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15 8 1"· 1:\, 978·98 1217· 112· 1 (rob~·: ,,!~. papo:r) 15 8 N·IO; '181·277· 112·5 Iphk .. ~1~. ""I> Operator a nd C lass Space Transform ation O perll tors fo r a Scala r Fllw;tion Eg llivalc nt Rcprcsmtatiolls In(!(]1 1iw,lellt a1 1d In w h1("ible Rep ro>sentations :U. l irredl1("ihle Represent lltions Sri till" T h t' IH!'lI t O rlhogollal Rela t ion . . . Complete ness o f RCprcsclll a tiollls . 3. 4.5 C'llaracter Tables of Finite Grou ps . 3.·I. G TIle Charwt t'r Ta ble o f ti lle Re pre:;ellta t ioll.'S of 5 U(2) 4.4 . 1 Euler An~ l e; 4.4 .2 Linear Represent ations of SU(2) 4...J. .3 8 phcricaillarmonics FUllct ions . TI ll' Lie T b eor elll~ C leh:;r.h  Gorrlllll C'od fki1'0l2:et· Cl'vslal logHl pbic Point Grou p: CrYstal Systems lind B ra vais Lattice 5.:1. 1 Restr ictio ns on Vec tors of Cn'stal La ttice. 5.3.2 Ttidil1ic Cnstlll System 5.3.3 ~l onod ill ic Ct'I's ta l SI'stcUl 5.3.'1 O r t. ho rhu mb ic Crystal Svslf' lTl 5.3 ..5 Trigonal anJ Hexagonal Crystal System. 5.3.6 Tet ragonal Crystal System .5.3.7 Cnbic Cryst al System S)2:a ce GrouJl 5.4.1 Symmetric E lcments . 5.4,2 Symbols of a Space Group 5. ·L1 1\ lell1Oo for Det.ermining the SPI\[!f' GranPI' 5.4 ...1 Exalll ple fUf the Space Groups ill T ype A . 5. ·1.5 Example for lhe Space Groups in Type B . .5.1.6 Analysis of the Symmetry of II Crystal Lillear Hepre;entlltio\)" of Spa['e Groupo; .5 ..5.1 Irreducible Represelltations of T 5.5.2 S taf o f \ VII.\"e Vect.on; a m! G ronJl o f
W il Y!:: ~rcri nr~
Representa tion ~ [ (l~rkes of Eleme nts in S . In ed uci ble Hepresent.ntiolls of Sfk l ) . T he Blndl Theorelll 5. 5.6 Energy Band in a Cn'sta l . E xercises
5. 5.3 5.5.'1 iiii.5
,
.5 [)
E E B MI II !H IO r,J G BOI IES G.L
/'du l tip li ~ati oll
o f Pe rm utat ions a l jons
6 II
P Cflllii t
6. 1.2 6.1.3
Cvde,; C lasses in a PCt"llllltatiolJ Gro uQ Altf'rnal ing Sll bgrouQ§ Tn u ls[,QI;il iOll o f T wo Nf'igh bored Objed", .
0. 1.'1
6.1. 5
185 185 187 L87 L89 L93
195 L95 L98 L98 L99 200
20·,
205 208 208 211 213 215 2lU 218 220 220 ')')')
224
,,, .. 227 228 ??fl
') :l l 23 L
?al 233
234 236 237
(' '' nl~1!t.,
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x,·ii
Youug Pattems . YOUllg, Tablea ux. and You ng; 0 2cratOis . 237 Yonng PlltierllS 237 Young Tahlf' allX 238 6. 2.3 Young Operators 210 6.2 ...J Fund amental P ro perty of Young 012cra tors 242 6. 2 .5 Prodnd s of Young, Operators . 241 2,16 Irreducihle Represelltations of S" 0. :3.1 Primit ive Idempotenls in the Group Algebra 246 o[ S" 24 9 6.3.2 O rthogon(ll Prilll iti\·f' Idf' JUplJteuts of S n 253 6.3.3 C(llcuill.iioll uf Repre~ell t (ltio ll tll iitr i ce~ fur S" 6. :t4 Calc u la t.ion of Charllcten; bv GraI!hic l\ let hod 257 6 ..3.5 TIl(' Permut at ion Gro up S~ 259 0.3.0 Inner Product of Irreducible Representatio ns 201 or s " n",,) Ort.hogon,,] n" p\""~"'llt "t i nn of S" 262 Outer Product of Irreducih le Represcnt flt ions of S" 268 0.5. 1 Represe ntations of S.,±", ilnd Its Subgroup 2G8 S" QS ", 65') ') 1 1 I,ittiewood Richardson Rll\!6.2 .1 6. 2 .2
6.:3
6.4 6.5
6 fi
L
E xerci se;
LW, GB OIIES AND I IE AI CEBBt!oS 7. 1
7.2
7 .3
7.4
and i t ~ St.rnct ul"e Con" t.a nts 7. 1. 1 The G lohal Propf'r tv of a Lie GrouQ . 7. 1.2 The Locill P roQert\' of a Lie G ro U12 7.1.3 T he Lie Algebnt 7. 1 A T he Killing Form a wl the Cartan Criterill The Regular Form of a Sf>misimple Lie Algebra 7.:2. 1 The Inne r P roduct in a SemisilIl12le Lie Algebra 7.2.2 The Curt aU Sublllgebril 7.:U Rf>glllar Comm utative Relations of C ellPra to rs 7.2.4 TIle Innf'r P roduct of Roots 7.2 ..5 Po:;it i ~·(' Roots and Si mple R oots . C la~sifklltioll of Simple Lie A lgehrll..~ 7.:1.1 Allg,le betwttn T wo Simple Roots 7.:3.1 Dvn kin Diagrmns 7.3.3 T he C,nt all r.lat rix Classical Sililple Lie A igeb r l;l.~ Lie
A l gf'hl"a.~
" 7 :1
') II 277 277 278 28 1 28;3 285 285 286 287
289 292 295
295 2Dti 302
;m2
xvi ii
Gro up Th eory f or Phys ici sls
7 ..J
7.6
7.7
7...1.1 The SU(N ) Group and its Lie Algebra 7...1 .2 Tht" SO(N) Group a.nd its Lie Al~ebra 7.4.:\ Tile USp(2l:) Groll ~ a nd it.s Lif' Algehra . ReprCSf'lltlll iolls of 1\ Simple Lie Algoc bnl 7. 5. 1 ileprf'sent.atiOl ls a nd \Veigbts 7.5.2 Weight Chaill a nd \Vevl Rd lect.ions 7.5.3 i\Iat hemal ical Propt"r tv of Rocpresentations 7. 5.4 FUlHlallJental Dominant \Veight s 7.5.5 TIle Casim ir One rat.or of Order 2 t\'lni!] Dat n of Simple Lit" Algf'brll.'1 7.6. 1 Lie Algebra A ~ and Lie Group SUr (+ I) 7.6.2 Lip AIgeiJrn Be nm\ Lie Group SO (2t' + I) . 7.6.3 Lie Algeura C{ and Lie Croup USp(2f) 7.G.4 Lie Algebra Dr and Lie Group SO(2f) 7.6 ..5 Lie Algebra C z . 7.6.6 Lie Algebra F .l 7.6.7 Lip Algebra E{) . 7.6.8 Lie Aigebra E7 . 7.6.9 Lip AIgf'hra F:s . Bluck Weight DiagTalJj~ 7.7.1 Chevalley Bases 7.7.2 Ort honormal Basis Stutes . 7.7.3 Method of Block Weight Diagram 7.7.'1 Some Representat.ions of A2 , SOllle Representatiolls of C~ I. I .D 7.7.6 P lanar Weight Diagrams Clebseh Gol'Jau Coefficients 7.S.1 Representat ions in the CG Series 7.8.2 l\'let huJ of DUllliuant Weight Diagnull . 7. 8 .3 Reductions of Direct Product Representations ill A2 Exercises

7.S
7.9
!L
llrl [IA BY GBQPE'S 8.1
Irred ucible Rep resent.ations of surN) 8 .1.1 Reduction of a Tensor Space 8.1.2 Basis Tensor:; in the Tensor Suhspace 8.1. 3 Cllcvallc\" Bw;e:; of Gellera('on; ill SU(N)
302 307 309 3 13 31:1 3IG 3 19 320 32 1 322 :323 324 325 325 .127 3"32R .329 :1:10 331 331 333 :335
,
3:37 338 3,12 343 3·1·1 345 :347 350
3f!3 :35:3 354 :l56 362
xix
Ineq uivalent a nd in . . .d lLci ble Reprcsentations Dimensions of Repre;pntat ions of S U(N) Snhciucpd R"presentlltions wit.h Re;pect.. to SUbgTOllpS . Ort honorma l lrr cci lJciblc Basis TJ'llsors 8.2.1 O rt.honormal Ba.~i.:,; Tell.so r~ ill 7Ij~1 8.2.2 Ort.bo l)ormal Bf\..~ js 19usors jn S" Direct P ro duct o f Tensor rtepresentatio ns 83 I O utl·r P ro d ll ci of Tl'll~o t1i 8.3.2 Covaril:l.1lt l\lld COlltnnariant TeJ1~or~ 8 3 3 T J"ilCeles.s t.... ] ixed Te ll ~o n; 8.3.'! Adjoint. Represellt(ltioll of Sl J{N) SU(3) Symmetry and \Vn.ve Funct.ions of Hadrons 8.·1.1 Q uant um Numbers of Quarks 8.4 .2 Pla nar Weig ht DiagL·ams 8.1 ..1 i\1&....~ Formnl"", . 8.4.'1 \Vave Functions of },[esons 8.4 ..') \ Vave Fu nctions of Bflrvolls
8.1 A 8.1. 5 8.1.6
82
8 .3
8...1
8 1)
9
Exer("i~e;
B EA. J. OBTH OGO NA.L C BOI !P S
9.1
Tf'n~or
9. 1.1 9.1.2
9. 1.3 q J.:I
9.2
RepreS PllLIt tiOllS 9. I ..') Adjoint. Representat.ion of SO(N) 9.1.6 Tensor RepresE'ntations of O(N) r Mat rix Gro ups 9.2.1 Property off jI,"latrix Grollps . !l '} .)
9.3
Rep re;;enl atimls or SO U'l) Te ns ors of SO( N) Irredncih je Bllsi" Tem;ors of SO(2f' + 1) Irreducible Basis Te nso rs of SO(2f) D ime nsjo ns of Irredllcible Tensor
The Case
II,,  '} (,
9.2.3 Tlle CasE' N = :u + I . Spinor Representa t ions of SOU\') . 9.3. 1 Cover in g, GrollpS o f SO(N) (1.:3.2 Fundamental Spino1'" o f SO( N) 9.3.;1 Direcluci ble Re pre!;euta tions of USp(2f ) \ 0.1. \ De (:mll po.~ i t io\l o f t he Tellsor Space of USp (2f)
461 46 \ ·163 4GB
10 1 'J Ort.honorm a l hTed llci ble Ba sis TellSors 10 .1.3 D imc t15iollS o f Irreducible Rcpresc uta Liou;; 10.2 Phy ~ici.\ l Applica tion 10.3 EXE'1"cis(';'; .
471
m
Appe ndix A
Identhies on Combina torics
473
Append ix B
Cowlriant and Contravariant Tensors
..\75
Appe lldix C
T11tJ Space Groups
477
B ib/iogm/! hll
'IBI
lnilc.r
487
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physicists
Chapter 1
REVIEW ON LINEAR ALGEBRAS
The main mathematical tool in group theory is linear algebras. In this chapter, we will review some fundamental concepts and calculation methods in linear algebras, which are often used in group theory. 1.1
Linear Space and Basis Vector
Let H(x) be the Hamiltonian of a system. Suppose that the eigenvalue E of H(x) is mdegenerate, µ = 1, 2, · · · , m,
H(x)ψµ (x) = Eψµ (x),
(1.1)
where x briefly denotes the set of coordinates for all degrees of freedom. ψµ (x) are linearly independent to one another. Any linear combination of ψµ (x) is an eigenfunction of H(x) with the same eigenvalue φ(x) =
m X
ψµ (x)aµ ,
H(x)φ(x) = Eφ(x).
(1.2)
µ=1
Conversely, any eigenfunction of H(x) with the eigenvalue E can be expressed as a linear combination of ψµ (x) like (1.2). Two eigenfunctions satisfy the following calculation rule, ! m m m X X X c ψµ (x) (c aµ + c bµ ) . (1.3) ψµ (x)aµ + ψµ (x)bµ = µ=1
µ=1
µ=1
The set of φ(x) is called a linear space L of dimension m, generated by m basis vectors of ψµ (x). φ(x) is an arbitrary vector in L and aµ is the µthcomponent of the vector φ(x) with respect to the basis vectors ψµ (x). 1
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2
physicists
Chap. 1 Linear Algebras
Generally, m objects eµ are said to be linearly independent if there do not exist m coefficients cµ which are not vanishing simultaneously such that m X
eµ cµ = 0.
(1.4)
µ=1
eµ satisfy the following linear formulas: eµ a µ + e ν a ν = e ν a ν + e µ a µ , ! X X X c eµ a µ + eµ b µ = eµ (caµ + cbµ ) , µ
µ
(1.5)
µ
where c, aµ , aν , and bµ are arbitrary complex numbers. The m objects eµ generate a linear space L of dimension m, which is the set of all possible complex combinations a of eµ a=
m X
eµ a µ .
(1.6)
µ=1
a is called a vector in L, eµ is a basis vector, and aµ is the µth component of a with respect to the basis vectors eµ . The space L is called a real space if the components aµ of all vectors a in L are real. A vector is called a null vector if its components are all vanishing. Two vectors a and b are said to be equal to each other if and only if their components are respectively equal, aµ = bµ . In linear algebras, the concepts of vectors and linear space are independent of the physical content of the objects. For a given space L and a given set of basis vectors, vector a is completely described by the m components aµ . Usually, the m ordered numbers are arranged as a columnmatrix a, a1 a2 a = . . (1.7) .. am
The columnmatrix a is another form to denote vector a. Sometimes, we do not distinguish two symbols a and a. A basis vector is a special vector where only one component is nonvanishing and to be one, ( 1 when µ = ν, (1.8) (eµ )ν = δµν = 0 when µ 6= ν,
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§1.2 Linear Transformations and Linear Operators
physicists
3
where δµν is the Kronecker δ function. n vectors a(1) , a(2) , · · · , a(n) are linearly dependent if there exists a linear relation n X
a(i) ci = 0,
(1.9)
i=1
where n coefficients ci are not vanishing simultaneously. Otherwise, they are linearly independent. In an mdimensional space L the number n of linearly independent vectors is not larger than m. In L, n linearly independent vectors generate a subspace L1 of dimension n. A subspace is called a null space ∅ if it contains only the null vector. The whole space L and the null space ∅ are two trivial subspaces. Usually, we only consider nontrivial subspaces. The sum of two subspaces L1 and L2 is a subspace, denoted by L1 + L2 , which contains all linear combinations of the vectors belonging to L1 and L2 . The intersection of two subspaces is a subspace, denoted by L1 ∩ L2 , which contains all vectors belonging to both subspaces. L is said to be the direct sum of two subspaces, L = L1 ⊕ L2 , if L = L1 + L2 , and one of the following three equivalent conditions is satisfied. (1) The intersection of L1 and L2 is a null space. (2) The dimension of L is equal to the sum of the dimensions of L1 and L2 . (3) Each vector in L can be expressed uniquely as the sum of two vectors, respectively belonging to two subspaces L1 and L2 . L2 is called the complement of L1 in L if L = L1 ⊕ L2 . L1 is also the complement of L2 . The complement of L1 in L is not unique. A space L can be decomposed as a direct sum of some subspaces more than two. 1.2
Linear Transformations and Linear Operators
A transformation gives a rule, with which a function changes to another function. An operator is the mathematical symbol for a transformation. An operator R(x) is linear if it satisfies R(x) {c1 φ1 (x) + c2 φ2 (x)} = c1 R(x)φ1 (x) + c2 R(x)φ2 (x),
(1.10)
where the coefficients c1 and c2 are constant. A linear operator describes a linear transformation. The operators used in this textbook are linear if without special indication. The multiplication R(x)S(x) of two operators
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4
physicists
Chap. 1 Linear Algebras
R(x) and S(x) is an operator, defined as a successive application to the function first with S(x) and then with R(x). Namely, if S(x)ψ(x) = φ(x), then R(x)S(x)ψ(x) = R(x)φ(x). Generally, the order of two operators in multiplication cannot be changed, namely R(x)S(x) 6= S(x)R(x). If an operator R(x) is commutable with the Hamiltonian H(x), [H(x), R(x)] ≡ H(x)R(x) − R(x)H(x) = 0,
(1.11)
the application of R(x) to the eigenfunction ψµ (x) of H(x) is still an eigenfunction of H(x) with the same eigenvalue, H(x) {R(x)ψµ (x)} = R(x) {H(x)ψµ (x)} = E {R(x)ψµ (x)} .
(1.12)
Thus, R(x)ψµ (x) belongs to the space L generated by m eigenfunctions ψµ of H(x) with the same eigenvalue E, and can be expressed as Eq. (1.2), X R(x)ψµ (x) = ψν (x)Dνµ (R). (1.13) ν
L is called an invariant space to R(x). The coefficients Dνµ (R) are arranged as a matrix D(R) of dimension m, called the matrix of an operator R(x) in the basis functions ψµ (x) of L, or simply called the matrix of R(x). Note that D(R) depends on the operator R(x), but not on x. The action of R(x) to any function φ(x) in L can be calculated by the matrix D(R). Namely, P P if φ(x) = µ ψµ (x)aµ , and R(x)φ(x) = φ1 (x) = ν ψν (x)bν , one has R(x)φ(x) =
m X
[R(x)ψµ (x)] aµ =
X
ψν Dνµ (R)aµ ,
νµ
µ=1
bν =
m X
Dνµ (R)aµ .
(1.14)
µ=1
Generally, a linear operator R describes a transformation of vectors in a linear space L satisfying R {c1 a + c2 b} = c1 R a + c2 R b.
(1.15)
L is invariant to R if the application of R to any vector a in L is still a vector in L, Ra = b ∈ L,
∀ a ∈ L.
(1.16)
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§1.3 Similarity Transformation
physicists
5
The matrix D(R) of R in its invariant space L is calculated from the application of R to the basis vectors eµ X R eµ = eν Dνµ (R). (1.17) ν
The action of R to any vector a in L can be calculated by D(R). If X X a= eµ a µ , Ra=b= eν b ν , (1.18) µ
ν
then Ra=
X
(Reµ ) aµ =
X
eν Dνµ (R)aµ ,
νµ
µ
bν =
X
Dνµ (R)aµ ,
b = D(R) a.
(1.19)
µ
It is worthy to emphasize the difference between Eqs. (1.17) and (1.19). In Eq. (1.17) a basis vector eµ transforms in the operator R to a combination of basis vectors, where the combination index of the basis vectors is the row index ν of Dνµ (R). Equation (1.19) is a component equation for vector a transformed by the operator R to another vector b, where the combination index of the vector components is the column index µ of Dνµ (R). Two equations are consistent because the basis vector eµ is a special vector, where only one component is nonvanishing but equal to one, X X (Reµ )ρ = Dρλ (R) (eµ )λ = Dρµ (R) = (eν )ρ Dνµ (R). (1.20) λ
1.3
ν
Similarity Transformation
For a given set of basis vectors eµ in a linear space L of dimension m, there is a onetoone correspondence between vector a and its columnmatrix a, and there is a onetoone correspondence between an operator R and its matrix D(R). However, the basis vectors in L are not unique. Any set of m linearly independent vectors can be chosen to be basis vectors. In this section we will discuss how the columnmatrix of a vector and the matrix of an operator change when the basis vectors are changed. Let e0ν be m linearly independent vectors with the components Sµν in the original basis vectors eµ ,
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6
physicists
Chap. 1 Linear Algebras
e0ν =
X
e0ν = S·ν .
eµ Sµν ,
µ
(1.21)
Since e0ν are linearly independent, S is a nonsingular matrix (det S 6= 0) and has its inverse matrix S −1 . X (1.22) eµ = e0ν S −1 νµ . ν
e0ν
Choosing to be new basis vectors, the components a0ν of the vector a and the matrix D(R) of the operator R can be calculated as follows: X X X e0ν S −1 νµ aµ = e0ν a0ν , eµ a µ = a= νµ
µ
a0ν =
X µ
Re0ν =
X
S −1
ν
νµ
(Reρ ) Sρν =
X
e0ρ Dρν (R) =
Dµρ (R)Sρν =
ρ
X
eµ Dµρ (R)Sρν ,
X
eµ Sµρ Dρν (R),
µρ
ρ
X
X
(1.23)
µρ
ρ
Re0ν =
a0 = S −1 a.
aµ ,
Sµρ Dρν (R),
D(R) = S −1 D(R)S.
(1.24)
ρ
The relation (1.24) between D(R) and D(R) is called a similarity transformation and S is the matrix of the similarity transformation. In literature, D(R) and D(R) are said to be equivalent to each other if Eq. (1.24) holds. Obviously, the matrix S has the same matrix form with respect to both the original set and the new set of basis vectors. If the new set of basis vectors is the same as the original one except for the order of basis vectors, the similarity transformation is called the simple one. Note that, the similarity transformation for two equivalent matrices is not unique. If X is commutable with D(R) and Y is commutable with D(R), both XS and SY satisfy the similarity transformation relation (1.24). Since S·ν is nothing but the column matrix of new basis vector e0ν in the original basis vectors, Eq. (1.24) can be written as X X D(R) S·ν = S·ρ Dρν (R), Re0ν = e0ρ Dρν (R). (1.25) ρ
ρ
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physicists
§1.3 Similarity Transformation
7
It is nothing but the definition of the matrix of R in the new basis vectors e0ν . The different choices of the basis vectors do not change the action of an operator on a vector. If b = Ra, one has b = D(R)a in the original basis vectors eµ . In the new set of basis vectors e0ν one has b0 = S −1 b = S −1 D(R)a = S −1 D(R)Sa0 = D(R)a0 .
(1.26)
Let L be an mdimensional space, L1 be its ndimensional subspace, invariant to the operator R, and L2 be the complement of L1 . Choose a new set of basis vectors in L such that the first n basis vectors belong to L1 , and the next (m − n) ones belong to L2 . Arrange the new basis vectors e0ν to be the column matrices of S, S.ν = e0ν . Through the similarity transformation S the matrix D(R) of R is changed to D(R). Since L1 is invariant to R, one has Re0µ =
n X
e0ν Dνµ (R),
1 ≤ µ ≤ n.
ν=1
(1.27)
Namely, the downleft corner of D(R) is vanishing, Dρµ (R) = 0
S
−1
when µ ≤ n < ρ,
D(R)S = D(R) =
D(1) (R)
M
0
D(2) (R)
!
.
(1.28)
This matrix D(R) is called a ladder one. Furthermore, if L2 is also invariant to R, one has M = 0, D(R) =
D(1) (R) 0 0 D(2) (R)
= D(1) (R) ⊕ D(2) (R).
(1.29)
This matrix D(R) is a direct sum of two submatrices, and is called a block one in the type [n, (m − n)]. Generally, a matrix is also called a block one if it can be changed into the direct sum of submatrices by a simple similarity transformation. In order to determine whether or not a matrix is a block one, one may separate the indices of the matrix into two parts and check whether the matrix entries with indices respectively belonging to different parts all are vanishing. If Λ is diagonal and XΛ = ΛX, X is a block matrix.
September 11, 2007
8
1.4
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Chap. 1 Linear Algebras
Eigenvectors and Diagonalization of a Matrix
In quantum mechanics, the eigenequation of a physical operator R(x) is R(x)ψ(x) = λψ(x). The eigenvalue λ is the possible observed value for the physical quantity. The eigenvalues describe the characteristic of the physical quantity and are independent of the choice of basis functions. In linear algebras, the eigenequation of an operator R is R a = λ a.
(1.30)
If L is an mdimensional space invariant to R and a ∈ L, one has X Dµν (R)aν = λaµ . D(R) a = λ a,
(1.31)
ν
The eigenequation (1.31) is a set of coupled linear homogeneous equations for m variables aν . The condition for the existence of nonvanishing solution is its coefficient determinant to be vanishing, det [D(R) − λ1] = 0.
(1.32)
Equation (1.32) is called the secular equation for D(R), and its roots are the eigenvalues of D(R). The secular equation is invariant in similarity transformation, so that the eigenvalues are independent of the choice of the basis vectors. It is easy to see from Eq. (1.32) that the sum of the eigenvalues of D(R) is its trace, Tr D(R), and the product of the eigenvalues of D(R) is its determinant, det D(R) (see Prob. 1). Equation (1.31) means that the eigenvector generates a onedimensional subspace which is invariant to R. If in L there exist m linearly independent eigenvectors a(ν) of R with the eigenvalues λν , respectively, one may choose a new set of basis vectors e0ν = a(ν) . Then, the matrix of R in the new basis vectors e0ν is a diagonal one. S −1 D(R)S = Λ,
S·ν = a(ν) ,
Λνµ = δνµ λν .
(1.33)
Therefore, the key for diagonalizing an mdimensional matrix is to find its m linearly independent eigenvectors. Since the eigenequation (1.30) is linearly homogeneous with respect to the eigenvector a(ν) , a(ν) can be multiplied with a constant cν . When the eigenvalue is degenerate, its eigenvectors can be made a nonsingular linear combination. This is the reason why the similarity transformation matrix S is not unique. The number of the
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§1.5 Inner Product of Vectors
physicists
9
P arbitrary parameters in S is equal to ν n2ν , where nν is the multiplicity of the eigenvalue λν . Those parameters play an important role in calculating a common similarity transformation for a few pairs of equivalent matrices, which is often used in group theory. Substituting an eigenvalue into Eq. (1.31), one is always able to solve at least one eigenvector. The eigenvectors for different eigenvalues are linearly independent. However, when an eigenvalue is a root of Eq. (1.32) with multiplicity n, it is not certain to obtain n linearly independent eigenvectors with the given eigenvalue by solving Eq. (1.31). The following matrix is the simplest example which has the eigenvalue 1 with multiplicity 2, but only one linearly independent eigenvector, 1 b 1 1 = , b 6= 0. (1.34) 0 1 0 0 It cannot be diagonalized by a similarity transformation. The eigenvalues are invariant in a similarity transformation. Therefore, two equivalent matrices can be diagonalized into the same matrix, X −1 D(R)X = Λ = Y −1 D(R)Y, such that the similarity transformation related to them is easy to be calculated, −1 D(R) = XY −1 D(R) XY −1 . (1.35) It can be proved that the sufficient and necessary condition for a matrix D(R) which can be diagonalized by a unitary similarity transformation is that D(R)† is commutable with D(R) (see Prob. 11). A matrix H is called Hermitian if H † = H. A matrix u is called unitary if u† = u−1 . Both a Hermitian matrix and a unitary matrix can be diagonalized by a unitary similarity transformation. A real unitary matrix is called real orthogonal. A real Hermitian matrix is called real symmetric. A real symmetric matrix can be diagonalized by a real orthogonal similarity transformation. Generally, a real orthogonal matrix can be diagonalized by a unitary similarity transformation, but not a real orthogonal one. 1.5
Inner Product of Vectors
There are three types of products of vectors, depending on the products to be a scalar, a vector, or a tensor. The product of two vectors is called the
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10
physicists
Chap. 1 Linear Algebras
inner product if the product is a scalar. In quantum mechanics the inner product of two wave functions is defined as Z hφ(x)ψ(x)i = (dx)φ(x)∗ ψ(x), (1.36) R where the sign denotes an integral for the continuous coordinate and a sum for the discrete coordinate. The form on the lefthand side of Eq. (1.36) is called the Dirac symbol. Generally, the inner product of two vectors habi in the linear algebras satisfies hc1 a(1) + c2 a(2) bi = c∗1 ha(1) bi + c∗2 ha(2) bi, hac1 b(1) + c2 b(2) i = c1 hab(1) i + c2 hab(2) i, hbai = habi∗ ,
(1.37)
haai = a2 > 0, if a 6= 0.
The inner product is linear for the second vector and antilinear for the first vector. The inner product becomes its complex conjugate if changing the order of two factors. The selfinner product of a nonvanishing vector is real positive, called the square module of the vector. Denote by a Hermitian matrix Ω the inner product of two basis vectors (1.38) heµ eν i = Ωµν , Ωνµ = Ω∗µν = Ω† νµ . Let a be an nonvanishing eigenvector of Ω with the eigenvalue λ, X X a= eν aν 6= 0, Ωµν aν = λaµ , ν
λ
X µ
aµ 2 =
X
ν
a∗µ Ωµν aν =
µν
X µν
a∗µ heµ eν iaν = haai > 0.
(1.39)
Hence, the eigenvalue λ of Ω is positive, namely, Ω is positive definite. The inner product of two arbitrary vectors and the matrix entry of an operator can be calculated with Ω, X X X a= eµ a µ , b= eν b ν , habi = aµ Ωµν bν , (1.40) µ
X ρ
Ω−1
ν
µρ
heρ Reν i =
X ρτ
µν
Ω−1
µρ
heρ eτ iDτ ν (R) = Dµν (R).
(1.41)
A vector is called normalized if its module is one. Two vectors are orthogonal if their inner product is zero. Two nonvanishing orthogonal vectors must be linearly independent.
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§1.5 Inner Product of Vectors
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11
The basis vectors are called orthonormal if heµ eν i = Ωµν = δµν .
(1.42)
In the orthonormal basis vectors the formulas for the inner product become simpler, X habi = a µ bµ , heµ Reν i = Dµν (R). (1.43) µ
There are a few different definitions for the inner product of vectors. Another inner product is defined to be linear for both factors, hc1 a(1) + c2 a(2) bi = c1 ha(1) bi + c2 ha(2) bi,
hac1 b(1) + c2 b(2) i = c1 hab(1) i + c2 hab(2) i,
heµ eν i = Ωµν = Ωνµ ,
(1.44)
det Ω 6= 0.
In this definition, the selfinner product of a vector may not be real. The inner product of the column matrices has been defined, namely X X a∗µ bµ , aT b = a µ bµ . (1.45) a† b = µ
µ
In comparison with Eq. (1.40) the inner product (1.45) means that the basis vectors for the column matrices are orthonormal. At last, we discuss the concept of the adjoint operator R † of an operator R. The conjugate matrix D(R)† of a matrix D(R) was well defined, D(R)† µν = Dνµ (R)∗ , (1.46) ∗ † † a D(R) b = [D(R)b] a = b† D† (R) a.
(1.47)
haRbi∗ = hRbai = hbR† ai.
(1.48)
The definition for the adjoint operator in quantum mechanics is the generalization of Eq. (1.47),
The adjoint relation between two operators is mutual. Note that the matrices of two adjoint operators are not necessary to be conjugate if the basis vectors are not orthonormal. Denote by D(R) and X the matrices of two operators R and R† , respectively X X Reµ = eρ Dρµ (R), R † eν = eρ Xρν , ρ
ρ
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physicists
Chap. 1 Linear Algebras
X ρ
∗ Dρµ (R)heρ eν i = hReµ eν i = heµ R† eν i = †
D (R)Ω = ΩX,
X=Ω
−1
†
X ρ
heµ eρ iXρν ,
(1.49)
D (R)Ω.
Conversely, if the basis vectors are orthonormal, two mutual conjugate matrices correspond to two operators adjoint to each other. Namely, in this case, a Hermitian (or an unitary) matrix corresponds to a Hermitian (or a unitary) operator. 1.6
The Direct Product of Matrices
If a quantum system consists of two subsystems, the wave function of the system is expressed as the product of two wave functions of the subsystems, or the combination of the products. Suppose that the two functional spaces L1 and L2 for two subsystems are respectively invariant to the operator R, Rψµ =
m X
(1) ψν Dνµ (R),
ν=1
Rφi =
n X
(2)
φj Dji (R).
(1.50)
j=1
For the composed system, the functional space L generated by ψ µ φi ,
1 ≤ µ ≤ m, 1 ≤ i ≤ n
(1.51)
is called the product of two subspaces, L = L1 L2 , which is (mn)dimensional. The space L is also invariant to R, i.e., h i X , R (ψµ φi ) = (ψν φj ) D(1) (R) × D(2) (R) νj,µi (1.52) ν j (1) (1) (2) D (R) × D(2) (R) νj,µi = Dνµ (R)Dji (R). The matrix D(1) (R) × D(2) (R) of R in L, which is (nm)dimensional, is called the direct product of two submatrices D (1) (R) and D(2) (R). The row (column) of the direct product matrix is denoted by two indices µ and i. The order of indices is usually arranged such that the second index i increases for a given µ, and then the first index µ increases. For example, the direct product of two 2dimensional matrices X and Y is X11 Y11 X11 Y12 X12 Y11 X12 Y12 X11 Y21 X11 Y22 X12 Y21 X12 Y22 X11 Y X12 Y X ×Y = = X21 Y11 X21 Y12 X22 Y11 X22 Y12 . X21 Y X22 Y X21 Y21 X21 Y22 X22 Y21 X22 Y22
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Exercises
physicists
13
The product of two matrices D (1) (R) × D(2) (R) and D(1) (S) × D(2) (S) is (1) D (R) × D(2) (R) D(1) (S) × D(2) (S) (1.53) = D(1) (R)D(1) (S) × D(2) (R)D(2) (S) . Thus,
D(1) (R) × D(2) (R)
−1
D(1) (R) × D(2) (R)
†
D(1) (R) × D(2) (R)
T
= D(1) (R)−1 × D(2) (R)−1 ,
= D(1) (R)T × D(2) (R)T ,
(1.54)
= D(1) (R)† × D(2) (R)† .
The trace and the determinant of direct product D (1) (R) × D(2) (R) are Tr D(1) (R) × D(2) (R) = Tr D(1) (R) Tr D(2) (R) , (1.55) n m det D(1) (R) × D(2) (R) = det D(1) (R) det D(2) (R) . If two matrices D (1) (R) and D(2) (R) depend on a continuous parameter α, one has i d h (1) D (R) × D(2) (R) dα (2) (1.56) dD(1) (R) dD (R) (2) (1) × D (R) + D (R) × . = dα dα
The direct product reduces to the product of a number and a matrix if one of the two factor matrices is onedimensional. Generally, D (1) (R) × D(2) (R) is not equal to D (2) (R) × D(1) (R), but their difference is only a simple similarity transformation. For example, when n = m = 2, the similarity transformation matrix for the two direct products is 1 0 0 0 0 0 1 0 (1.57) 0 1 0 0. 0 0 0 1 1.7
Exercises
1. Prove that the sum of the eigenvalues of a matrix is equal to the trace of the matrix, and the product of eigenvalues is equal to the determinant of the matrix.
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14
physicists
Chap. 1 Linear Algebras
2. Calculate the eigenvalues and eigenvectors of the Pauli matrices, 0 1 0 −i σ1 = , σ2 = . 1 0 i 0 3. Calculate the eigenvalues and eigenvectors of the matrix
0 0 R= 0 1
0 0 1 0
0 1 0 0
1 0 . 0
0
4. Calculate the eigenvalues and eigenvectors of the matrix 0 0 1 R = 1 0 0. 0 1 0 5. If det R 6= 0, prove that both R† R and RR† are positive definite Hermitian matrices. 6. Prove: (1) if R† R = 1, then RR† = 1; (2) if R−1 R = 1, then RR−1 = 1; (3) if RT R = 1, then RRT = 1. 7. Find the independent real parameters in a 2 × 2 unitary matrix, a real orthogonal matrix, and a Hermitian matrix, respectively, and give their general expressions. 8. Find the similarity transformation to diagonalize the following matrices: √ 1 √1 − 2 √ (1) 2 √0 − 2, 2 1 1
(2)
cos θ − sin θ sin θ cos θ
.
9. Find a similarity transformation matrix M which satisfies M
−1
0 −1 0 0 − cos θ sin θ sin ϕ cos θ 0 − sin θ cos ϕ M = 1 0 0 . 0 0 0 − sin θ sin ϕ sin θ cos ϕ 0
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Exercises
15
10. Find a similarity transformation matrix M which satisfies the following three equations simultaneously
0 −i 0 1 0 0 M −1 i 0 0 M = 0 0 0 , 0 0 0 0 0 −1 0 1 0 0 0 0 1 M −1 0 0 −i M = √ 1 0 1 , 2 0 1 0 0 i 0 0 0 i 0 −1 0 i M −1 0 0 0 M = √ 1 0 −1 . 2 −i 0 0 0 1 0 11. Let R=
1 0 0 −1
S=
,
1 2
√ −1 − 3 √ . 3 −1
Find the common similarity transformation matrix X satisfying
1 0 X −1 (R × R) X = 0 0
0 −1 0 0 2 0 1 0 2 X −1 (S × S) X = 2 0 0 0 0
0 0 0 0 , 1 0 0 −1 0 0 0 0√ . 3 −1 − √ 3 −1
12. Find the similarity transformation matrix X ing three matrices simultaneously, 000100 000100 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 , , 1 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 001000 010000
to diagonalize the follow
0 0 0 1 1 1
00 00 00 11 11 11
111 1 1 1 1 1 1 . 0 0 0 0 0 0 000
13. Show the general form of an m×m matrix, both unitary and Hermitian.
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Chap. 1 Linear Algebras
14. Prove that any unitary matrix R can be diagonalized by a unitary similarity transformation, and any Hermitian matrix R can be diagonalized by a unitary similarity transformation. 15. Prove that R and R† can be diagonalized by a common unitary similarity transformation if R† is commutable with R. Further prove that the necessary and sufficient condition for a matrix R to be diagonalized by a unitary similarity transformation is that R† is commutable with R. 16. Prove that any matrix can be transformed into a direct sum of the standard Jordan forms, each of which is in the form when a = b, λ Rab = 0 or 1 when a + 1 = b, 0 the remaining cases.
physicists
Chapte r 2
GROUP AND ITS SUBSETS
Croup tllL'Ory is a powerful tool for studying the symmetry of a physical ,;y,;telll. In tili,; chapler. the lIIatil ellla~ical definit.iun of a group will be ahstracted from t he common property of the ~pts of the symmetric transforma lions of physical systems. Somt' simple example'! are e xplained to give the readers a concrete understanding 011 groups. The group table, which is and 1>.1 are t he wave functions with the even and odd parit.ies. rcspectivcl.~·. No m atter how t he detail propel"t~' of a system i". ollly if t.he s)'Htelll iH invariant ill Ihp spat.ill.l illvt'rsioll , till" wave fllll!"tioll of any st(l.tionary state of the s.\"Stem (the eigenfunction of the Hamiltonian) can be chosen with a definite parity. Hamely. the parity is conserwxl and can be used to cla.;sify the wavc fuudions of statiollary st.ates of t he sysi.;>lJl. Furt hermore, tIll' pleclrir: dipole tTl;lllsicio]l bet.we€J1 t.wo states wit.\! thp
§2.2 Group and its M"Uiplication
Tabl~.
same parity is s uppressed. T his is t he select.ion rule for the electric dipole transition in q uanium mechil. llic~. 'The simplf' f'xample shows thflt some prf'cise in for mfltion of a systf'm can be obtained from its symmetry although the Schrociinger equation is hard to be solved.
2.2
Group and its Multipli cation Table
A trall~fon l Hl.tion i,. cl.llh~ d t he ~Yllllllf't.ric tran~ formation of a system if it preserves the system invariant.. The set of the symmet.ric t. ransformat ions of a s~'stem charact erizes the symmetr.v of t he s~'stem. There are some ("omnlOn properties ill t h e seb of ~Yllll lletl"ic t r3i1sfonnat iol1s . In p hy~ics. tllf' IIm lt.iplicfltion RS of t.wo t.TIIl\sfol"lllll.t.iollS R IIno S is usnfllly of'fhlPO liS successive applicat.ions first. by S alld then by R. T hus. t.he multiplication of two symmetric transformations is still a symmetric onc of the S;.·stCIIl. The lIlnhiplicatiulI rnle saiislie;; the a.~sociali"e Iil.w. The idenlical (rausforIl1fltion E. which presen'el3 evprything il1\·ariant.. ~~ fl ~ymmetric one of flny system. For allY symmetric tramfo rmation R of the system. the multiplication ER is equal to R. The illVetbC of a symmetric transformation is also a sywlllet ric one of the syst.elll. Those prop{'ftie>; are eRwutitd (l.uri eOUlIllOI\ for tllP Sf't of s.vmnl("tric transformations of evpr." system. and can he used to dt'tille a group.
n
D e finition 2.1 A group G is a set of elements satisf:ving the following four a.xiotn~ with respect to the givcn tlIultiplication !"Ulc of clcments. (a) Tltt' set is close[l to this 111l11t iplic(l.t.i[lll.
RSEG,
V R aud S E G.
(2 .2)
(b) The multiplication rule of clements satisfies the associati\'c law,
R(ST) ([') T he
~f't
~
( RS)T,
V R, Sand T E C.
(2,3)
contfl.ins an irientiea I f'leme11t E E G sat.isfying
ER = R.
'r/ REG.
(2,4)
(el) For a11.\' elen1('nt REG, the set cont ains its inverse R 1 satisfying
IrtR = E,
'r/ REG.
(2,5)
20
Chap. 2 Group and Its Subsets
In Definition 2.1. the clements can be an,\" objects and the multiplication mle of e lements can he defined arbit.rarily. The main thillg' for Ii group i~ tlHlt. tilt:' set. sll.tishf'S four fixioms with resppl't to the given 1l11l1t.iplicat.ion rule. T he set of symmetric transformations of any s~'stem wit.h rt?Spect to the llluitiplication rule of transformations satisfics four axioms. so that thc set is a gl'OUp. called t.he s.)"mmetric group of t he system. In most· cases in physics, the deillellts of a group arc trallsfol"lnations. opcrators, 01' matrices. If wit.hout. spf'cifi('atio]!, the 1I1111t,ipl icatioIl rul(· i~ d('/iIlP SHllle. COllvl'rsf'ly. t.wo i;;olllorpliic groli P" lIlay haVl' different gnmp tables if t.he enumerations are unsuitable. Rearnmg,ement. theorem restricts strongl.\· the number of nonisomorphic groups with a given order g. For ex,l.lu ple, rearrangement t heorem cOlllpletely deterlllilleti the group tables of a group with t he orde r 9 = 2 or .'1. HS given in Table;; 2.1 and 2.2.
The g roup t a ble of V 2
Table 2.1
'1 2
C
0"
, , a
Table 2 .2
",
"
The group table of C 3
, , ,
C, w w'
w w'
w
w'
w
w'
w'
,
,
w
v 2 , c(l.lled the inversion group of order two, is the symnlf't.ric group of a system whkh is invariant iu tIle spatial inversiotl 17. V 2 is isomorphic outo C 2 constructed by two Hlunbers 1 and 1 with rei;]Jcct to the lllllitiplieatiull rille of lllllJlbers. Grollp C J iH COllstrllcted hy th rl'l' ('olllplex llI111lbers I' = I. w = exp(  i27T/3). and ",/ = exp(i2;;j:l) with re>pect to the multiplication rule of complex numbers. Ill\'2 uue has f = 171 . Each ei(o')llellt in C J b equal to a I'U\\'(o'1" ofw, l>ecan~(o'''';' = ",,2 and f' = ",..1. Generally, a tillite group ("o]l"tnwh.>d hy t ill' powers of one element R is called the cyclic !!,TOUp C ,,,,' , R'\'lj ,
(2.10)
where ]\,' is the order of C N . C,\,' is ilU Abelian group. The gTOUp table of call ue fil\(o'd as folluws . Arrallge the eiel!lellt.,; ill the ot"der: E , R, R~, .... R N The elellll'uts ill the /irHl row of the tahle are t.lll' ,;fllne 11." tho~1'
eN
J
•
§2.2 Group and its M"Uiplication
Tabl~.
23
in its top line (right mult.iplication clements), rC8pC('tivcl.I'. The clements in each row al"e obt ailled by Ill uvillg t he elemelilb ill the precedillg ruw left ward by one ho:: , flllfl moving the leftm()l';t elemf'ut to the r ightmost position. A typical physical example for the cyclic group C N is the N  fold proper axis in t.he theory of c rys tals. A pure rotation in threedimensiollal space is called a proper rotation. allli a proper rotation lIlultipliell by a spat ial inversion (1 is railed t.he impropf'l" fOt.flt.ioll. If Ulf' rotation R aronnd a givf'n axi~ thwuglJ the il.llgle 2rr/N is a ~Yllllllctric tn\ll~fonnatioll of a cry~tal. tllf' cydic group CN I!;eneratffi by R belong to t.he symmetric )l;roup of t.he crystal, the axis is its .iV fold proper axis. and R is its Nfold proper rotatiou. Thc Ifold pwper rotatiou is the identical trausformation E. and the group C l iH t. rivial. The dirpdioH of UIP 2fold proppr axis doe:< HOi. maCtef because R = R  1 • The direct.ioll of the Nfold proper a,is with I\' > 2 is essence for determining the generator R. Siuee the ulJluuer of clelm;lJt~ iu a fiuite gruup b Ihllited, the powen; uf t.o D' ill t.h .. h"n~fonll"t;ioll D. NllllWly, ill t.lllllellt.s H fi nd S helonging to thf' sallie le ft coset is /i iS E 11. and t.hat. for them bdonging to t he same right coset is RS  1 E fl . For a finite group. the number of clements in auy coset is equal to the order II of t he subgt·oup H . and the group G is equal tu t he uuion uf the subgroup H and its left {:(lspt :; (or right (·o..;ets) , to
'Ha
§2,$ Subsets
C =
;
"
)' = 1
)= 1
GrollP
U Rjll = U 11 R j ,
(2. 15 )
T hus , the ordcr h of a subgroup H is fI whole number divisor of the ord~r 9 of the group G (LagTange Theorem), 9 = hd. d is called the index of a subgroup, awl the Humber of t he left crn;et~ (or right cosets) of H in G is d  I . A gronp G riops not contain allY 1I0ntriviai snhgronp." if its orripr 9 is a prime number. It leads that two grou ps are isomorpllie if their orders are of the same prime number. [t is t:llliy to fillJ .. lit: COM'ts of a subgroup of a fi ni te group G ill lel'lm; of its grou p tahle. In t.he group tRblp, the set of plement.'; in pach row of the columns related to the elements in the subgroup is the subgroup itself or its left. coset (Rj II ). and the set of clements in each colulIlll of "he rows ["elak,,] to the e lements of the subgroup is the subgroup itself OJ' its right {'(Y.';pt (H R)). Por examp!p, it cfln hI' S ff'n from Tahir' 2.6 that tlw sllhgrollp {E, A} in the group 0.3 COllTaillS two left coset.s, {D , B} and {F. C}, and two rig ht coset,; . lD, C} and {F, B}. The left cosets of the subgroup in O:J lire not "'1""\ to its ril!,ht c o"e(". Allotber suhgrO lljl tE, D, Fr wit I. iudpx 2 has only one coset {A, B, C}.
2,3.3
Conjugate Elements and th e Clas s
T he p]plw'nt. 8' = TSr  1 is said to hI' ('onjuga.te t.o S in a group G where 5, 5' , and T all belong to C. The conjugate rela tion of t wo elements is mutual. If two clemcnts arc both conjugate to a third clement, thcy tlrc a\;;o cOlljngate to each other:
Tlw snhs!,t, of all nmt,nally cOlljngalp Co in G
elplll~nt.s ill a gro np (,' is ('alle!\"
ria.""
(2.16) where n(o) is t he number of elements cOlltained in the cla~s Co. There is no COllllllOl! elelllelit in two d ifferellt classes . The identical clemcllt E itself f01'111;; a dll.;;s.
Thprefofp, t.ilp other
da.~s
doeH 1101. cOlllll.ill t h e idpllt.ical
eh"ment and is not (l subgroup. For an Abelian group, each element itself forms a class, and the number g, of classes in an Abel ian group is equal to the order 9 of t he group. FOI' a gil'elJ deliH'lIt T E C. T5 j T 1 f TSkT  l if S;
i= Sk,
so thaI.
Chap. 2 Group and Its Subsets
TC a T  1
_
C".
V T E G.
(2.17)
Theo renl 2.2 Dellute by /I (u) tbe llu!lIber uf eletJIenu; ill a class Co. uf a finite group G with I.he oroer g. ami oenolt' hy 11/(",,) t he numher of T E G satis(yillil, TBjT 1 = SI.' , where B j and S" are two given elements in then 9 = n(o )m(o). a nd m(o) is independen t of Bj and 51.. The key for the proof if< to defille A. snhgro1!p H v,'hose eieillellts are commutnhle with an ..l"lllent Sj in Cc>' Any .. If'nlellt in t.h .. left coset TH satisfies T8)T  I = Sk. T he detailed proof is left as exercise (see Prob. 1,,\). T hc inverse Sj l of an clcment Sj ill a class Co. is conjugatc to the inl'ClbC SkI uf IIllot.her t'iemellt S1' E Cet :
e,.,.
T hus , the set of STJ also comtructs a class . called tIle reciprocal class
d. 1)
of Cn • c1 1) COlltll illS the same number of elements llH C n • T he reciprocal relation of two classes is tllutual . If a class C" contains the inverse S:; 1 of its delll'mt Sj , thell Co = Ci,I ). "1I.11",(] t lu? self~l"ecil'l"o H'" liIeans the mauytooue corl"csponJeuce bctweel! Ii ill G ami Ii' in G'. T lwre are different manytaone correspondences hetween elements of two groups. A group G is homomorphic outo another p'oup G' if and only if there exi1;t~ a mauytoone eorret;pouJelJ(;e whidl is ill\'ariaut to the ulUltiplieatioll rule of e lements. O lJe CI.I111101 lIIake cOlJclllsion lhat two gf()\l~
Chap. 2 Group and Its Subsets
arc NOT homomorphic only based Oil that there is a manytoone correspoudellce hetween twu groups which is uot iuvlI.riaut to the multiplicfll.iun rnle of el('ITI are tlw Ifold proper axes of the cube. but the 2fold axes of the tetrahedron. Denote b:.' T JJ , /1  x. y. ': , the rota~iOllti around the coordinate ax are denoted by AJ aud B j • 0 :; j :; 5. Aj are located above t he x y plalle. The za xis i" ill the difl~di(jll from t ht' origin 0 to ti lt' vertex Au. The ,,axis is in the direction from 0 r.o the midpoint of t,h(' ('dg,e A 2 Bf ,. The regular icosahedron contains six 5fold axes. tell ;}fold axis. and fifteelJ 2fold axes. Thtl 5,fold axe; are along, the d irections fro lil B j to AJ with tile gt'lJern tors T J • 0 :; j :; 5. Ont' 5fold axis (j = 0) i,; Hio llg tilt'
§2.5
Propc~
S!lmmctric
G ro ~ p3
oj Regular Polyhcdro fM
"
positive :ltxis. T he polar allglcs of t lu'" remaining 5fold aXL'S all a rc 0 1 • thei r ltZimlitha l a n:;.le; al'e 2(j  1)r./5. re; pect.i\·ely. T he 3 £01.1 a xes are alo ng t he linp.:> connf'ct ing the centf'rs of two opposilf' t r iallF;1es wi t h t he ge ne rittol"S RJ . I :5 j:5 10. T he polar angles ofthe 3fold a xes a re O2 when 1 ::; ) ::; 5. and 03 wllc n 6 :5 j :5 10. Tilcil" azimut hal a ngles res pecti vely are (2)  1};;/5. The 2fold axes a re along t he l ines conllecting t ilt'" midpoints of twu oppus it e edgcs witll t he gencmto r:;; S)" 1 :5 j:S;: 15. T he ].X>Ln angles of flU' 2 fol,1 I\Xt'!< 1\1"t' O~ wlwn I ::;: j :5 5. Oa whell 6 :5 j :5 10. 1\11(\ rr/ 2 wJ]('11 IJ :5 j::; 15. T heir azimu t hal flllgies are 2(j  1) .. /5 when I "$ j 5. (2)  l) rr /5 wheII (j S j :5 10. and (l)  :1)rr/ 10 when 11 :5 ) :5 15. fL>:;].J= t ive i.\". All proper axt'li are nonpular. allJ allY twu axt'li wi t h Ihe sallie fold a ff' e'1nimiellt to eac h olher. T hO/je polar angleli are caku laleU in Proh. [2 of Chfl p. u t.n llt' i iL.. Il1lil. t ht' Tadius H of ib CiJ"{;lI111circII'
12
and lhe rad iu... r o f its illSCri bcd circle arc (also ~
R
(
5 I .j5 '1' 8) = 0.951 1.
E
S.
;';12
5,
E
E
S,
s,~
S,
S.
S. E 8'2 S,
S, 8" E
S.
8,
S,
Sl~
S.
E
R. R,
R. R,
m
R,
R'f1J
R.~
R'" • R..
,
n',
R'
R,
R,
Table 2.13
S,
A,
S,
.\~
A, A, A.
S~
A,
1l Ilu clCHll'1I1 with order 6 ill I. it L'" incuuvl'niclIl to '·Olt."t rllc t .. l.'OI' direct proflur't of two groups ont' OftPll lTIPt't,s ill the real problems is that the groups HI and Hz are two sets of operators rcsp~'{;ti\'cly affecting two different subsystcms so that the clclllents in the twu group~; are t:omnmtab le, RJSJ, = SJ,Rj . Deline two grolll'" HIS I ;:::; HI lind R\H2 ~ H 2 , when" HI and 51 lire the irient.icill plenH'nts of HI and H 2 , respectively. The element R1S I is the only COlIllllon element ill two groups H 1 S[ and R I H 2 . T he set
(2 .29) satisfies fonr axioms llnder thp mllltiplicatioll rnles of plements in If I lind Hz . where the identical element is R 1 S l . This group G is lhe direct product of two groups. Hl and H 2 •
2.6.2
Tm proper Point Groups
An improper point group G contains both improper and proper rotations. Au improper l"otatiou S' is a product of a proper rotatiolJ Saud a spat ial ill\'er~ioll u. (1 is COlllll1utHule witll Hlly rotation aud iL~ ~q]( a]"e is ('qunl to
45
§2.6 imprO[>"T Poi nl Group,
the identical clement E,
s'
= uS = Suo
(2.30)
The "lllJl lllen l~ i" defilled witl l tht> prodnet of d igitH, is isomorphic onto (l group, composed of all real numhers where the Illultiplication rule of elements is defined with the addition of digits. 3. If H I and H2 are two subgroups of a group G, prove that the COllllllon elemen ts in H1 and H2 abo form a subgroup of C. 4. ProvO" t.hflt, a gronp whosO" ordO"r .rJ is a pr ime nlllnhO"r mnst. hO"
group C q .
fI
c}'c\if'
5 . Show t hat up to isomorpllism. t llcn' a rc only IWO diffc rent fourt llo rd(r group,;: T ile cycl ic gronp C~ itud t ile fom·thorder inver·doll gTon p. 6 . Show t lll!.t up l\) isolllorphi.slll _ t here are ouly two d ifferent sixtllorder gTOlIJlS: TIll" ryeli r grou p ('6 Imcl I Ill" symmf'tI"ir g·ronp 0 :1 of it rf'lI,nlclollgs tu ..: Jifl"o.:l"\;ut ur the MlllIC). shull' tlmt the llUlul.>er "'(tI ) of eirlll(·llts P EG siltisfy ing 5, = pS'; p  l is a/II (a).
Chap. 2 Group and Its Subsets
15 . Prove that, being the product of two subsets, the product oftlVo classes ill a group G must be a SHill aggregate of a few whule cla.",~a;. Namel.\', the slim agAregate conta.ins all elements conjugat.e to any product of two elements belonging t·o the two cla~ses. respectively. 16 . Calculate the multiplication rable of the group T by extending the llluitiplication table of t.hc subgroup C 3 = {E. R I . Rf} of T. 17 . TIl(· lllult.iplicittioll tlthle of tlle fiuitl' group G is a..~ follows. E
.,
E
E
D
A R C D
F
F
R C
, ,, ", ,.
K
II N
K
.II N
,
B
F
D J
F
B E
C
D
.II
A
K A
N
E
N II
.II
M E N F
N II
N K
,
A
K
, , " ,
A E F
J B
II F A
K E
C
,
D
.11
M N K
J
,
C
K
, ,
D C
, ""
D
C
,.
,
E
, , A
F
r
B
C
J
D
B
,
N
.II
:Y
.II
N
K
J
C
F
D
, " " , , " , ,, , "
F C
A
D R
C R
F
,
N F
J
F
A
K
D
E
.II
K
,'"
K
B
n
B
D B
C
C J F
F
A
E
A K
D B
N E
K E
.11
(a) Find rIlE'" inverse of each element in G: (b ) Point ont thl' ele nl{'nts which Cflll commutr with any element in G: (c) List. the period and order of each element; (d) Find the dcments in each class of G: (e) pil)!1 ,11 illvariltut. sllbgroups ;, G. Fur eltch iuvarifl.lIt subgruup. list it.s coset." and point. out ont.o Wllich group it" C"(uot.ient. .e;ronp is isomorphic; ( f) 2'lJnke a judgment whether G i~ bOlliorphic OlJto the tetraheJl"al ~ylJlnU'trif' group T , or isomorphic Ollto tlU' rpgnlar sixsiO(·rI polygon symmetric group D fi •
Chapte r 3
THEORY OF LINEAR REPRESENTATIONS OF GROUPS
The t heory of linear representations of groups is the fou ndation of group theory. III this chapter we will introduce the defiuition of a represelltatioll of a grmlp, ~ t udy tIlt" eoncept.R and propocrties of inequiva.lent. and irrocdueible representat ions, discuss the methods for find ing all inequivalent and irreducible representations of a group, and de monstrate the fnndamental steps for the applicH.tiulI oj" group th~ry to physie; through /til example.
3.1 3 . 1.1
Linear Representations of a Group D efin i t i on of a Linear R e prese ntation
If a given group G is isomorphic or homomorphic onto a group composed of matrices, t.he matrix group, which describes t he property of G at lca~t partly. is called a iillear repre;etltatioll of G, or IJI'ielly a represelltatioll. D e finition 3 . 1 A matrix ?,roup D(G), composed of nons ill?;ular m x In matrices D(R). is called an mdimcnsional representat ion of a group C. or briefly a rep resentation of G, if G is isomorphic or homomorphic outo the lUl\trix gronp D(G). T IJP ml\trix D(R) , to wJlich an eJenwnt R in G maps, is called the representation matrix of R in the represeUTation, and Tr D(R) = X(R) is the character of R, T he representation matrix D( E) of the identical element E is a unit. matrix, D(£)  1. The rcprC'Sentation lIlatricC'S of R and its inverse R I are Illlltlially illverse luatril'es, D(R I ) = D(R)  l. Tl le repre~{·lltat ioll D(C) is said to be faithful if G is isomorphic ont.o D(G), An.\C gronp has an idelltical representation, or called the trivial one, where D(n)  1 for every element R ill G. A ltJatrix group is iu; OW11 representation. c. 'rIS.'I EC".
(3 .17)
A class operator is commut able with evC'!'y clelllent in C and every dass C()IJ\,p.r~t'ly, j f X i~ CotllllHltahle with every eieilleut. ill G,
operator ill £.
(3 .18 ) [rom Eg . (3.17) X has t o be a linear combination of the class operators
1 "
X ~ L
0,
L F(S,I L
9 (> = 1 S , EC"
F" = n(o)  l
L
T SiT  J =
L
("'F" ,
(> = 1
T EG
(3.19)
F(S)).
S , EC"
h1 compnrisoll with Eq. (:UR), P(Sj) depends the eiemf'nt 51' F(S)
OJ]
t,\w
cl(lRS
Co bllt. not on
= Po. Since the product C",C,J i~ commutablE' with
every eicllleut. T in G, one has 0,
CnC d
L
f(o "J, rl Cy .
[(0.3.1')  f(,t.3.o.il
13.2U)
1= 1
where I( n , ,3, ..)') is a nOllnegative integer and cau be calculated direct ly from the group table (i;~'C Prob. 15 of Chap. 2 in [!.la and GlI (2001 )]) . The dll.~~ operat.()l1< are iillf'ariy illdl'pPtJdellt awl "pall a lillear ~pa('p £ ~ of dim em ion .fir , called the class space, Equation (:320) shows t.hat the dass space C" is closed for the multiplicat.ion of its wctors so that C c is an alg,AJl'a, called the cia.';t> algebra.
3.2
Transformation Operators for a Scalar Function
Denote simply b.\· x all the coordinates of degrees of freedom in a quuntum aud b~' ~'{.r) the scalur wave fuuctiuu . R is a liueur trausformatiull
s,~·stcm.
of tIle
Hy~tPlll.
wllidl lIl 11.)" he e itller 11. Hpal:e time trallHformat ioll HIlCh
a.~
a
r.nmslation, a rotation, or an inversion etc .. or an intemal trans format.ion. such as a rotation in the space of the isotopic spin. and .~o on. Under the iran~ forIllatio tl R . .L: i~ dJaugeu to x' = R.r aud the wave fUllctioll ~') (.T) il; changed to \./ (.r' ). III order to ~how e xpl icitly the depPlldplI{"p
§3.2
T"IR~fo""a1IO"
OJWnltOr8
fo~
(I
S calor ru r..:i ,o ..
of the wa ve function on the transfol"lnation R. we introdu(;e an operator PR , t/(.r'l == Pnv(.r' ). Bei ng, 1'1 sCHlar wave fUnctioll , tlw \'aluc of the transformed waw fu nct ion Pn1/.' at tllf' point RI s honld Ilf' equal to the value of the original w;we funct ion t ' at t he point r , nomdy
x ~ x' = RI . d,!;)
:r = R l.r' ,
(3.21 )
...!!.... I;','(.£')  PndR.r:)  d.c).
Rt'pilwillg Ihe arglllllt'1I1 ( R;,.) ill PR l,··(IlJ·) wit h .r. OIl!'
h1\.~
(:1.22) t.."(.r) a nd PH dx) a re two different fUllctio ns of I. Equation (3.22) shows the relation between the values of two functiolis at. llifTercnt points. At tile f>a lllc time, t he relfllion gin~~ the methud to calculate the trallsftol"1l1cd fu nct.ion f'RlA .r ) from tht' original function "'·(.T). NlIllWly. first rf'plft('E' the argume nt J" in ¢(x) with R  I.'f , and then. re!l,Md t',( R  II) lIS Il flllleti oll of I . whic h is nothing but tile lI"anSfO rllled fUlll'tiOIi Pl/ ¢(.c). OIo\'imL~ly,
Pn is 11 lillea r o pe n. h.r
Pn {a ~I{J") I
[email protected](:r)} = a~ l( n  l.r) I bOC R Ix) = UPHl!.'(X) I bP/I(xj. (3.23) Eqnl!.tion (3.22) l'ihow~ a OIletoOll{' "'orr~ponden('e hdwCCll t.he oprmtOI" PR and the t ransformation R. This correspondent!' is inVMinlit ill the product of tmnsformations. .1" 
S
x" = Sx' = (SR)x.
~" '(JJ) ~ ~·o(.r)
3i!!.
tY'(J''') "::"
PS PR~"'(X") =
P."
=
L,
q, ),D),,,(R) , (3 .34)
D(R) forms another rcprC3cntation of C. Both D(R) and D(R) arc the matrices of the same op(~rator P R in the same lillear space C, but with different. basis vcctors. Two representations are said to be equivalent.
Definition 3.2 A repr(';'
R EG
gT " ,
XI''' (X  I)
1n
L
= gT. "I'
"
If the represen tation is not selfconjugate, the Ilrst line of the abo\'e equation is equlil to zero owing to Them'em 3.3. If D ( H) is renl, X is a unit Illotrix and T = I. Couver~ely, if X is a symmetric unitary matrix , x t = X · = X  I, we are going to show that X can be decomposed to y2 , where Y is a symmetric unitary matrix. In fact , if Xa = Aa, thell X I a* = A Ia*. Boti] a and a* are the eigenvectors of X with the l:iame eigenvalue. Nnmeiy, the eigenvectors of X can he chooeu to be real , and X can be diagonalized through a real orthogonal similarity transformation M. Let M I XAI = 1'" = ['2. Thus, Y = Mf'Al  I is unitary, X = y2, and y  I = A4r*l\I  l = Y*, Since V is II. symmetric unitary matrix and y  2D(R) y2 = D(R) ., y  I D(R)Y = YD(R)*y  1 (y 1 D(R)Y),. D (R) is a real representation. D Coro llary 3.5.1 group satisfies
T he character in a n irreducible representation of a finite
1 { 1,  ~ X(R') ~  1, gREG 0,
3.5
real representat ion, selfconjugate, but not rcal reprt'Sc ntation, not selfconjugate representation. (3.71)
S ubduced and Induced Representatio n s
Discuss a finite group G of order [I containing ge classes . A class C" in G consists of n(a) elcments. Vj(G) is an mjdimensional irreducible reprcsentntion of G. The cha racter of S E Co in DJ(G) is denoted by XJ( S) or X~. H ={ E = T 1 , T 2, . "n } is a subgroupofG with index n = .q/ h. The eosets of 11 in G are denoted by fl,.l1, 2 ~ 1" ~ n. Assume that R,. have been chosen and RJ = E such that any element in G can be expre;sed as RrT t uniquely. The cll!..~s Cfj in the suhgTOUp II contains rr(J3) elements. If (H ) is an Inkdimensional irreducible represen tation of f1 , and t.he character of T t E CI' in d'(ll ) is denoted by r(T1 ) or X~ . The set of repre;entatiol! matrices !Y( T t ), T t E H , forms a repre;en
Chap.!J Theory of Representations
74
tation DJ (H) of H , which is called the ~ubduced reprCSClltatioll from au irreducible re presentation DJ(G) of G with respect to the subgroup H. Generally, t he subduced representat ion is reducible with respect to the subgroup H
(3.72)
Denote by
tlJI'
the Ink bases ill the represen ta tion space of l f (H )
Define an extenned s pace of dimension nffik with the hases 1/Jr!, = PRr W!" where 1/;11' = tlJI, . The extended space is invariant with respect to the group G , and corresponds to an nml.;dimcnsional representation l:l. k{ G ) of Gin
the following way. For any given element S in G and for each R F , SRr can be expressed as RuTt, where 1J a.ne! t are completely determi ned by Sand 1' .
Since
one obtains Xk(S) =
E
.6..~I1.T!'(S) ,
13 .73)
'"
This rcprc:;;c ntation .6. k· (GJ is callcd t hc ind uccd rcprescntation from t hc irreducible re presentation Dk (f/ ) of t he subgroup H with respect to G. In general, t.he induced representat.ion is red ucible with respect to G:
b"
~~L 9 S EC
, x'IS)" x' (S)
~ ~L 9
, n(n)
(xtJ' X; ,
(3 .74)
where the character of S E Co. in the represelltatioll 6 "'(C) is denoted by Xk (S) = X~. In general, some elements in the class Co belong t.o the subgTou p [J , and some in Co do not. T he elements in Co belonging to 11 constitute a few whole classes of the subgroup fl , denoted by Cfj . It is pos.~ible that no eleilleut in C" belongs to the s ubgroup H. For th i~ cl;L';e
§S.5 Subducw and Induced Repres£f1ta l;ans
we say that no CfJ exists. From E(l. (3.73), the diagonal element of 6 k (8) appears only when T = tI , i,e. , SR r = RrT t . Thus , Xk(S) is no nvanishing only when the class C" contains a few elements belonging to the subgroup H . Denoting by KfJ t he number of differcnt R .. satisfying R;I SRr E CfJ, , olle HiS
k",I:
X", =
KfJ XfJ '
L, f3
From Prob. 14 of Chap, 2, the number of elements R in G satisfying R ' Sn = Tt is m(a) = g/n(a) . Expressing R by RrTx and letting T t E CfJ , one has R;ISRr = T'rTIT~ 1 E CfJ . On the other hand, the numbcr of Ty in the subgroup fI sati~fying TyTtTy 1 = T t i~ m ({3) = h/n(fJ) . If Rr Tx satisfies S(RrT,,) = ( RrTx )Tt , then RrTxTy satisfies this formula too, However, the latter does not make any new contribution to the characters Xk(S) because R ;:I SR .. = T>: TyTtTy ITx l = T.rTtT;J;I . Therefore,
m(et) m(fJ)
" fJ =  
~
9 n (fJ) h n(a)'
, = ,9() 'L.. " (R)" Xo. n f' XfJ ' Ina {J
(3.75)
It is easy to show from Eq. (3,75) that the multiplicities bj l: in Eq, (3.74) is cqual to the multiplicitia; ajl: in Eq. (3.72): bjk
= !. L
n(a)
9 "
(:if,r X~ = ~ L
n({3)
,
(4)* xt = ajl: .
(3.76)
xb,
III fact, the class C" in G contains a few classes CfJ of H , X::' = and the different classes Co. correspond to the different classes Cp. An element in C"" which docs not belong to II , makes no contribution to X ~. T herefore, the sum over the classes Co. in Eq . (3.76) is equivalent to the SUill over the classes C {J in H . The formula (:1.76) is called the Froheni ' L~ theorem. In terms of the method of induced representations , the character tables and the irreducible representations of the groups O 2,,+1, O 2,, and I can be calculated (see Prohs . 15, 16, aild 18 of Chap. 3 in [M a liml Gu (2004)]) . We list the results as follows. Table 3.10
The character table of D 5 (p = 0,
E
2C,
2C,
5q
A B
I I
I I
I I
I  I
_p_ l
0 0
E, E,
, , 2
_ p_l
,
(V5 
1)/ 2)
The group O2,,+1 contains one (2n + I ) fold axis, called the principal axis, and (2n + J) I.,>q uivaleut 2fold axes. located in the plane perpeudicular
Cha p. !J Theory of R epresentatio n s
76
to the principal axis . T wo generators of O 2,,+1 ar c the (21l + 1) fold rotation C2,,+1 and one of the 2fold rotations C2, C 2n +I C2 = C 2Cin1+ l. The order of 02n+1 is.fJ = 4n+2. The number of the classes in D2»+ 1 is 9c = »+2. The group O 2,,+1 has two inequivalent represent.ations of one dimension , DA and DB, and n inequivalent irreducible represcn tations of two dimensions, DEj , 1 ::; j ::; 11 : Table 3.11 (..\ =
eD, A
The character table of D 7 fJj = V + ..\ j, j = 1,2 , 3)
i21f 7 / ,
E
£,
,
E, E,
2 2
B
,, 7C;, ,
2eT
:lC T
2C,
"'" "
~
"
,, ,,
,,
'"
"
'H
"
,n n
(3.77)
The character table of D3 is listed in Table 3.7. T he character tables of 0 5 and 0 7 arc listed in Tables 3.10 and 3.11. The group 02" contains one (2n)fold a.''!:is, called the principal axis , and (2n) 2fold axes , located in the plane perpendicular to the principal axis. T he 2fold axes are divided into two sets, each of which contains n equivalent 2fold a.'{cs. They form two elasses, respectively. T wo generators of 0 2" /:Ire the (2n) fold rot/:ltion CZ n and Olle of the 2fold rotatiOllS C 2, C2nC~ = C 2C;,1 = cg . T he angle bet.ween two 2fold axes corresponding to C~ and C~' is IT / (2n). T he order of O 2,, is 9 = 4n. T he number of t he elasses in 02n is 9c = n + 3. T he gI"OUp OZn has foUl" inequivalent representations of one dimension , DA1 , DA1 , DB" and DB1, and (n  1) inequivalent irreducible represelltlltiollS of two dimellsion.s, DE" 1 ::; j ::; n  1.
(3 .78)
The character tables of 0 4 and 0 6 are listed in Tables 3.12 and 3.13. along the principal axis through For the group 0~" +2, the rotation
Ci,7tJ
§S ,5 Subducw and Induced Repres£f1ta! ;ans
77
IT angle does not belong lo the subgroup O 2 ,, +1 , and is commutable wi th any dement in 04n +2. T hus , 0 4 ,, + 2 is a di rect product of two snhgroups:
C 2  {E , C 42.."+I} +2 '
Table 3.12
The character table of D 4
D,
E
2C,
C,
A,
1 1 1 1
1 1
 I  I
1 1 1 1
2
0
2
.4~
B, B, E, Table 3.13
2C~
2C;'
1
1
 I
 I  I
1
 I
1
0
0
The character table of D 6
0,
C
2C,
2C,
C,
3C 2
"
1 1 1 1
1 1
1 1 1 1
1 1
1
1
 I
 I I
.'h
B, B, E, E,
2 2
I  I 1
 I
 I  I
3C,
 I  I 2
 I
1
0
2
0
0 0
1
The proper symmetric grou p I of a regular icosahedron contains six 5fold axes with the generators T j , 0 :S j :S 5, ten 3fold axes with the generators Rj , 1 :S j :S 10, and fifteen 2fold axes with t he generators Sj , 1 :S j :S 15. All axes a re nonpola r and any two axes with tht;l S IlIllt;l fo ld are equivalent to each other , T he order .q of I is 60. The number .Q c of the elasses in I is 5. I does no t contain any nontrivial invariant subgroup. T he character table of I is given in Table 3.1 4. The represelltatioll matI'ices of the geuerators of I are calculated iu P roh. 12 of Chap. 4 of [i\h and Gu (2004)]. (Also see [Deng and Yang (1992)].) The c haracter table of I (p =
Table 3.14 1
.4 T, T, G H
E
12C~
12(...'i
,,, 1'_', ,', , I I 1
1
0
1
0
20C~
1 0
lSC~
1
0
 I  I
1
0
 I
1
(J5 
1)/ 2)
78
3.6
Chap. !J Theory of Representations
Applications in Physics
At the beginning of Chap ter 2, before the study of group t.heury, we raised a sim ple example to see how to obtain some precise information of the system th rough analyzing its symmet ry. Now , we have studied the fu ndament.al concepts on group theory and the theory of represcntat.ions. It is time to discuss the typical applications of group theory to phys ics.
3.6.1
Classification of Static Wave Ftmction s
T he fi rst s tep ill the application of group theory to physics is to find the symmetric t ransformations of a given quantum system with the Hamiltonian H {x) . A symmetric transformation R preserves t.he Hamiltonian invariant,
[PR , H(x) J ~ 0,
H {x)
(3 ,79)
PH is the transformation operator for scalar wave functions corresponding to t he symmetric transformation R. The set of the symmetric trans formations i~ the symmetric group G of the system. Second, find the inequivalent ilTeducible representations and their characters of the symmetric group of the system. T his is a tas k of gro up theory. Usually, one chooses the convenient forms of the irreducible representation matrices D j( R ) such that the represcntation mat rices D j(A) of as much generators A of G as possible are diagonal. Of course, t he representatioH mat.rices DJ {B) of the remaining generators B are not. diagonal if G is not A belian. T he representatioll matrices of any element in G can be calculated from those of the gcnerators . Third, if the energy level E is rn degenerate, there are Tn linearly independent eigenfunctions ¥>I'(X) of H (x) with the eigenvalue E: p. = 1,2 , ... , m .
(3 ,8U)
WI' (x) ~pan all 1JIodimem; iOll al fUlld ional s pace C. AllY fUlld.ion 4>(x) ill C is the eigenfunction of H (x) with the eigenvalue E , a nd any eigenfunction of ll (x) wi th the eigcnval ue E belongs to C. Duc to Eq. (3.79), PnW,,(x) is all eigenfunction of H (x) with the same energy E. Namely, C is invariant in the action of the symmetric operat.or PRo \Ve can calcuillt.e the mntrix D (R) of PR in the basis function t/J,,(x) : m
PRt/!,,(X) = ¢1,(R 1x) =
L 1/ = 1
1/! ,,(x)Dvl,(R) .
(3 .81)
§.9.6 Applicatians in Physics
The set of D (R ) forms a representation of the symmetric group G of the system, called the represe ntation corresponding to the energy E. The character x (R) = TrD{R) of R in D{R) is easy t.o calculate. D(G) descrihes the transformation rule of the eigenfunctions of H (x) with E in the symmetric transformations. Cenerally, the re presentation D (G) is reducible and not in the convenient form. Through a simihuity transformation X , D(G) can be red uced into the direct sum of irreducible representations,
x ' D(R)X ~ EEl
ajDj(R),
x( R) ~
j
L
a,x'(R).
(3 .82)
j
T he multiplici ty aj of the irreducible representation D J(G ) ill the reducible representation D (G ) can be calculated from the orthogonal relation (3.51 ): I" . I" . aj =  ~ x;l{ R)* X( R ) =  ~ n(o)x{,*Xa . gREG
Y
(3.83)
"
T heil, X ClUJ be calculated frum Eq . (3 .82) ill the following way. When R. in Eq. (3.82) is taken to be the generators A where Di( A ) are diagOllal , X is the similarity transformation matrix t.o diagonalize D {A ). Namely, the column matrices of X are the eigellvectors of D (J1 ) with different eigenvalues, respedively. The solution X contains some undetermined parameters. T hose parameters will be partly determined in s ubstitut.ing X into Eq. (3.82) , where R is taken to be the remaining generators B. Since the matrix on the righthand side of Eq. (3.82) is a block matrix, X , which satisfies Eq . (3.82) for all gellerators, still cont/:lins some undetermined parameters whose number is L j oJ Those paramet.ers should be chosen to make X as simple as poss ible. From Eq . (3.82), the row index of X is t he same as the column index of D(H), denoted by IJ, /:Illd the COh.llllll iudex of X i!; the !;allle a!; the COJU lllll index of the block matrix on the righthand side of Eq. (3.82), which is enumerated by three indices j, p, and ·r . T wo indices j and p are denoted the ineducible representation D j( R) and its row and the additional iudex r is needed when a.j > I to di:;titlguiJ;h different 1)1 in the reductiOIl (3.82) . New basis funct.iolls P:;"'(x) are the combinatiollS of t/!I'(x) by X:
PIAx) =
L ,
PR 1, there are aj sets of basis fUllctions wf,..(x) which belong to the representatio n D j of G and are distinguished by the parameter 1" . Any linear combination of the functions oJ">t,,(x) with the same j a nd p is the eigenfunction belonging to the pth row of D J (3 .85) where the combin otion coefficient;; Y/~ are independent of p . The combina.tion matrix y i is related to the undetermined parameters in X. The physical meaning of the function I/I/,r (x) depends on the given group G and the chosen representation Di. Since the representation rnat.rices Vi(A) of some generotors A in G are d iagonal, I/If,..(x) is the common eigenfunction of the operators PA as given in Eq. (3 .84).
3.6.2
Clebsch Gordan Series and Coefficients
If a quantum system consists of two subsystelns (sec §1.6), the wave function of the system is expressed as the product of two wave functions of t.he s ubsystems or their combinations . Suppose that two functional spa.ces r.J and C k of two subsystems are the representatio n spaces of irreducible reprcsentations D j(G) and D k(C), respectively, then
PHtq.(x) =
L
·~(x) D~I'(R),
PH¢~(Y) =
"
L,
rpi (y) D t( R).
The functional space C of the composed SyHtem iH t he direct product of Cj and Ck , and is spanned by I/It~ (x,y) = t/if,(x)rp~(y) where 1 ::; II. ::; Inj and 1 ::; v 1I1k. In the symmetric transformation, ljtt~(x,y) transforms according to the direct product representation
:s
(3 .86)
The representation can be reduced through a s imilarity transformation Cjk
(cjkrl
[Di(R ) x D~·(R)] cPe =
EB
.,
aJDJ(R ).
(3 .87)
§.9.6 Applicatians in Physics
The series on the righthand side of Eq. (3 .87) is called the Clebseh Gordall series. Tl1killg; the trace of Eq. (3.87), one has
"
,
. r(R.)x (R) = L
"
I LJ X
"(R).
(3 .88)
e
The multiplicity aJ as well as the matrix ik can be calculated from Eqs. (3.88) and (3. 87) . Similar to the discns.;ioll in the preceding subset(x)J = =
L
(Pnlq)~(x)I1Pf,(x))
,
=
xif"
Pl/IjJ~(x) =
¢i(x) Di ... (R ).
one has
(q)~(x) I ~(x))D~I'( R) =
L,
L, L
,
X;;D~I'{R)
Dt(R l)·(¢~(xll1/>f.(x»
From the Schur Theorem 3.3, X,~j, =
=
()kj6,,1' (,pkl l ~ ),
L,
D~),(R)X;:,.
(3 .91 ) where (.pk llt,M) is
called the reduced matrix entry, which is a constant independent of the suiJ,;cript I). TIllis, the Theorem is proved,
(3 .92) The functions belonging to two inequivalent irreducible unitary representations of Pc arc orthogonal to each other, so arc thcir combinations. 0 In quantum mecbanics, most pbysical observable;; are calculated through matrix entries. \Vhen the static wave functions belong to given rows of given unitary irreducible representations, the \Vigner Eckn.rt Theorem simplifies the problem of calculating 1II,;"Inj matrix entries (¢>~(x) l t,&t(x)) to a problem of calculatiug only one reduced matrix entry. Furthermore, the static wave functions in the real problems ustwlly are hard to be solved such t.hat. even one matrix entry cannot be calculnted because the wave functions arc unknown. Howevcr, some protise informatioll of the system can be obtained through analyzing its symmet ry. Some matrix ent ries are known to be vanishing (selection rule) , a nd t.he rati05 of the matrix entries can be obtained by eliminating the reduced matrix entry as a parameter, although the parameter cannot be calculated. We will exph)itl this method by examples later. If a set of operatore L~(x) of mechanical quantities transforms as follows in the ~ymmet,ric tnlllsfonnatio m, PR j
PRL ~(x)Piil =
L, Li(x)Dil'(R) ,
(3 .93)
83
§.9.6 Applicatians in Physics
L~(x) are called the Irreducible tensor operators, then ,
L L1(x)# [D k(R) x Dj(R) j .\1".fJl' · (3 .94) " L~( x )VJ~ (x) can be combined by tbe Clehsch Cordan coellicients to Fl/r (x) PRL~(x)t/;~(x)
=
which belong"S to the kith row of D":
Fi/ r(x)
=
PnFl/r(x)
L
L~(x)v,::"(x)C:f,,JlIh'
'" = L
L~(x)1/J~(x)
Fil'r(x)Dil'A1(R),
(3 .95)
fIf'
=
L
Fil,'(x)
Jtllr
[(CkJ)  ']
. JUr,PI'
There are (1nj'1nk1nj) m atrix entries (¢{' (x) IL~(x)ltbj',(x)) for the mechanical quantities L~( x) between two st.atic wave funct ions ¢{' (x) and 1/J{.(x). The \Vigner  E ckalt Theorem greatly si mplifi es the calculation problcmlS ,
(3 .96)
munely, the information of those matrix entries related to the symmetry of the system demonstrates Itself through the Clebsch  Cordan coefficients, and the remaining information is given in a few !educed matrix entries (¢i'I ILkll""""),, which are independent of the row indices 1/, (I, and I). The number of the reduced matrix entries is equal to the multiplicity of the irn.."f\llcible represeutatioll Dj' in the reduction of the direct product represent.atioll Dk x D j .
3.6.4
Normal Degeneracy and Accidental Degeneracy
If an mdegenerate energy E of the original Hamiltonian Ho(x) corresponds to a representation D( G) of the symmetric group C , the degeneracy is eall~\ normal if D(G) is irreducible and called accidental if D(C) is reducible. We begin with the energy level of the original system to be normal degeneracy. Introduce a ''symmetric pertmbation" )"H I (x) whkh docs not disturb the symmetric group of HO(:I;). ThllS, Ho(x) lind Hl (X) are both commutable with the symmetric transformatioll operators Pn :
IPR , Ho(x )1 =
0,
(3 .97)
84
Chap.!J Theory of Representations
T he perturbation is introduced smo othly as the parameter), increases from 1 continuously. Denote the eigenfunctions of Ho{x) with E by tPf,(x) belonging to the pth row of an irreducible representatioll Dj(G) :
o to
(3.98) For the first approximation, the energy shift t1Ei is calculated by
(3.99) Namely, in the first a pproximation, the eigellvalue does not split . In fad, it Clllluot split ill a rbitrarily high approximation. Otherwise , if it ;;pli lJi int.o, say two eigenYlJlllCS BI and E'l with m) and m2 eigenfunctions (mJ + mz = m), the mj eigellfunctions of EJ transform among themselves under PR and correspond to a representatioll of dimension mi . The representation cannot contain the original irreOucible representation DJ (G ) because 7nJ < 7n. Then, those 7nJ eigenfunctions of EJ wonld be orthogonal to all m eigenfunctions of E , and could not be obtained continuously from any combination of the original eigenfunctions. Thc conclusion is that a symmetric perturbation cannot split an energy with a normal degeneracy. Now , we turn to an e nergy E of the original Hamiltonian with an accidental degeneracy, where the representation corresponding to E is reducible, but docs not contain an irrL'ilucible represcntation with the multiplicity aj > I. The energy shifts under the symmet ric perturbation, bllt spli ts only between eigenfunctions belonging to d ifferent irreducible representationl;
(3 .100) T his conclusion is also nonperturbation, I f the representation corresponding to E contains an irreducible represe ntation Di(e) with the multiplicity (lj > 1, the sets of eigenfunctions belo llging to Di(e) are combined
(3 . 101) where T, oS = 1, 2, ... , aj. By making \L~e of the symmetry of the system, the method of group theory greatly simplifies the calculation, If the original Hamiltonian Ilo(x) and the perturbation Hamiltonian Hdx) have different symmetries, olle mlly choose the common symmetric tralll;forlllutionl; of Ho(x) and H\ (x) t.o form t.he SYlIlmet.ric group of the
85
§.9.6 Applicatians in Physics
system so that the perturbation Hamiltonian IJ dx) is the "symmetric perturbation" . It is common viewpoint that the energy level of the original system is normal degeneracy if G contains all symmetric transformations of the system . The accidental degeneracy is related to the existence of some undiscovered symmetric transfor mations of the system [Zon and Huang (1995) 1.
3.6.5
An Ezample of Application
\Ve arc going to ra.ise n physical example to demonstrate the ,;tep,; of application of group theory to physics . Discuss a qun utum system with a square well potential in two dimensions. T he Hamiltonian equation is (Ii = 2m = 1) rP"lj1
d21/1
lltb =  dx'l  dy'l + V (x,y) =
{goo
V~ = E VI
'
(3 .102)
when Ixl < 7r , lui < 7r, the remaining cascs.
F irst, it is evident that t he symmetric group of the system with the square well potential in two dimensions is the group D 4. T he character ta ble of D4 is listed ill Table 3.12 . Two generators of D4 are t he . ,le,,1 is the same as that generated by e a , e~ C = ea C.
by e;;
The ore m 3.8 The direct s um of n left ideals C" respectively generated by the orthogonal idempotents e~ (see Eq . (3. 11 2)) is equal to the gI"OUp algebra C if and only if the ~\1Il1 of e a i ~ equal to the identical element E:
Proof
"
"
a= l
a= l
If E is equal to t he sum of e a , t
tE
(3.121 )
L" tea .
Th us , any vector of vectors tea helonging t(l C a , respectively. Conversely, if C is the direct sum of C", E can be decom posed uniquely into a sum of vectors belonging to C,,, respectively, where the vector belonging to C a is E e" = ea . o =
t in C can he decompOliOO uniquely into a
3.7.3
Tw o ~8i de
=
Slim
Ideal
The left ideal and the right ideal, generated by the Slime idempotent ea , are not the same as each o ther generally. If they are the same, Ce" = ea C is ca lled a twoside ideal generated by a n idempo tent €a,
Ce" = ca C = I",
txE T",
ViE C and xEI" .
(3.122)
§S. 7
11T1~~!
The eigenvector of C2 with the eigenvalue 3( _1 )1' is that wit.h t.he eigenval ue 0 is [ J~:l + 2 4>,~~} J and
 11 = J6{E
t· D t F + A + B +C}, j
, D + F  A  B C},
{(I)
(E) 1 (l) } 1 boo = 2V3 00 = 2V3{2E  D  P + 2A  B C},
b\7·)= CI1 When
1/
=I
11. ,
{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 (3) '51,(4)) a 110 + V'' 00 / V'f2 L: ·
, ,, ", ,, , I'
e
0
,0,
0 0 0
0 0 0
," _{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
; 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. = ±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;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: ,
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 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.
 k,
,"(wi ~
, I:
e;/'W =
, L I'=j
>4,
Chap.
Xi(w)X'(w) ~
Th "",,dimensionai Rotati"" Group
.)(,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' ~
(52
Chap.
, ) 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)'
+ 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 (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 tJ.< )
exp { i27r (RJ. 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 = 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 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 ,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 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,
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
~;:.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.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,: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 £ 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
:+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 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
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 
aI'
= 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 .