions for the Coulomb potential are given in [Gu a1. (2002)]. The Levinson theorem for the Dirac equation in (N + dimensional spacetime is given in [Gu et a1. (2003e)]. 9.5
The SO(4) Group and the Lorentz Group
The Lorentz group is an important symmetric group in physics. T Lorentz group is a noncom pact Lie group, whose representations can calculated from those of the compact Lie group SO(4). The study of Lorentz group provides a typical example in mathematics for studying
The SO(4) Group and the Lorentz Group
§9.5
noncompact Lie groups.
9.5.1
Irreducible Representations of SO(4)
The Lie algebra of 80(4) is D2 ~ AJ EB A j • In this subsection the 80(4 group is evidently decomposed into a. direct product of two 8U(2) groups Based on this decomposition, the parameters of SO( 4) are chosen and th irreducible representations of SO(4) are calculated analytically. Six generators of SOt 4) in its selfrepresentation are given in Eq. (7.85) Through the suitable combinations one has
0 (±)
Tj
1
1 ( 0 0
= 2: (T23 ± T J4 ) ="2
(9.141
±i
and those by the cycle (1 2 3). Tl ±) can be expressed as
(9.142
The generators are divided into two sets, each of which satisfies the com mutative relations of generators in SU(2), 3
[T"(±) , T(b ± )]  ~. 'L"
cabc
T( ±) c
(9.143
,
c=1
This property becomes clearer if one makes a similarity transformation N N 1 Tl+) N
= (u a/2) N
~
x 12,
;,
NITl) N
= 12
x (aa/2) ,
U~ 1, ~,)
Note that Eq. (9.22) is a generalization of Eq. element R in SO(4) becomes
(9.144
(9.144).
An arbitrar
446
Chap. 9 Real Orthogonal Groups
R =
t
exp (i
wabTab)
a
~ (wi+)T~+) + w~)T~)) }
exp { i
(9.14
exp{iw(+)il,(+) .T(+)} exp{iwHn() 'TH}
.
N {u(n(+), w(+))
X
u(n() ,wH)} Nl,
where W (±)
1
 w

23
±
w
 w(±)n(±)
14 
I'
w~±) = w31 ± w24 = w(±)n~±) ,
w~±)
= W12 ± w34 = w(±)n~±)
,
Thus, R is expressed evidently as the direct product of two unimodu unitary matrices of dimension 2. R keeps invariant if two unitary matric change their signs simultaneously. Namely, Eq. (9.145) gives a onetwo correspondence between the element R in SO(4) and the element u u' in SU(2) x SU(2)', and the correspondence preserves invariant in t multiplication of elements, so that
SO( 4)
~
SU (2) ® SU (2)'.
(9.14
The parameters of the group SO(4) are chosen to be those of SU(2) SU(2)' where the group space of SU(2)' is shortened by half such that the is a onetoone correspondence between the set of parameters a.nd the gro element at least in the region where the measure is not vanishing. Name the group space of SO( 4) is
o < w(+) < 27r ,
o :::; wH
0:::;
7r :::;
e(±) :::;
7r,
:::; 7r,
!p(±) :::;
(9.14
7r,
where e(±) and !p(±) are the polar angle and the azimuthal angle of n(±), spectively. Due to the shortening, two end points of a diameter correspo to one element so that the group space of SO( 4) is doublyconnected R(n(+) ,w(+); n() ,w()) R(n(+) , w(+)·" n()
= R( n(+), (27r 
w(+)); n(),
7r) = R(n(+) (27r  w(+))· 1
n H 7r)
) '
(27r  w()))
•
(9.14
§9.5
44
The SO(4) Group and the Lorentz Group
The covering group of SO(4) is SU(2) x SU(2)'. The weight function fo the group integral of SO(4) is
(87f 4 ) 1
dR =
sin 2 (w(+) /2) sin 2 (w H
/2) sin e(+) sin e()
(9.150 x dw( +) dw H del +) del ) dip (+) dipH.
An irreducible representation of SO( 4) is a direct product of represen tations of two SU(2), DJk (n(+l,w(+l;n(l,wH) = Dj (n(+l,w(+)) x Dk (n(),w(l).
(9.151 Djk is (2j + 1)(2k + I)dimensional. Its row (column) index is denoted b two letters (f..Lv) and its generator I~~ is expressed in terms of I~ of SU (2 I ajk (+) = Ija
X
1 2k+l,
I ajk() = 12J+l
3
Jk  'L~ " I ab 
Ik
a'
3 fabc
(IJk(+) c
+ IJk(l) c
c=l jk I a4
X
 '~ "
fabc
(IJc
X
1 2k+l
1x kc ), +1 2j+l
c=l
I aJk (+)

I ajk ()  IJa
X 1 2k + 1 
12jH X
Ika'
(9.152 The reduction of the direct product of two Djk can be calculated by th CG series for SU (2) such as j1+12
EB
(CJt12)  1 (DhO(R) X DhO(R)) (Cjd2) =
DJo(R),
J=IJ,121 k , +k2
(C kl k2)1 (DOkl(R)
X
D Ok 2(R)) (C kl k2)
=
EB
DOK(R),
(9.153
j,+12
DJI k, (R)
In fact,
X
Ig
DJ2k2 (R)
k (±)
C:::'
EB
are related directly with the Chevalley bases,
H 1  2I3jk (l
El = Iik(l
,
+ iI~k(),
H2 E2
= 2Ii k (+l, = Iik(+l + iIg k (+).
(9.154
448
Chap. 9 Real Ortho90nal Groups
The highest weight in
is M = (2k, 2j), and its Young pattern is
Djk
DJJ ::: [2j,0], Dj k ::: [( Djk:::
D]~: ::: Djk :::
j
+ )(j + k), (j  k)],
= k,
> 0 is integer, k  j > 0 is integer, j  k  1/2 > 0 is integer, k  j  1/2 > 0 is integer. j  k
[()(j + k), (k  j)],
[+s, (j + k  1/2), (j  k  1/2)]' [s, (j + k  1/2), (k  j  1/2)]'
(9.15
The identical representation of SO( 4) is DO 0, and its selfrepresentatio is equivalent to D& ~. The fundamental spinor representation of SO(4) D~ 0 EB DO 1: and its generators (see Eq. (9.49)) are
5 23 = (0"2
X
0"2)
/2, /2,
5 31 =  (0"1
X
5 12 = (0"3
1) /2,
X
5 14 = 5 34
= (1
0"])
(0"1 X
5 24 =  (0"2
0"2)
/2,
O"J) /2,
X
x 0"3)
(9.15
/2.
Through the combination (9.141),5;; are the generators of D~ 0 and DO respectively. In fact, through a similarity transformation X, one has
(9.15
X15+ X = a
9.5.2
(O"a/ 2 0
0) 0
'
Singlevalued Representations of 0(4)
The grou p space of SO (4) falls into two pieces depending on whether the d terminant of the element R is 1 or 1. The piece with det R = 1 construc an invariant subgroup SO(4). As discussed in §9.1.6, the representative e ement in the coset is usually chosen to be T which is a diagonal matr where the diagonal entries are 1 except for T44 = 1. The transformatio rule of generators in the action of T can be calculated from that in th selfrepresentation, namely from the multiplication rule of the element.s 0(4),
where
TTnbT 1 = Tab,
TTa4 T 1
TT~ ± )Tl = T~4'),
T
=
Ta4,
(T(±))2 T 1
=
(9.15 (T('f))2,
419
The SO(4) Group and the Lorentz Group
§.9.5
(9.159)
The tensor representation Djk of SO(4), where j + k is an integer, is singlevalued. Denote by PR the transformation operators of 0(4) for scalar functions and by Lab its generators. Similarly, and (L(±)) 2 can be obtained by Eqs. (9.141) and (9.159). If the basis function rptt belongs to the representation Djk of SO(4),
D;
PH rpjk
IlV
=
(9.160) f.J.1
v'
rpt~ is the common eigenfunction of Li±) and (L(±)) 2,
= V rpjk ,LV' (LH)2 rpjk = k(k + 1) rpjk /LV·
L() rpjk 3 IlV
. j.1./}
In the transformation of
T,
(9.161)
.
PT rpt.~ is still the common eigenfunction of
L~± ) and (L(±)) 2, but the eigenvalues are interchanged between J.L and v and between j(j + 1) and k(k + 1), P
PT
rpjk = (J5kj I
/LV
L(+) (J5kj 3 VJ.1 ( L( +))2
VJ.i.'
= v(J5kj LIM
k(k
vJ.L
=
L() (J5kj 3
VJJ,)
(J5kJ =
(J5kj
+ 1)(J5kj
v~,
VIl
2
rpjk
j.i.V)
= l/(J5kj /"""
.
(LH) (J5~~
vil'
= j(j + 1)(J5~~,
where P; = PE = 1 is used. (J5~~ belongs to the representation Dk) of SO(4). When j i k, neither of the two representation spaces of Djk and Dkj of SO(4) is invariant in 0(4), only t.heir direct sum is invariant. Namely, from two irreducible representations Djk and Dkj of SO( 4) one induces an irreducible representation t"jk of 0(4). In addition to the indices J.L and v, a new index a = ± has to be added for the row (column) indices of t"jk to distinguish the two representation subspaces,
450
Chap. 9 Real Orthogonal Groups
where R E SO(4). The representation by changing the sign of t,.jk(T) lea to an equivalent one. When j = k, define ./,jj± '" IJ!jj 'f/J..1V
J..lV
± pjj j..J.V)
P ./,jj± = ±./,jj± To/~V
'f/Vj..J
.
Thus, 1/Jti,,± span two different spaces both of which are invariant in 0( Namely, from one irreduci ble representation Djj of SO( 4) one induces t inequivalent irreducible representations t,.jj± of 0(4) , t,.jj±(R) = Djj(R) ,
9.5.3
R E SO(4),
The Lorentz Group
The Lorentz transformation A is a transformation between two inert systems in a fourdimensional spacetime (see Eq. (9.118)),
(9.16 The matrix entries of A satisfy the condition
Aab and A44 are real, Aa4 and A4a are imaginary,
a and b = 1, 2, 3.
(9.16
This condition preserves invariant in the product of two Lorentz transf mations. The set of all such orthogonal matrices A, in the multiplicati rule of matrices, constitutes the homogeneous Lorentz group, denoted 0(3,1) or L h · The orthogonal condition (9.164) gives 3
detA = ±1,
A~4 = 1 +
L
I
A a412 2: l.
(9.16
a=l
These two discontinuous constraints divide the group space of Lh into fo disjointed pieces. The elements in the piece to which the identical elem belongs constitute an invariant subgroup Lp of L h , called the proper Lore group. The element in Lp satisfies detA = 1,
(9.16
Since there is no upper limit of A 44 , the group space of Lp is an op region in the Euclidean space, and Lp is a noncompact Lie group. T representative elements in Lp and its three co sets are usually chosen to
§9.5
The 50(4) Group and the Lorentz Group
the identical element E, the space inversion the spacetime inversion p:
0",
the time inversion
= diag (1, 1, 1, 1) E Lt = Lp, detA = 1, 0" = diag (1,  1, 1, l)EL~, detA = 1, detA = 1, T = diag (1, 1, I, l)EL~, P = diag(l, I, 1,  1) E Lt, det A = 1, E
451
T,
and
A44 2: 1, A44 2: 1, A44 :::; 1,
(9.168)
A44 :::; 1.
Four elements constitute the inversion group V 4 of order 4. 9.5.4
Irreducible Representations of Lp
Discuss the generators in the selfrepresentation of Lp. Let A be an infinitesimal element of Lp:
A
=1 
1
= det A = 1 
= 1  iaX T , XT = X, TrX = O. AT
iaX, 1 = AT A = 1  ia (X + XT) , iaTrX,
Thus, X is a traceless antisymmetric matrix. Expand X with respect to the generators Tab in the selfrepresentation of SO(4): 3
A= l  i
3
L
WabTab 
a
1 i'" (0 ~
i
L
W a 4 T a4
a=l
T(+) aa
(9.169)
+ n*TH) aa'
a=1
where
Wab
is real,
Wa4
is imaginary, and 1
Oa
=2
3
L
EabcWbc
+ W a 4·
(9.170)
b,c=l
Except for the imaginary parameters, the generators in the selfrepresentations of SO( 4) and Lp are completely the same, so are the generators in the corresponding irreducible representations of two groups. The finitedimensional inequivalent irreducible representations of Lp are also denoted by Djk(Lp), whose generators are the same as those for SO(4) (see Eq. (9.152))
Chap. 9 Real Orlho90nal Groups
452
3
I~~ = ~
Eabe
{Ii k (+) + Ii k ()}
,
11,:
=
1~k(+) _1~k(),
c=l
1 ajk (+) = I·& X 12k+l ,
Ijk(l a 
(9.
1k 12j+1 X a'
where I~ and I~ are the generators in the representations of SU(2). representations Djk(SO(4)) and Djk(Lp) have the same dimensions are simultaneously singlevalued or doublevalued. The reductions of direct product representations are also the same. However, the global p erties of the two groups are very different because the parameters W Lp are imaginary. The finitedimensional irreducible representation of L p , except for the identical representation, is not unitary. There infinitedimensional unitary representations of Lp [Adams et al. (1987 The transformation generated by Tab is obviously a pure rotation longing to the subgroup SO(3). But now, its transformation matr fourdimensional,
00)
lP  sin lP sin lP cos lP 0 0 0 0 1 0
COS
. R(e3, lP) = exp {2lP T 12} =
(
o
0
(9.
0 1
The transformation generated by T34 with the parameter W34 = iw Lorentz boost along the direction of zaxis with the relative velocity v
A(eJ,iw)
= exp{i(iw)T34} = v = tanhw,
(O~ 10 ~h
.? h )
cos w 2sm w o 0 isinhw coshw coshw = (1 v2)1/2,
(9.
sinhw = v (1 v2)1/2,
where the natural units Ii = c = 1 are employed. In comparison with Lorentz boost in the physical textbook, the transformation matrix beco its inverse because the viewpoint of transformation of the system is here, instead of that of the coordinate frame in the usual physical textb Similar to the Euler angles in SO(3), a new set of parameters of Lp i pected such that each element A of Lp can be expanded as a product of s pure rotations and a Lorentz boost along the zdirection. From a giv matrix in Lp, calculate w from A44 = cosh w. Extracting a factor i si from A a4 , one obtains a unit vector n((),lP) in the threedimensional sp
The 50(4) Group and the L07'eniz Group
§9.5
4
whose rectangular coordinates are (i/ sinh w )(A 14 , A 24 , A34):
= coshw, iA 24 /sinhw = sinBsin
iA 14 /sinhw
= sinBcos
iA34/ sinhw = cosB.
(9.17
Thus, the parameters Band
R(
e"
~
)R(
e"
B)A(
e"
iw)
(~)
{R(e3,
= A(
R(CL,!3, 'Y) = R(e3,CL)R(ez,!3)R(e3,'Y),
(9,17
where
o :S w < 00,
o :S B :S n:,
n::S
n::SCL:Sn:,
O:S!3:Sn:, n::S'Y:Sn:,
(9.17
The geometrical meaning of the expression (9.175) is evident. Two rot tions in the twosides of A( e3, iw) transform two frames before and after t Lorentz transformation A such that two zaxes are changed to the directi of the relative motion, and the remaining axes are changed to be parallel each other, respectively. After two rotations, the Lorentz transformation is simplified into A(e3, iw). The representation matrix of A in Djk (Lp) is easy to be written. D to Eq. (9.170), Da is real for a pure rotation and imaginary for a Loren boost,
Di k (CL,!3,/') = Di(CL,,8,'Y) x D k (CL,!3,'Y), D j k(e3, iw) = exp(wI5) x exp( wI;),
(9.17
where I~ and I~' are diagonal. Thus ,
jk Df.JV (If) B w , 0' ,fJ(.l , 'V) \J..L'v ' y ) ) I X
= ""' e i (J1+v)'Pd jJl..p (B)d kl/T (B)e(pr )w ~
pr j k ei(p+r)o.dPfl' (8)d TV' (B)e i(J1'+v'h. I
I
(9.17
454
Chap. 9 Real Orthogonal Groups
9.5.5
The Covering Group of Lp
For a matrix Lie group G, there exists an exponential mapping of G if every element can be written in an exponential function of matrix where t exponent belongs to its real Lie algebra. Equation (9.145) shows that the exists an exponential mapping of SO(4). Based on Eq. (9.145) the cover group of SO(4) is found. In mathematics, there exists an exponential ma ping of a Lie group G if every element in G belongs to a oneparameter L subgroup. The exponential mapping exists for a compact Lie group, but not necessary to exist for a noncompact Lie group. For example, the set twodimensional unimodular complex matrices, in the multiplication r of matrices, constitutes a group SL(2, C). SL(2, C) is noncom pact and exponential mapping does not exist, because the following elements ca not be written in an exponential form in the condition that the expone belongs to the real Lie algebra of SL(2, C),
(1o 2) =
_eO'J + i0'2
1
'
( 1  i 1 ) 1 1 + i
= _ e(J
I
+i0'3 .
In this subsection we are going to find the covering group of Lp first, a then, to show that every element in Lp can be written in an expone tial function of matrix where the exponent belongs to its real Lie algeb namely, there exists an exponential mapping of Lp although Lp is nonco pact and some elements in Lp cannot be diagonalized. The selfrepresentation of 80(4) can be changed to D~ ~ by the sim larity transformation N (see Eq. (9.145)), so can that of Lp because t generators in the two representations are the same.
A = A(zp,e,w,a.,/3,,) N {u( zp, e, 0) x u( zp, e, O)} {exp (W0'3/2) x exp ( W0'3/2)}
. {u(a.,/3,,) x u(a.,/3,,)} Nl N {M x (0'2M*0'2)}
(9.17
N J ,
M = u(zp,e,O) exp (W0'3/2)U(a.,/3,"Y)
E SL(2,C),
where D 1 j2(a.,/3,,) = u(a. , /3,,) is the element of SU(2). The determina of M is 1 so that M belongs to 8L(2 , C) . Equation (9.179) gives a m from each element A in Lp to at least one element M in 8L(2, C). If o element A in Lp maps two elements M and M' in SL(2, C) through E (9.179), then,
The SO(4) Group and the Lorentz Group
§9.5
4.5.
Namely, M M'l is a constant matrix c1. Since det lvI = det M' = 1, on has c2 = 1. There is a onetotwo correspondence between an element A in Lp and two elements ±M in SL(2, C). The correspondence preserves invari ant in multiplication of elements. Since the orders of Lp and SL(2, C) both are 6, the onetotwo correspondence shows that SL(2, C) is homomorphi onto Lp (see Prob. 18 of Chap. 9 in [Ma and Gu (2004)]). SL(2, C) is the covering group of Lp. We are going to show that every M in SL(2, C), if neglecting the possibl minus sign, can be written in an exponential function of matrix where th exponent is traceless. The product of two eigenvalues of M is equal to l M can be diagonalized if its two eigenvalues are different,
y1lvIY
=
( eiT 0
( TO) } ' = exp { i 0 T
e0iT )
Y E SL(2, C).
(9.180
M = exp (iTY0"3y1) and T is a complex number. Except for 1, M can be changed to a ladder matrix if its two eigenvalues both are 1,
Z~l M Z
=
(12) 0 1
= exp {(02)} 0 0 '
Z
E SL(2,
C).
(9.181
M = exp {Z (0"1 + i0"2) Zl}. If two eigenvalues of M are both 1, on can diagonalize  M instead of M, because ±M correspond to the same A in Lp. Therefore, removing the possible minus sign, M = exp( iB) wher B is a traceless matrix and can be expanded with respect to the Paul matrices,
M
= N
{exp (in· u/2) x exp (in' . u/2)} Nl
exp (in. T(+)) exp
Wab
(in' . TH)
exp
{i~ (J2aT~+) + J2~T~)) }
exp
{i L
WabTab 
a
where
(9.182
(9.182) into (9.179), the arbitrary element A in Lp i
Substituting Eq. expressed as
A
= exp (in· u/2).
is real and
Wa4
it
wa4 T a4} ,
a=l
is pure imaginary,
(9.18.3
456
Chap. 9 Real Orthogonal Groups
1
wab
3
= 2L
f:abc (Dc + D~) ,
(9
c=1
Thus, every element in Lp is written in an exponential function of trix where the exponent belongs to the real Lie algebra of Lp. When parameters are infinitesimal, Eq . (9.183) returns to Eq. (9.169). representation matrix of A in DJk is exp
{
. 2
L
jk· 2
wab1ab 
a
exp{  i t a=!
} L3 wa4Ia4 jk
a= l
naI~} XCXP{it
(9
n;I!} .
a=l
In principle, two sets of parameters can be transformed to each ot.he though the transformation is quite complicated. In fact, from the pa eters (cp, w, a, /3, ,) one calculates M by Eq. (9.179) and determ the parameters Wab and W a 4 by Eqs. (9.182) and (9.184). Conversely, the parameters Wab and W a 4 one calculates !VI by Eq. (9.182) and Eq. (9.179), and then, determines the parameters (cp,e,w,a,/3,~() by (9.174) and (9.175).
e,
9.5.6
Classes of Lp
The classes of Lp can be discussed from the classes of SL(2, C) owin Lp ~ SL(2, C). When two eigenvalues of !VI E SL(2, C) are differen is conjugate to an element given in Eq. (9.180). Since D3 = 27, on W12 = cp = 7 + 7* and W34 = iw = 7  7* . ±A1 belong to two class SL(2, C) but correspond to one element in Lp belonging to a class o characterized by the parameters cp and w, or by the representative ele A(cp, O,w, 0, 0, 0):
A
=
(~~:: ~~~n: ~ ~) o 0 coshw isinhw o
0
i sinhw
coshw
oS w
(9
<
GO.
!VI = ±l constitutes two classes in SL(2, C) but correspond to one cla Lp. If two eigen val ues of M both are 1, except for 1, !VI is conjugate element given in Eq. (9.181). The set of M constitute another cla SL(2, C). Two classes of SL(2 , C) correspond to one class of Lp. Sinc
§9.5
The 50(4) GTOUp and the Lorentz GTOup
457
where OJ = 2i, O 2 = 2, and 0 3 = O. The representative element in the class ofLp is A(1f,1fj4,w,1f,31fj4,O) with coshw = 3,
(
1
o 2
o
1
2 2i
0 1  2i
o o
+. 2i )
(9.187)
A is a proper Lorentz transformation matrix with all four eigenvalues to be 1. (see Prob. 18 of Chap. 9 in [Ma and Cu (2004)]).
9.5.7
Irreducible Representations of Lh
The irreducible representations of Lh can be obtained from those of Lp and the properties of T and p of two cosets L~ and Lt. Another representative element (5 = Tp in the coset L~ is calculable. Since p is commutable with every element in Lh and p2 = E, it takes a constant matrix in an irreducible representation of L h . The property of T has been given in Eq. (9.158). From the singlevalued representation Djk(Lp), its induced representation with respect to Lh can be calculated like those of 0(4). Introduce four irreduci bie representations of the inversion group V 4 of order 4, V(l)(T) = V(2)(T) = V(l)(p) = V(4)(p) = 1, V(3)(T) = V(4)(T) = V(2)(p) = V(3)(p) = 1.
When j
= k,
(9.188)
there are four induced representations I:!,.jj).. (L h ), A E Lp ,
I:!,.jj)..(A) = Djj (A), 1:!,.;Z:v'I',(T) I:!,.jj)..(p)
= V()..)(T)0I'I"OVV"
= V()..) (p)
1,
1::;
(9.189) A ::;
4.
When j i= k but j +k is an integer, only the direct sum of two representation spaces DJk(Lp) and DkJ(Lp) is invariant for the group L h , and there are
458
Chap. 9 Real Orthogonal Groups
two induced representations ~jk±(Lh)' ~jk±(A) =
jk±
Djk(A) ffi DkJ(A), )_
~ ILVOl,V' IL' /3 (T  6( 0I)/36 1L1L, 6vv' ,
0:,
(9.19
f3 = ±1,
where two representation spaces of Djk (Lp) and Dkj (Lp) are distinguish by the subscripts 0: and f3. Since the diagonal entries of ~jk± (T) are all the representation' by changing the sign of ~jk± (T) leads to an equivale one. For the doublevalued representation Djk (Lp), we only discuss t Dirac spinor representation D(Lh), which is the induced representation D~ 1 (Lp). Introduce 4
"ilL =
2..=
AlLvl'v ,
v=J
Since A E Lh is an orthogonal matrix, "ilL also satisfy the anticommutati relations (9.38). From Theorem 9.1 two sets of I'lL and "ilL are equivale and they are related by a unimodular similarity transformation D(A), 4
D(A)II'ILD(A) =
2..=
AlLv~(v,
det D(A)
= 1.
(9.19
v= l
The set of D(A) forms a multiplevalued representation of Lh where t generators are ~
11L1/ =
4
(9.19
blLl'v l'vl'lL) .
Introduce the charge conjugate matrix C T C  I 1'" C = I'lL. ,
C t C=l,
CT=C,
det C
= 1.
(9.19
Due to C I 1ILvC = 1~v, a new constraint is added to restrict the repr sentation to be doublevalued,
(9.19 Remind that the righthand side of Eq. (9.194) is not equal to D(A)*. physics, Eq. (9.194) becomes Eq. (9.195) for the elements in Lh
(9.19
459
Exercises
The representation matrices of some representative elements are
D(T) = ±1'41'5,
(9.196)
D(p) = ±h5 .
D(A) satisfies Eqs. (9.191) and (9.195). There is a twotoone correspon dence between ±D(A) and A in L h . The set of D(A) is called the Dirac spinor representation which is the covering group of L h . For L p , the Dirac spinor representation is reduced to two irreducible representations of di mension 2, both of which are isomorphic onto the covering group SL(2, C) of Lp. The Dirac spinor representation is not unitary and satisfies (see Prob. 17 of Chap. 9 in [Ma and Gu (2004)])
t
A44
1'4 D (A) 1'4 = IA441 D(A)
9.6
1
(9.197)
.
Exercises
1. Calculate the dimensions of the irreducible representations of the SO(8)
group denoted by the following Young patterns: (a) [4,2], (b) [3,2] (c) [4,4], (d) [(+)3,2,1,1], (e) [(+)3,3,1,1]. 2. Calculate the ClebschGordan series for the subduced representations of the following irreducible representations of the SU(N) group with respect to the subgroup SO(N), and then check the results by their dimensions for N = 7:
[2],
[3],
[2,1]'
[4],
[3,1],
[2,2]
[2,12] ,
[5],
[4,1] ,
[3,2] ,
[3,1,1]'
[2,2,1]'
[2,1 3 ] ,
[6] ,
[5,1],
[4,2] ,
[4,1,1],
[3,3] ,
[3,2,1]'
[3,1 3 ] ,
[2 3 ] ,
[22,12] ,
[2,14].
3. Calculate the ClebschGordan series in the reductions of the following direct products of the irreducible tensor representations, and check the results by their dimensions for the SO(7) group:
(1) [2]
@
[2],
(2) [2]
@
[1,1],
(3) [3]
@
[2, 1].
4. Calculate the orthonormal bases in the irreducible representation space [2,2] of SO(5) by the method of the block weight diagram, and then express the orthonormal bases by the standard tensor Young tableaux in the traceless tensor space of rank 4 for SO(5).
460
Chap. 9 Real Ortho90nal Groups
5. Calculate the orthonormal bases in the irreducible representation s [2,0,0,0] of SO(8) by the method of the block weight diagram, and t express the orthonormal bases by the standard tensor Young tabl in the traceless symmetric tensor space of rank 2 for SO(8).
6. Calculate the spherical harmonic functions y£,l in an Ndimensi space.
7 . Calculate the dimension of the irreducible spinor representation o
SO(7) group denoted by the following Young patterns:
(1) [8,4,2]'
(2) [8 , 3, 2]'
(4) [8 , 3,1 , 1]'
(5) [s, 3, 2, 2] .
(3) [s, 4, 4],
8 . Calculate the Clebsch Gordan series for the direct product of the t sor representation [A] and the fundamental spinor representation [s SO(2£ + 1) or [±,,] of SO(2£), where [A] is a onerow Young pattern a onecolumn Young pattern.
9. Calculate the basis states in the fundamental spinor representation ofSO(7). 10. Calculate the basis states in the fundamental spinor representa [±s] of SO(8).
11. Expand the eigenfunction of the total angular mom entum with highest weight in terms of the product of the spherical harmonic f tion y£,l(x) and the spinor basis x(m).
12. Discuss the classes in the SO(4) group and calculate their chara in the irreducible representation Djk.
13. Calculate six parameters of the following proper Lorentz transfo tion A, and write its representation matrix in the irreducible repre tation Djk(A) of the proper Lorentz group Lp:
A(rp,B,w,a.,(3,,)
=
(
0 o1 V3/2
0 i(sinhw)/2 0 1/2 V3(coshw)/2 iV3(sinhw)/2
00
0 (coshw)/2 isinhw
coshw
14. Prove that the Dirac spinor representation satisfies
t~ _ A44 ()I , 4D ( A ) Y4  IA441 D A .
J
Chapter 10
THE SYMPLECTIC GROUPS
In §7.4.3 we have introduced the fundamental property of the USp(2 group and the Sp(2e, R) group. In this chapter we will study the irreducib representations of USp(2e) and their applications to physics.
10.1
Irreducible Representations of USp(U)
10.1.1
Decomposition of the Tensor Space of USp(U)
The element u in USp(2f) is a transformation matrix in a (2f)dimension complex space, Xa
~ x~
=L
(10 .1
UabXb,
b
where the index a is taken to be j or ], 1 ::; j ::; a
The
U
= 1,
e, in the following order
I, 2, 2, ... , f, £.
(10 .2
matrix satisfies
(10 .3 where I Jab= { 01
J
= Ie
x (i0"2)
when a = j, b = ], when a=), b=j, the remaining cases,
= _J 1 = J T ,
det J
(10.4
= l.
Thus, the selfrepresentation of USp(2f) is selfconjugate
u*
= JJuJ.
(10 .5
1(;2
Ch(11). 10
'I'lL , Sym plec li c
C:" Otl7'"
The tensor Tal ... a n of rank n and the basis tensors to USp(2€) are defined as T
a' ... an ~
(OUT)al ... a n
L
=
{Ja,n n
with If'SPC :c
'Ua,b , · ··'Ua"bnTb, ... bn,
b, ... b n
{Ja, ... a n
~
OU{Ja, ... a n
L
=
(lO.G
{Jbt ... bn 'Ub, a, ... 'Ubna n ·
bt ... bn
Due to Eq. (10.5) the contravariant tensor is equivalent to the covariant on In comparison with the tensors of SO(N), there also exists two invarian tensors of USp(2€): Jab and fat .. a2l' In fact, denote by J an antisymmetr tensor of rank 2 whose component is equal to Jab, (OUJ)ab
=L
'Uac'UbdJcd
=
('UJ'UTtb
= Jab·
(10.7
cd
Jab is invariant in USp(2€). Through a contraction of two antisymmetri indices j and J, the trace tensor of a tensor of rank n for USp(2€)
L
Jara3Tai ..
arlarar+l···a~la"a.::l+l···'
a,..u"
is a tensor of rank (n  2),
L
J ara , 'U at b , ... 'U an b n T bt ... bn
ara"bl ... b n
==
L
Ua1b r ·· ''Uur_Ibr_l'Uar+lbr+l''
.Ua~_lb~_lua"+lb.!l+l··
,Ua"Ib n
b, ... bn X (.hrb,Tb, ... br_tbrbr+J ... bs_tb,b, +J'..) .
(10.8 The subspace of the trace tensors is invariant in USp(2€). A tensor spac can be decomposed into the direct sum of a series of subspaces of traceles tensors
L
JabT. .. a ... b ...
= O.
(10.9
ab
For example, similar to Eq. (8.56), a tensor of rank 2 is decomposed into sum of a traceless tensor of rank 2 and a scalar
The Weyl reciprocity holds for the tensors of USp(2e) so that the tells space can be reduced by the projection of the Young operators. Deno by the space of traceless tensors of rank n of USp(2f). Projecting by Young operator y1"] , one obtains a traceless tensor subspace yL"]r = where the basis tensors are taken to be the tensor Young tableaux. \Ve first study the condition whether the traceless tensor space is null space or not. Without loss of generality, assume that the row numb of [A] is r, and there arc m pairs of digits j and J in the first column of a arbitrary tensor Young tableau in Fixing the unpaired digits in th first column and the digits in the remaining columns, one changes the pai of digits from 1 to e, where some tensor Young tableaux may be vanishin owing to antisymmetry of indices. The number of linearly independe tensor Young tableaux is the combinatorics of m among [f  (1'  2m) ]. Th traceless condition is written as Eq. (10.9), where the (m1) pairs of digi are fixed and only one pair of digits runs over from 1 to e. The number traceless conditions is the number of possible values of the (m  1) pairs digits, tha t is the combinatorics of (ml) a mong [f(r2m)] . The tracele tensor space 1;"1"1 is a null space if the number of traceless conditions is n less than the number of independent tensors, namely, [.e  (r  2m)]/2 < m and then, r > e. Thus, the traceless tensor space is a null space if th row number r of the Young pattern [A] is larger than e. The other invariant tensor of USp(2.e) is the totally antisymmetric te sor ca\.a2c ofrank 2f. This tensor violates the traceless condition (10.9). is irrelevant to the reduction of th e tensor space for USp(2£). The tracele tensor subspace yL"]r is minimal because there is no further constraint construct a nontrivial invariant subspace in yL"Ir. Thus, the irreducib representation of USp(2f) is selfconjugate and denoted by the Young pa tern [A] with the row number not larger than I!.
r
01"
01"]
01"].
01Al
10.1.2
OrthonormallTreducible Basis Tensors
The generators in the selfrepresentation of USp(2.e) are given in Eq. (7.10
where 1 :::; d :::; 3, 1 :::; j < k :::; f, and T}r) are the generators in the se representation of SU(.e). The generators satisfy the orthonorma.l condi t io Tr(TATn) = (jAB· The Chevalley bases of USp(2f ) ca.n be cakula t 'ci b Eq. (7.141),
464
Chap. J 0 The Symplectic Groups
where 1 :::; jJ < vectors Oa are
e.
He
= TiP x 0'3,
Et
= TiP x
(0'1
Fe
= TiP x
(0'1  i0'2)
+ i0'2) /2,
/2,
(10.11 The nonvanishing action of the generators on the ba.s
= 01"' HI" 0li =  0li' EI"0I"+1 = 01"' FI"01" = 01"+1, HI"01"
HI"0I"+l
= 01"+1,
= Be, EI"0'jI = 0I"+J' FI" 0;+f =  Oji, HeOe
HI" 0 I"+l = °1"+1' HeOe = Be' EeOe FeOe
= Be, = °e·
(10.1
Namely, Oa is the common eigenvector of HI" and He, but its arranged ord is not convenient for the calculation of the raising and lowering operator The minus signs in the actions of E" and FI" are also not convenient. Defin (10.1 namely the order of 4>Ci) 1 :::; 1, 2,
C\' :::;
e,
2e, is
e, e 1,
2, 1.
Thus, Eq. (10.12) becomes H JL 4>I'
= 4>1"'
HI"4>2t1" = 4>2£1"' He4>e = 4>[, EI"4>I"+1
Ee4>f+1
= 4>1"' = 4>e,
FI"4>2 t 1"
= 4>2 e I"+I,
HI" 4>,,+ 1 = 4>1"+1, HI"4>UI"+l
= 4>2£1"+1,
He4> £+ 1 = 4>£+ 1, E ,,4>2eI"+1
= 4>2eI"'
(10.1
FI"4>1" = 4>,,+1, Fe4>e
= 4>e+ l,
where 1 :::; jJ < e. The basis tensor 4>o.\.o.n is the direct product of the ba.sis vectors 4> The action of a generator on the basis tensor is equal to the sum of i action on each basis vector in the product. The standard tensor Youn tableaux yl']4>a'lG n are the common eigenstates of H J/) but generally n orthonormal and traceless. The eigenvalue of HI" in the standard tens Young tableau yl'\]4>c'!.O:n is the number of the digits jJ and (2£ in the tableau, minus the number of the digits (jJ + 1) and (2£ 
jJ)
fille
jJ
+1
§ 10.1
frr·educible Representations of USp(2E)
4
The eigenvalue of He in the standard tensor Young tableau is equal the number of the digit £ in the tableau, minus the number of the dig e+ 1. The eigenvalues constitutes the weight 'Tn of the standard tens Young tableau. Two standard tensor Young tableaux are orthogonal their weights are different. The action of FjJ. on the standard tensor You tableau is equal to the sum of all possible tensor Young tableaux, each which is obtained from the original one by replacing one filled digit J.l wi the digit (J.l + I), or by replacing one filled digit (2£  J.l) with the dig (2e  J.l + 1). The action of Fe on the standard tensor Young tableau equal to the sum of aJl possible tensor Young tableaux, each of which obtained from the original one by replacing one filled digit e with the dig (£ + 1). The actions of EjJ. and Ee are opposite. The obtained tensor You tableaux may be not standard, but they can be transformed to the sum the standard tensor Young tableaux by the symmetry (8.22). The row number of [A] in the traceless tensor subspace AJ is not larg than i!, [A] = [A I, A2, ... ,Ae]· Through a similar proof as that for Theore 8.3, there is one and only one traceless standard tensor Young tableau JjAJ which is annihilated by every raising operators EjJ. [see Eq. (7 .115 This traceless standard tensor Young tableau, where each box in its n row is filled with t.he digit ex, corresponds to the highest weight in JjAJ, an the highest weight M = LjJ. wjJ.MjJ. is calculated from Eq . (10 .14),
0l
M 1< = AjJ.  AjJ.+l,
Me = At,
1 :::; J.l
< e.
(10.1
It means that the irreducible representation of USp(2e) is denoted by t Young pattern [A] with the row number not larger than £. The remaini basis tensors in JjAJ can be calculated from the standard tensor You tableau with the highest weight by the lowering operators FjJ. in the metho of block weight diagram, where the multiplicity of a weight can be obtain by counting the number of the traceless standard tensor Young tablea with the weight in JjAJ. The calculated standard tensor Young tablea are traceless ane! orthonormal. Obviously, they are normalized to what t highest weight state is normalized to. The irreducible representations of Sp(2£, R) can be obtained from tho of USp(2£) by replacing some parameters to be pure imaginary (see t discussion below Eq. (7.106)) but preserving the generators invariant. The basis tensors in the representations [1, 1,0] and [1, I, 1] are calc lated in Probs . 5 and 6 of Chap. 10 of [Ma and Gu (2004)]. The tens in [1,1,0] is antisymmetric and can be decomposed into a traceless tens and a trace tensor (scalar), as shown in Eq . (10.10). The represent.ati
466
Chap. 10 The Symplectic Groups
[1,1,0] contains a single dominant weight (0,1,0) and a double dominan weight (0,0,0). The traceless standard tensor Young tableaux with th weight (0,0,0) are
which are orthogonal to the trace tensor
Please notice the different definitions in the basis vectors CPo: between Eq (10.13) of this textbook and Eq. (10.26) in [Maand Gu (2004)]. Namely, C changes a sign for USp(6) . The representation [1,1,1] contains two sing dominant weights (0,0, 1) and (1,0,0). In the following the basis tensors in the adjoint representation [2,0,0 of USp(6) are calculated. The tensors in the representation are symmetri so tha.t all basis tensors are traceless. The simple roots rJ.l. of USp(6) ar expressed with respect to the fundamental dominant weights W v ,
There are two typical tensor Young tableaux
They are normalized to 4 and 2, respectively. The highest weight in (2,0 , is (2,0,0). The block weight diagram a.nd the basis tensors of (2 , 0,0] o USp(6) are listed in Fig. 10.1. Please compare Fig. 10.1 with Fig. 9. Since [2 , 0,0] is the adjoint representation of USp(2£), all positive roots ca be written from Fig. 10.1. In addition to the three simple roots given i Eq . (10 .16), the remaining positive roots arc
03
= WI + W2  W3 = rl + rz, = WI  W2 + W3 = rl + r2 + r3,
05
=
01
W2
=
rl
+ 2r2 + r3,
02
= WI + W3 = r2 + r3,
04
=
06
= 2Wl
+ 2W2 = 2r2 + r3, = 2rl + 2r2 + r3 ·
2wJ
(10.17 The calculations related with the multiple weight (0,0,0) are given 3 follows .
§J O. J
Irreducible Representations of USp(2f)
.121
I
o::::r:::u .121
2
1
3
1
2
.12 1
1 .121
1
I
1
3
1
I
1
4
1
o:::::r::IJ
.12 1
2
1 • 1 .12 1
I
1
5
1
IT:I:3:J
.121
3
1
2
1
6
1
.12
5
1 .121
Ci::ITI
.12 1
3
1
6
1
ITITI
.121
4
1
6
1
.121
1
6
1
6
1
5
1
6
c:::c::r:TI + 1 2 1 5 1 Ji73 (  c:::c::r:TI + LI:ITI + 2 LIJ::TI ) c = .fi73 (I I 1 6 1  LI:ITI + LIJ::TI) A
=
B =
Fig. 10.1 The block weight diagram and the basis tensors in [2,0,0] of USp(6).
1(2, 1,0)) = F21(l, I, I)) = J21 1 1 5
I,
l(l,2,l))=F1 1(l,l,l))=F3 1(l,O,l))=J21 2 41, 1(0,2,2))
= Jl72F2 1(1, 0, 1)) =
1
3
1
3
I·
From three states, a.n A1triplet, an A 2 triplet, and an A3triplet are co structed, respectively. There are three standard tensor Young tableau with the weight (0,0,0) so that the weight is triple . Assume
1(0,0, Oh) =
Jl72FJ
1(2, 1, 0» =
FJ
11 5
= 1 1 1 6 1+ 12 1 5
I,
468
Chap. 10 The Symplectic Groups
+ a2 1(0,0, Oh), F3 1(0,2,2)) = b1 1(0,0,0)1) + b2 1(0,0, Oh) + b3 1(0,0, Oh), E] 1(0,0, Oh) = E1 1(0,0, Oh) = E2 1(0,0, Oh) = 0, F2 1(1,2, I)) = a1 1(0,0,0)1)
ar
where + a~ = 2 and bi + b~ basis state 1(1,2, I)), one has
E]F2 1(1,2,1)) =
= F2E]
+ b5 =
V2a1
2. Applying E]F2
=
F2 E] to th
1(2,1,0))
1(1,2,1)) = F2 1(1,1,1)) = 1(2,1,0)).
Thus, a1 = J1fi. Choosing the phase of the basis state 1(0,0, Oh) suc that a2 is real positive, one has a2 = Applying E]F3 = F3E] an E2F3 = F3E2 to the basis state 1(0,2,2)), one has
J372.
E1F3 1(0,2,2))
= V2 b1 1(2, I, 0)) = F3E]
E2 F3 1(0,2,2)) = ( J1fi b1 = F3 E 2 1(0,2,2))
+ J372 b2 )
= .j2 F3
1(1,0,1))
1(0,2,2))
= 0,
1(1,2, I))
= V2 1(1,2,1)).
Thus, b1 = 0 and b2 = 2/../3. Choosing the phase of the basis sta 1(0,0, Oh) such that b3 is real positive, one has b3 = fi73. Then, 1(0,0, Oh)
= fi73 { F2
1(1, 2, I)) 
J1fi
1(0,0,0)1) }
V4f3 F2 12 1 4 1  /173 {Ill 6 1+ 1 2 1 5 I} = V1f3 {Ill 6 1+ 1 2 1 5 1+ 21 3 1 4 I}, 1(0,0, Oh) = J372 {F3 1(0,2,2))  J473 1(0,0, Oh)} = J372 F3 13 1 3 1 fi73 {  1 1 16 1+ '12'1'51+ 21 =
= 10.1.3
fi73 {Ill
6
3
14
I}
1  1 2 1 5 1+ 13 14 I} .
Dimensions of Irreducible Representations
The dimension of an irreducible representation [A 1 of USp(2f) can be ca culated by the hook rule. In this rule, the dimension is expressed as quotient, where the numerator and the denominator are denoted by th symbols y~.\l and y,~.\l, respectively: (10.1
§10.1
Irreducible Representations of USp(2e)
·
We still use the concept of the hook path (i, j) in the Young patt.e
[A], which enters the Young pattern at the rightmost of the ith row, go
leftward in the i row, turns downward at the j column, goes downward the j column, and leaves from the Young pattern at the bottom of the column. The inverse hook path (i, j) is the same path as the hook pa (i, j) except for the opposite direction. The number of boxes contained the hook path (i, j) is the hook number h ij of the box in the jth colum of the ith row. yPl is a tableau of the Young pattern [A] where the box the jth column of the ith row is filled with the hook number h ij . Defi a series of the tableaux y~~J recursively by the rule given below. y~)..l is tableau of the Young pattern [A] where each box is filled with the sum the digits which are respectively filled in the same box of each tableau y~ in the series. The symbol y~)..l means the product of the filled digits in so does the symbol y~)..l. The tableaux Yk~l are defined by the following rule:
(a) y~~l is a tableau of the Young pattern [A] where the box in the jth column of the ith row is filled with the digit (2C+j  i). (b) Let [A(1)] = [A]. Beginning with [A(1)], we define recursively the Young pattern [A(a)] by removing the first row and the first column of the Young pattern [A(aI)] until [A(a)] contains less than two rows. 1 to be a (c) If [A(a)] contains more than one row, define tableau of the Young pattern [A] where the boxes in the first (a  1) rows and in the first (a  1) columns are filled with 0, and the remaining part of the Young pattern is nothing but [A(a)]. Let [A(a)] have r rows. Fill the first (r 1) boxes along the hook path (1, 1) of the Young pattern [A(a)], beginning with the box on the rightmost, with the digits A~a), A~o), . .. , A~a), box by box, and fill the first A;a) boxes in each inverse hook path (i, 1) of the Young pattern [A(a)], 2 ::; i ::; r, with 1. The remaining boxes are filled with O. If a few 1 are filled in the same box, the digits are 1 summed. The sum of all filled digits in the pattern with a > 0 is O. The calculation method (10.18) is explained through some examples
yk
yt
470
Chap. 10
Ex. 1
The Symplectic Groups
The dimension of the representation [3,3,3] of USp(6).
6 4 2
10 5 3
11 9 4
4 3 2
3 2 1
d[3,3,31 [USj!'J(6)]
5 4 3
= 11
x 5x 3x 2
= 330.
Ex. 2 The representation of onerow Young pattern [n] of USp(2£ The tensors in the representation are symmetric, so that all standard tenso Young tableaux are traceless. dl n l[USp(2£)]
= dl n l[SU(2e)] =
(n
+ 2£ n
1)
(10 .1
.
The representation of onecolumn Young pattern [In]
Ex. 3 USp(2 e).
2£ 2£  1 Ilnl _ Yp 
1 1
2£ + 1 2£
1
2£  n + 3 2f  2n + 2
+ 2£  n + 2 2£  n + 1
n+ 1
dl1ndUS p(2£)]
=
(2£ + 1)!(2£  2n + 2) n!(2£  n + 2)!
(10 .20
Ex. 4 The representation of tworow Young pattern [n, m] of USp(2e
yln ,mJ = I P
2f
. 2f
I' .. I 2f + m
I . ' ..
+ ~l I
= I 2/~ 2I
y~n,ml
. 2f
+m
 1
I 2f + n 
2
I 2t
+n
 I I
2
10m I I
I
2£ 2f
+m +m
 1  3
I
U+n2 12t+n+m  l
= =1n=m=+=l=1===I=n===7=+=2=r1nm'I.. ."'1'1I,
§10.2
Physical Application
d
[US (2£)]= (nm+1)(2£+n+m1)(2£+n2)1(2£+m3) p (n + 1)!m!(2£  1)!(2£  3)! (10.2 For the groups USp(4) and USp(6), one has
[n,m]
+ l)(n + m + 3)(n + 2)(m + 1)/6, (n  m + l)(n + m + 5)(n + 4)(n + 3)(n + 2) x (m + 3)(m + 2)(m + 1)/720.
d[n,m ][USp(4)] = (n  m d[n, m][USp(6)] =
10.2
Physical Application
The Hamiltonian equation of a classical system with £ degrees of freedo is dpj (10 .2
dt
Arranging the coordinates qj and the momentums Pj in the order,
x"
= (ql, PI, q2, P2, ... , qi, Pi),
(10.2
one obtains the coordinates Xa in the (2£)dimensional phase space. T Hamiltonian equation can be expressed in a unified form dx _ jaH
dt , ax'
(10.2
called the symplectic form of the Hamiltonian equation. If dX a satisfy Hamiltonian equation (10.24), after the symplectic transformation
(10.2 dZa still satisfy the Hamiltonian equation (10.24), dZ a
dt
Th e method of RungeKutta is commonly used in the numerical c culations by computer. This method does not reflect the characteristic the equation of motion so that the calculation error will be accumulated. t.he calculation is repeated in a tremendous number, the accumulated er
Chap. /0 Th e 8Ynl7,/ect'ic Groups
'172
will make a big deviation of the calculation data from the real orbit, example, the calculation in cyclotron reaction and in satellites. If each s in the numerical calculation reflects the characteristic of the Hamilton equation, say each step satisfies the symplectic transformation, _
Za 
(0)
Za
+T
~
L
8H(x)
J ab 
b
nuZb
(10
,
where T is the length of the step, the accumulated error will decrease grea The group leaded by Professor Feng Kang studied deeply this probl Please see his paper [Feng (1991)] in detail. 10.3
Exercises
1. Prove that the determinant of R in Sp(2£, R) and the determinant o in USp(2e) are both + l.
2. Count the number of independent real parameters of R in Sp(2£, R) u in USp(2e) directly from their definitions (7.95) and (10.3).
3. Express the simple roots of USp(2e) by the vectors Va given in (7.79) for the SU(e+ 1) group, and then, write their Cartan Weyl ba of generators in the selfrepresentation of USp(2e).
4. Calculate the dimensions of the irreducible representations of the USp group denoted by the following Young patterns: (1) [4,2],
(2) [3,2],
(3) [4,4],
(4) [3,3,2],
(5) [4,4,3].
5. Calculate the orthonormal bases in the irreducible representation [1,1 of the USp(6) group by the method of the block weight diagram, then, express the orthonormal bases by the standard tensor Yo tableaux in the traceless tensor space of rank 2 for USp(6).
6. Calculate the orthonormal bases in the irreducible representation [1,1 of the USp(6) group by the method of the block weight diagram, then, express the orthonormal bases by the standard tensor Yo tableaux in the traceless tensor space of rank 3 for USp(6).
7. Calculate the ClebschGordan series for the reduction of the di product representation [1,1,0] x [1,1,0] of the USp(6) group and highest weight states for the representations contained in the series the method of the standard tensor Young tableau.
Appendix A
Identit ies on Com binatorics
There are two binomial identities for a complex z and a positive integer a
(1
+ z)
n _

n
a ~
(l+z)"=
z
a
a _ a .I
()
()
n'
n
 n!(a  n)!'
(A.I)
~ (_1)nzn(a+~l).
Hereafter, the summation index runs over the region where the denominator is finite. Since (1
+ z)o+b = 2:=
zm
(a: b) =
Til,
= 2:=
Zn
(~) 2:=
Zi
t
n
(1
+ z)O(l + z)b
C) = 2:=
Zm
L ( ~) n
m
(m ~ n) ,
one has
(A.2) 2:= {p!(v 
r
+ p)!(U  p)!(r _ p)!} ]
=
(U + v)! l1!V!r!(u + v  r)!
Since (1
+ z)ab = 2:=
C
1)
a:,m 
(_I)mzm
TTL
= (1+z)a(I+z)b=2:= z71(~) n
2:= f.
(l)tzeC+~l)
474
Appendix
one has
~(I)P(~) (~=~)=(V~u),
L
(I)P(v  p)! _ (v  u)!(v  r)! P p!(u  p)!(r  p)!  u!r!(v  u  r)! ·
(A
Since (l+z)ab = (1
=~
(_1)m z rn (a+b:ml)
+ z)a(1 + Z)b
=~ (_1)nzn(a+~I) ~ (I)eZeC+~I) = ~ (_1)mzrn~ (a+~I) C+:=~I), one has
~ (U+~I) (V+~=~  l)
L P
(u
+
+
r  p  I)! _ (u pi (r  p)! 
p  1)!(v
=
(u+v~rl),
++ v
r  1)!(u  1)!(v  1)1 r! (u + v  1)!
(A.
Appendix B
Covariant and Contravariant Tensors
Let G be a group whose elements Rare N x N matrices. R can be looke like a coordinate t.ransformation in an Ndimensional space,
Xa ~ x~
=
L
RadXd .
(B.
d
A covariant tensor field T(x)a, ... an of rank n with respect to the group contains n subscripts and Nn components, which transform in REG as
[ORT(x)la,
an
=
L
Ra,d, ... RQndnT(RIx)d,.dn·
(B.2
d, ... d n
A covariant tensor field becomes a covariant tensor if its components a independent of coordinates Xa' A covariant tensor field can be expande with respect to the covariant basis tensors (h, b2 .. b n
(B.3
T(x)a, ... an
=
L
T(X)b, .b n (lh, .. bn)a,.a n = T(x)a,(1n .
(B.4
b, .. . b n
The coefficient T(X)Ql. an is equal to the tensor component T(x)a,an values, but transforms in R like a scalar. The tensor transformation carried out by the basis tensor lh, ...b n ,
[ORT(x)la,
an
= [PRT( x )lala n =
OR{h, .. bn = QR()b,.bn =
L
d, ... d n
T(R1x)a, ... an ,
()d, ... dnRd,b " , .Rdnb n ,
(B.5
476
Appendix
because
=
[ORlh1 ... b"L1an
L
R atdt ··· Rand"
[lht ... bnldt ... d"
dt ... d"
L
= R atbt ·· . Ranb n =
[Odt ... dJat ... an
R dtbt ·· .Rd"b".
d1···d n
A contravariant tensor field T(x)a t ... an of rank n with respect to th group G contains n subscripts and NIt components, which transform i REG as
[ORT(x)tta"
L
= d
t .. d
n
A contravariant tensor field becomes a contravariant tensor if its compo nents are independent of coordinates Xa. A contravariant tensor field ca be expanded with respect to the contravariant basis tensors Obt ... b" ,
(B.7
T(x)
=
L b] .. .b
T(x)b t
.. bn
obt .. bn
n
T(x)a t .an
=
L
T(X)b t
.. bn
(Ob t .. 6" ) at
.. a n
= T(x)a t .. an
(E.S
b, .. .b" Th~
coefficient T(x)at.a n is equal to the tensor component T(x)at.Q" values, but transforms in R like a scalar. The tensor transformation carried out by the basis tensor Obt .. b"
[ORT(xW t .. an OROb1 .. b n
= [PRT(xWl ... a" = T(RlxYt ... a",
= QRObt ... b = n
L
(R  1 )b t dt ··· (R  1kd"
Od t
..
dn . (E.g
dt ... d n
A mixed tensor field T(x)~~···.~: of rank (n, m) with respect to the grou G contains n subscripts, m superscripts, and N n + m components, whic transform in REG as
[ORT(x)l~tt·.~::
(E.IO
Appendix C
T he Space Groups
230 space groups are listed with both Schroenfiies notations (Sch) and international notations for space groups (INSG). The star in the ordi number denotes that the space group is symmorphic. The subscripts the symbol for INSG are moved to a bracket for convenience. For examp the symbol F ± 2Ho2~H for the space group D~~ is replaced with F
2( HO)2'(OH).
Ordinal
*3 4 *5 *6 7 *8 9
Ordinal
*16 17 18 19 20 *21 *22 *23 24
Table C 2 Monoclinic crystal system Sch. INSG Ordinal Sch. INSG P2 P±2 *10 Ci Cih C 22 11 P2(0~0) P ± 2(0~0) C5h C 23 A2 A± 2 *12 Cih P2 C s1 13 Cih P ± 2( ~ ~0) 14 P2( ~OO) C; P±2(~2:0) C3s A2 15 cgh A±2(2:00 ) C·sl A2( ~OO)
qh
Table C 3 Orthorhombic crystal system Sch . Ordinal Sch. INSG INSG Dl Cl2v P22' P22 *25 2 D2 26 P2(00~)2' P22'(0~0) C5v 2 D32 27 P22'(00~) P22'(~ ~O) C:lv D42 C4 28 P22' (~OO) P2(00~ )2' (~ ~ 0) 2" 29 D 25 A22'( ~OO) P2(00L )2' (~O~) C5" D62 A22' 30 P221(OL ~) cg" D72 F22' 31 P2(00 L )21(~OO) C;" D8 P22t(L ~O) 122' 32 2 cg" D 9 33 12(00~)2'( ~ ~O) civ P2 ( 002:I ) ~2 hI )"I ?1) 2
478
Appendix
Ordinal 34 *35 36 37
Table C.3 Sch.
Orthorhombic crystal system (continued) INSG Ordinal Sch. INSG P22 (~ 1 ~) 55 P±22'(~ ~O) D~h DlO C22J 56 P±2(~ ~0)2'(~0~) 2h DII C2(00~)2' 57 P ± 2(00~)2'(0~0) 21. DI2 e22'(00~ ) 58 P ± 22'( ~ ~ ~) 2h DI3 A22' 59 P ± 2(00~)2'(0~ ~) 21. D14 A22' (O~O) 60 P±2n ~0)2'(~ ~O) 2h DI5 A22'(~00) 61 P±2(~0~)2'(~ ~O) 2h D16 62 A22' (~ ~O) P ± 2(00~)2'(~ ~ ~) 2h D17 F22' 63 A±22'(~00) 2h DIS 64 A±22'(~~0) F22' 2h Dig 122' A ±22' *65 2h 20 D 2h 66 A ± 2( ~00)2' 122' (~ ~O) D21 67 122' (~OO) A ± 22'(0~0) 2h P±22' 68 A ± 2(~00)2'(0~0) D~~ D23 F±22' *69 P±2(~ ~0)2'(0~~) 2h D24 70 P±22'(OO~) F±2(t *0)2'(0:j:j) 21. 25 D 2h 1 ± 22' P±2(~ ~0)2'(0~0) *71 D26 72 P ± 22'( ~OO) 1±22'(~ ~O) 2h D27 73 P±2(~ ~0)2'(1 ~~) I ± 2(~0~)2'(~ ~O) 2h D28 74 1 ± 22'( ~OO) P ± 22'( ~O~) 2h P ± 2(~0~)2'(00~)
CIO 2" CII 2" C 12 2" C13
2v
CI4 2" CI5
*38 39 40 41 *42 43 *44 45 46 *47 48 49 50 51 52 53 54
2v 2v l7 C 2v CiS 2v clg 2v 20 C 2v C21 2v C22 2v CI6
(* t t)
D~h D~h D~h D~h
D~"
Dg"
D~h D~h
Ordinal *75 76 77 78
Sch.
*79 80 *81 *82 *83 84 85 86 *87 88 *89 90 91 92 93 94 95
C5
CI 4 C2 4 C3 4 4
e4 4
e 46 51 4
52 4 CJ"
Cih CI" Cth CS h cg h Dl4 D2 4 D34 D44 D54 D6 4 D74
Table C 4 Tetragonal crystal system INSG INSG Ordinal Sch. P4 96 P4(00*)2'(22 0 ) D~ D9 142' *97 P4(00t) 4 DIO 98 P4(00z) 14(00* )2' 4 P42' P4(00~ ) *99 Cl v 14 100 P42' (~ ~O) clv 101 P4(OO~ )2' (OOD [4(00t) CIv P4 102 P4(00~)2'(~ ~~) eJv P42' (00 ~) 14 103 CS v 104 P±4 P42'(~ ~~) C~v P4(001 )2' 105 P±4(00~) exv 106 P ±4(~00) P4(OO~ )l( ~ ~O) C~v 142' *107 P±4(0~ ~) C~v CIO 142' (OO~) 1±4 108 4" CII 109 14(00 )2' (~OO) [ ± 4( ~) 4v C12 P42' llO 14(00* )2' (~O~) 4v P42' P42'(~ ~O) *111 D~d 112 P42'(00~) P4(00~ )2' DL P42'(110) 113 P4(00* )2'( ~ ~O) D~d P42'( [ [ 1) 114 P4(00~ )2' Did 222 P42" P4(00~)2'(~ ~O) *115 D~d P4(00:}.)2' 116 P42" (00 ~) D~rl
tt
t
Appendi.x C
The Space Croups
'
Table C.4 Tetragonal crystal system (continued) Sch. INSG Ordinal Sch. INSG P42"( 0) 130 P±4(200)2'( ~02) D~d D~h P42//(1 131 P ±4(00~)2' D~d D~h _2 I2 1) 2
Ordinal
117 118 *119 120 *121 122 *123 124 125 126 127 128 129
D~d
142//
J42/1(00~)
IO D 2d Dll 2d DI2 2d D!h DJh
142' 142'(0&~) P±42' P±42'(00~)
P ± 4(~00)2'(0~0) P±4(200 )2'(02 ~) P±42'(b 10) P ±42'(f t1) P ± 4(~00)2'doo)
D~h D4. 4h
D~h
D6 4h Dlh
Ordinal
d43 144 145 *146 *147 *148 *149 *150 151 152 153 154 *155
Ordinal
·>168 169 170 171 172 173 .<174 *175 176 *177 178 179 180 18 1
132 133 134 135 136 137 138 d39 140 141 142
D10
4h Dll 4h DI2 4h DIJ 4h Dl4 4h D I5 4h Dl6 4h D17 4h D18 4h DIg 4h
D1~
P ± 4(00~)2'(00~) P ± 4(0 1  )2'(0 1 0)
P±4(of 1)2'(OE~) P ± 4(00E)2'( 110)
P±4(00f)2'd t~) P±4(01 ~)2'(100) P±4(0# f)2'(#0~) I ± 42' 1±42'(00~) I±4(111)2'(100) J±4($; I)2'dO~)
Table C 5 Trigonal crystal system Sch. Ordinal Sch. INSG INSG CI P3 P32 *156 CJv 3 P32/1 C 32 P3(00~ ) *157 cjv C3 158 P3(00~ ) P32'(00~) cgv 3 4 P32//(00~) 159 C 3v R3 Cj Cl3; R32' P3 *160 CSv 2 R32'(1 1 1) 161 C 3i R3 C~v oJ 2 2 Dl P32// P32/1 *162 Ddd 3 D2 P32' 163 P32"(00~) Did 3 D3 P3(001 )2" P32' *164 D~d 3 D4 P3(001 )2' 165 P32'(00~) D~d 3 D5 P3(00i)2" R32' *166 D5d 3 D63 167 P3(001 )2' R32'(~ ~ ~) D~d R32' Dj
T ab le C 6 Hexagonal crystal system Scil. INSG Ordinal Sch. INSG C6 P6 182 P6(00j)2' D~ C26 P62 P6(00~) *183 CJv CJ 184 P62' (00&) P6(00~) CJv 6 185 C 64 P6(00& )2' (OO~) P6(00~) C~v 186 P6(00&)2' Cg P6(00~) ctv P62" C 66 P6(002) *187 Ddh P6 188 P62/1(00~) D~h Cdh P62' P±6 *189 C~h D~h 190 P±6(00~) P62'(00~) CJh D~h Dl P62' P±62' *191 6 D~I' D26 P6(001 )2' 192 P ± 62'(00~) D6h D3 P6(00g )2' 193 P ± 6(00& )2' (00 &) D~h 6 D4 P6(00Y)2' 194 P ± 6(00~)2' D~h G P6(00~)2' Dg
480
Ordinal
*19.5 *196 *197 198 199 *200 201 *202 203 *201\ 205 206 *207 208 *209 210 *211 212
Appendix
Sch. TI
T2 T3 T4 T5 Tl h
T2h T3h T4h. T/~
T6h T7h
oj
02 03 04 05 06
Table C 7 Cubic crystal system INSG Ordin al Sch. I NSG P3'22' 213 07 PJ'4( '4 )2 11 '4 13'4 (1_1) 2"(_11) F3'22' 214 08 4 ,1 4 ,1 4 4 TJ 13'22' P3'42" *215 d T2d F3'42" *216 P3'2(!0~)2'(~ ~O) T3d 13'42" *217 J3'2( ~0~)2'( ~ ~O) T4d P3'22' 218 P3'4(! ~ !)2"(~ !~) 219 Td5 F3'4(00 ~ )2" (00 ~ ) P3'2(~ !0)2'(0~!) T6d F3'22' 220 13'4( )2" ( * P3'42/1 0 h1 *221 F3'2(* *0)2'(0* 13'22' 222 0 2h. P3'4( ~00)2" (OO~) 223 0 3h. P3'2(~0~)2'(~ ~O) P3'4(~ ~ ~)2"(~ ~~) 224 J3'2 ( O!0)2'(00~ ) 0 4h P3'4(0~ 1)2"(! !O) F3~ 42" P3'421/ 0 5h *22.5 226 0 6h F3'4(00~ )2"( 00~) P3'4q ! !)21/(~! !) F3'42 1/ 227 0 7h F3'4(0* *)2"(* 228 0 It8 F3' 4(! 1 )2" * !) F3'4( )2" * [3~ 12" [3'421/ 0 h9 *229 P3'4(l. 1 l)2"(1 ~ l ) /3' 4 ( 1 ;l. :l) 2" ( ~ 1 1) 230 0 10
i J (1'4
*)
*** (* t)
**)
**t
±
(*
to)
B ibliography
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