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. Next consider a rotation by () in the plane defined by ez and ew (see Fig. 3.1 on page 64): (
~r
e (}
)
c~s () - sm ()
= (
sin () ) ( ~ z cos () ew
)
•
Hence
e r e x sin () cos e/> + ey sin () sin e/> + ez cos () e (} ex cos () cos e/> + ey cos () sin e/> - ez sin ()
-e x sin e/> + ey cos ¢
(3.35)
We have the cyclic vector products (3.36) and also (3.37) For spherical polars,
v -- {) __
1{)
__
1
{)
(3.38)
+ e (r;. {)() + e 4> r sin () {)e/> '
= e r {)r
so the angular momentum is represented by
l
=
r Ap
-inrer A V .to { _ -1,,,,
.to { _
-1,/1,
_
{)
erA e (} {)() {)
_
__
1
__
{)}
+ erA e 4> sin () {)e/> 1
{)}
e 4> {)() - e (} sin () {)e/>
(3.39)
'
where we have used Eq.(3.37). With the help of Eq.(3.35) we read off the Cartesian components of the angular momentum:
-in {-sin¢~ {)() -iii { cos c/>
-
cose/>cot(}~} {)e/>
:0 - sin c/> cot 0:c/> }
-in~ {)¢ .
(3.40)
Hence L± = L1 ± iL2
{) + i cot () {)e/> {)} = he ±04>{ ~ ± {)()
•
(3.41 )
Orbital Angular Momentum
54
From Eq.(3.10), we have
and
(3.42) On the other hand
2 L32 - nL3 = - n 2{8 8(p2 - Z.8} 8¢ ; and on adding these equations together, we obtain
2
L =
-n
2{ 1
8. 8 sin 888 sm 8 88
+
1 82 } sin2 8 8¢2 .
(3.43)
Using Eq.(3.34), we can now write (3.44) This last result can also be obtained by squaring Eq.(3.38), but that is rather painful and the above method is easier. The most general solution ofEq.(3.26), in the polar representation Eq.(3.40), namely
IS
(3.45) where Ti (cos 8) is an arbitrary function of 8, which we have written as a function of cos 8 for convenience. Inserting this form into Eq.(3.27), with use of Eq.(3.43), we find
55
Spherical Harmonics
In terms of the variable z = cos (),
(3.46)
this takes the form
d (1 - z2 )-d d [ -d Z z
+ ff( + 1) -
m2 2 1,- z
1Tlm( Z) = 0,
(3.47)
which is called the associated Legendre equation. Given that it is a secondorder ordinary differential equation, it has two independent solutions, and if we only use Eq.(3.26) and Eq.(3.27), there is no convincing way of deciding which particular solution we need. Nevertheless, by using Eq.(3.41), we can pick out the required solution of Eq.(3.47). In configuration space, Eq.(3.14) becomes
e±i~
[±! +
icot 8
:4>1TF'(cos 8lei=~ = J(e Of m)(e ±m + ll1{±1(zl ei(m±1)~
and this reduces to
This is a first-order, homogeneous differential system, and it has a unique solution, up to normalization. It will prove helpful to define (3.48) in terms of which we find
, ddz Rtrn(z) = =fJ(f=fm)(f±m+ l)RFm±l(Z) ' with z = cos (). In particular, we see that
so that
RiAz)
= ci
CF being two z-independent constants. (sin 0) m Rim (cos ()) so that
From Eq.(3.48) we have
= Ti (cos ())
= (sin ()) -m Rem (cos ()) ,
(3.49)
56
Orbital Angular Momentum
and in particular
Ri.-l(cos 0)
(sin 0)21. Ri,-A cos 0) = ci (sin 0)21.
Ril(COS 0) ,
(sin 0) 21 Rt,A cos 0) = ct(sinO)21.
We have, for m
(3.50)
> -£,
(_~) R+
1
.J(£ - m + 1)(£ + m) C
1
dz
(£ - m)! ( d) l+m 2 l (21)!(£ + m)! - dz (1 - z )
by iteration, with use of Eq.(3.50). Similarly, for m
Rim(z)
=
V(l+ m c+
l
(z)
l,m-l
(3.51)
< l,
~ 1)(£ _ m) (~) Rl,m-l(z).
(£ + m)! (2£)!(l- m)!
We have thus two expressions for
.
(~)l-m dz
(
1- z2 l
) .
(3.52)
Tr, namely (3.53)
obtained by iterating m down to -£, and (3.54) obtained by iterating m up to i. Since these expressions must be equal to one another, we can find a relation between the two constants. Evidently c- (£ - m)! (1 _ z2)m
l (£ + m)!
(_~)l+m (1- z2)l = c+ (~)l-m (1- z2)l. dz
l
dz
By equating the coefficients of the highest power of the polynomials on both sides, namely zl+m, we find ct = (-l)fci. Use of this result, and inspection of Eq.(3.53)-(3.54), reveals that (3.55) The Legendre polynomial for integral, non-negative l is defined by 1
P1(z) = 211!
(
d
dz
l )
2
l
(z - 1) ,
(3.56)
57
Spherical Harmonics
and the associated Legendre function of the first kind by
= (1 ~ z2)m/2 (~) m
Pt(z)
Pt(z),
(3.57)
< i. From Eq.(3.53), we have that, for l > 0 and
for integral, non-negative m o <m< l,
(l- m)! m (2l)!2 i l!{l + m)!Pi (z).
(3.58)
On looking back at Eq.(3.47), we see that the associated Legendre functions of the first kind satisfy the second order differential equation d [ -d (1 Z
Z
2
d
)-d Z
+ l(l + 1) -
m2 ]
1- Z
2
m
Pi (z)
= o.
(3.59)
The standard normalization and phase convention are as follows: 1
«() A-.) = (_l)m [2141T + 1 {l - m)!]2 pm( ()) im,p im , (l + m)! i cos e ,
y;
tp
(3.60)
when 0 < m < l, which is consistent with Eq.(3.45) and Eq.(3.53). To cover negative m values, we use Eq.(3.55), which implies 1
«() , A-.) =
l:.T .I im
tp
[2l + 1 (l + m)!]2 41T (l _ m)!
p-m{ ()) im¢ i cos e
,
(3.61)
for -l < m < O. The spherical harmonics have been normalized so that (3.62) Proof of the orthonormality of the spherical harmonics The factor 8mm in Eq.{3.62) is obviously correct, for l
la
h
dq,e;(m-m')¢
= 211"6mm , .
(3.63)
So we can set m = m' in what follows. Multiply the differential equation (3.59) by Plf1'(z) and subtract the result from the expression obtained by interchanging land l'. Integrating the result, we find
[f(f + 1)
~ t(t + 1)]/, dzPt(z)P;'(z)
=
(3.64)
Orbital Angular Momentum
58
A partial integration suffices to show that the right side vanishes: the boundary terms contain a factor 1 - z2, which is zero at the boundaries, and the terms in the integral cancel against one another. Thus when f f:. f' the integral on the left side must vanish. This justifies the factor 8u' in Eq.(3.62). It remains to show that the normalization is correct, i.e., to show that (3.65) for m
~
0, where we have defined
or
{I dz [Pi(z)]2 .
=
i-1
To calculate this integral we use Lemma 1 below, which gives om+1
om _ i
i
- (f-m)(f+m+1) '
and on iterating this up to m = f, we find
of
om+2 i
=
(f - m)(f - m - l)(f
+ m + l)(f + m + 2)
(f+m)! i (f - m)!(2f)! Oi .
(3.66)
From Eq.(3.56) and Eq.(3.57) it is easy to show that pi(z) i
=
(2f)! (1 _ z2)i/2 2if! '
(3.67)
and so it follows that Oi i
= {I
)J2 = i -1 dZ [pi( i z
[(2f)!]2 A 2if! i)
(3.68)
where 2 i
1
Ai
=
/ -1
dz (1 - z)
21.+1 (f!)2
= (2f + 1. )., '
where use has been made of Lemma 2. Hence
om _ i
2 (f+m)! - 2f + 1 (f - m)!
(3.69)
Thus we vindicate Eq.(3.65). The above proof can be immediately extended to m < 0 by means of the equivalence Eq.(3.55). END OF PROOF
59
Spherical Harmonics
Lemma 1
1 1
Cr
-1
cm+l
dz [Pi(Z)J2
=
(£ _
m)(~ + m + 1) .
Proof From the expression (3.57), we have
and this leads to
We integrate both sides of this equation and perform an integration by parts:
The integrated parts are all zero. In particular, m [z [Pi(Z)]2]
~1
= 0, trivially
so if m = 0, and in view of the fact that Pi(±l) = 0 for m > 0 (see Eq.(3.57)). We now use the differential equation (3.59) to simplify the first term under the integral, obtaining
1 1
-1
dz [Pi(z)]2 [£(£ + 1) -
(£ - m)(£ + m
m2
1- z
2-
m
+
m 2z2 1- z
2]
+ l)Cr"
Lemma 2
Ai
-1 =
1
-1
2 i 2i + 1 (£!? dz (1 - z) = (£ ), .
2
Proof The integral can be evaluated by noting that
Ai + Ai-l =
/1
2
dZ(z -1)
-1
i-I 2
1
z =-2£
/1 -1
+1
.
2
i
1
dZ(z -1) =-2£Ai'
Orbital Angular Momentum
60
where a partial integration was performed. Hence Ae
=-
2f 2f + 1 Ae-l
e
=
2ff! (-1) (2f + I)!! Ao
=
e 22e +1(fl)2 (-1) (2f + 1) 1 .
END OF LEMMATA
3.3
Zeeman Effect
If the Hamiltonian of a system commutes with the angular momentum operators,
[I, H]
(3.70)
= 0,
then the angular momentum is conserved, i.e., it is time-independent. This property is shared by the three-dimensional harmonic oscillator and the nonrelativistic model of the hydrogen atom, and indeed it is true for any system with a spherically symmetric potential (Exercise 3.4). On condition that the eigenvector If, m) of L2 and L3 is also an eigenvector of the Hamiltonian, Hlf, m)
= Ee,mlf, m) ,
(3.71)
Ee,m cannot depend on m. This follows from Eq.(3.70), for
Howeyer, from Eq.(3.14}, L± If, m) is proportional to
If, m ± 1), and hence
HL±lf,m) = Ee,m±lL±lf,m) ,
and so Ee,m = E e,m±1, on condition that m < f resp. m > -f. This degeneracy is in general broken if the system is exposed to a magnetic field, whence m is sometimes called the magnetic quantum number. The effect was investigated by Zeeman, and we will explain its simplest form here, returning later to more complicated instances of the effect. Suppose that a particle of mass M and charge e executes a circular trajectory of radius r, under the influence of a magnetic induction, jj. Classically the magnetic force is
.... e .... F = -v /\B, e
and so ~
"""+
e-+
~
e-+
-+
. . ."
-+
F·r = -v /\B·r = - r /\v·B, e e where r is the position and v the velocity of the particle. Introducing the momentum, p = M v, we have
....
F .r
=-
e
Me
..
r /\p. B = -
e........
Me
L .B .
(3.72)
61
Exercises
For circular motion the centripetal force is M v 2 / T, in the direction of ....
-r, so
= - M v 2 = - 2E ,
F .r
E being the kinetic energy of the circulating particle. On combining this with Eq.(3.72), we find
E=;i.B, where
. . _ __e_L f-L -
21v[ c
(3.73)
'
which is interpreted as the magnetic moment of the orbiting particle. In the quantum mechanical treatment of this system, we add to the original Hamiltonian, H, a term j1 . jj to account for the effect of the magnetic field, where j1 is still given by Eq.(3.73), but L is now the quantum operator that represents the angular momentum. Let us treat the case of a constant, uniform magnetic induction, and for convenience we choose the x3-axis parallel to this field. Then the total Hamiltonian is
H' = H -
eB L
2Mc
3,
and the eigenvalues of this operator are given by H I If,m) =
(
El,m - emnB) 2Mc If,m).
(3.74)
Hence the energies, El,m, originally degenerate with respect to m, are split, each level being shifted from its neighbor by the same increment. This equal separation, which is generally observed as a splitting of spectral lines, constitutes the phenomenon called the normal Zeeman effect.
3.4
Exercises
Problem 1 Show that any operator that commutes with two components of the angular momentum operator commutes also with the total angular momentum operator. Problem 2 The position, the momentum and the angular momentum of a particle are represented respectively by the operators if, P and L. Show that
(1)
(2)
L 'p=o=L'if L 1\ L = inL
Orbital Angular Momentum
62
Problem 3 Show that (1)
I /\ if + if /\ I
(2)
[L2, if] =
2ihif -2ih(l/\ if - ihif) =
Problem 4 Given a Hamiltonian
where V depends only on q2 and is such that it ~6sses a Fourier transform, show that [I, H] = o. ~o~~~~(es Problem 5 The wave-function of a particle has the form fer, fJ) cos ¢>. Give the possible results of a measurement of the X3-component of the angular momentum, with the corresponding probabilities. Problem 6 Ii, m} is an eigenket of L2 and L3 with eigenvalues l(l + 1)h2 and mh. Show
Problem 7 A particle in a spherically symmetrical potential is in a state given by the wavefunction
where a and (3 are constants. (1) Take a = f3 = 1. What are the possible results of a measurement of L2? Calculate the corresponding probabilities. What are the possible results and the probabilities associated with measurements of Lx, Ly and L z? (2) Take a = 0 but f3 = 1. Answer the same questions for this case.
Problem 8 Suppose that the wave function 'If;(T') is an eigenfunction of Lx and L y • (1) Does this not contradict the uncertainty principle for noncommuting operators? (2) What are the possible results of a measurement of L z ? (3) Show that 'If; is spherically symmetric.
63
Exercises
Problem 9 Consider the following matrix representations of of the of the angular momentum operator.
X1-
and X2-components
-1,
o (1) Calculate L 3 , using the algebra of angular momentum operators. (2) What is the corresponding value of i!, the angular momentum quantum number? (3) Calculate (L 1), (Lr) and D..L1 = ([L1 - (L 1)]2)1/2. Here () refers to matrix elements taken with respect to a normalized eigenvector of L3 belonging to the eigenvalue 1. (4) Consider the state (I = 1, J2). If is measured in this state and the result + 1 is obtained, what is the state after the measurement? How probable was this result? If L1 is now measured, what are the outcomes and their probabilities? Does this second measurement change the state?
Lr
HI,
Problem 10 The generalization of angular momentum in configuration space from 3 to 4 Euclidean dimensions involves the differential operators Ljk
= -i(x j 8 k
-
Xk8j) ,
where j and k take on the values 1,2,3,4, with 1i = 1. Introduce
(1) Discover the commutation relations between (2) Show that the operators
J
and
K.
each obey the standard commutation relations for an angular momentum, and that they commute with one another.
64
Orbital Angular Momentum
Polar Coordinates: Summary w
rsinO
x
rsinOeos¢
y
r sin 0 sin ¢ = w sin ¢
z
reosO
r2
8
x2
= weos¢
+ y2 + z2
tan- 1
[
= w2
+ z2
[:1
';X'z+ Y'] = tan- 1
.. .. '
.... (" ... ' ..
~.~:~
...... .
-..
. ..
"'-:;'"
"--------p... pp"."ecp .
~
Figure 3.1
:.. ........... -.... .
...........
.... .
..
.. .
'.
.- ..
,...:-:-: ...
... ... ... ... .......... ..,
2
1 {j2
1 {
V = --r + r {)r2 r2
1 {). {) 1 {)2} smO- + - -sin 0 {)O {)O sin2 0 {)¢2
Chapter 4
Central Potential
When the potential depends only on the radial variable r equation,
[-::: V2 + Vir)] .p(T)
=
Ii I, the Schrodinger
= E.p(T) ,
(4.1)
can be written
where use has been made of Eq.(3.44). A particular solution has the form (4.2) and from Eq.(3.27) we find that ui(r) must satisfy (4.3) with.£ a non-negative integer.
4.1
Asymptotic Solution for 'r --+ 0
When r
~
0 and .£
~
1, ui"() r
~
.£(.£ + 1) Ui () 2 r , r 65
Central Potential
66
;2.
on condition that VCr) is less singular than Two independent solutions are r.e+l and r-.e. The second is not of physical interest, since it would lead to (4.4) which diverges for £ ~ 1. This is not compatible with the interpretation of 1-zP(i)1 2 as a probability density. For.f. = 0 the argument is more subtle. In the case that V(r) is analytic at the origin, so that V(O) is well-defined, we may write
Uo" (r)
- -2 k Uo (r ) ,
~
where Ii is a constant. Of the two independent solutions sin kr and cos kr, the second one yields a wave-function
-zP (i)
f'J
cos kr .
(4.5)
r
The reason for excluding this is not because of difficulties with the probability interpretation - the integral (4.4) is finite for .f. = 0 - but rather because Eq.(4.5) does not satisfy the Schrodinger equation at r = O. Indeed, Eq.(4.5) implies
V
2 cos kr - 1 r
21
+v-r
_k2 -zP(i) - 47r8 3 (i). To obtain this result, we have used the following identity in distribution theory: (4.6) This result can be proved by introducing a so-called Green's function, namely
I!
G(r) = (2 1r )3 where
E
eiif ·r
d3 q q2
.,
-ZE
(4.7)
--+ O. Then
(4.8) and
1
G(r) = -4 • 7rr Clearly, Eqs.(4.7)-(4.9) are equivalent to Eq.(4.6).
(4.9)
Asymptotic Solution for r -+ 0
Proof of Eqs.( 4.7)-( 4.9) For any integrable test-function,
fer),
67
Eq.(4.7) implies
(4.10) Here (4.11) is the Fourier transform of fer). Now by the fundamental theorem of Fourier analysis, the inverse of Eq.(4.11) is
fer) = (2~)3 /
d3qe-iifOr
i(in .
We see then from Eq.(4.10) that / d 3 rf(f)V 2 G(r)
= - f(O),
which proves Eq.( 4.8). In polar coordinates we can write Eq.( 4.7) in the form G(r)
=
{OO q2dq rTf dO sin 0 (2Tf d¢ exp(~qr c.os 0)
(2 1 )3 7r
J
1 (2 )2 7r
Jo
1
-(27r)2
J
0
J
0
q -
0
'l,f
{OO dq. 2q2. {I dz eiqrz
loo
1
(21r)2ir
0
q q2
Z€
J-1
e iqr -
e- iqr
dq--:--- - - - - q2 - if iqr
{OO
J
-00
q e iqr
dq (q - iE)(q
+ if)
.
( 4.12)
This integral can be performed by closing the contour of integration in the upper half-plane of the complex variable q. The only pole of the integrand inside the contour is that at q = iE, and its residue yields G(r)
END OF PROOF
=
21ri 'l,E (21r )2ir 2iE 1 41rr
( 4.13)
Central Potential
68
Perhaps the reader feels cheated by the first part of the above proof, on the grounds that it relies on a sophisticated result in Fourier theory. She or he can find a more general, and moreover direct proof of this result in Chapter 8 (see Eq.(8.5) et seq.). When VCr) ~ -;- for small r, as is the case for the Coulomb or the Yukawa potentials, the .e = 0 radial equation looks like
for small r. The most general solution has the form 00
uo(r) = ao [1 -logr]
+ aIr + 2: [av + bv logr] rV, v=2
where ao and al are arbitrary, while av and bll are determined from them by Eq.(4.1). For the same reason as in the case that VCr) is analytic, this does not yield a solution of the three-dimensional Schrodinger equation, unless ao = O. Thus in all cases of interest, the behavior of the radial function as r --+ 0 is as follows: (4.14)
4.2
Asymptotic Solution for r -+
00
For large values of r, Eq.(4.3) becomes /I
ue (r)
2mE
+ -2-ue(r) ~ 0, ;"
on condition that V (r) --+ 0 sufficiently rapidly. In the case that the total energy is negative, we set 2 fi,
2mE
= ---
;,,2
>0
'
so there are two asymptotic solutions, e±KT. Of these, only the negative exponential is acceptable, since the positive one is not square integrable and so could not give rise to a probability density. However, after the regular solution has been chosen at the origin, ue(r) is fixed, apart from normalization, and in general when it is continued from small to large values of r, it will be a combination of both asymptotic solutions:
3D Oscillator in Polars
69
Only for those special values of the energy (and so of K) for which Be = 0 do we have an acceptable bound-state solution. Whether one or more bound states exist depends on the form of the potential, V (r ). When the energy is positive, we set 0
k 2 = 2mE
n2
> ,
and now the continuation of the regular solution to large values of r can be written
where now both solutions are allowed: they are not square integrable, but they are acceptable scattering solutions. With a particular choice of overall normalization, this is rewritten
ue(r)
rv
2ik [e- i (kr-i1r/2)
_
Se(k)e i (kr-i1r/2)]
,
where Se(k) is called the scattering function. It depends of course on the form of V(r). The reader is referred to Chapter 8 for a further study of scattering.
4.3
3D Oscillator in Po lars
Let us reconsider the isotropic harmonic oscillator in three dimensions (refer to Section 2.3). The potential is V(r) = Ar 2 , and with the substitutions p=
ue(r)
=
1/2mA
n
r
2
1+1
1
e-'I P p2- Ve(P) ,
the radial equation Eq.( 4.3) becomes
pvt+ (£ + ~
-
p)v~
+ H2€ -2£- 3)ve = 0,
(4.15)
where
E
€=-
nw
with
w=ff..
Try now the series solution 00
ve(p) =
L akpk. k=O
(4.16)
Central Potential
70
The differential equation Eq.( 4.15) implies 00
L ak [k(k + £ + ~)pk-l + H2€ -
2£ - 3 - 4k)pk] = 0,
k=O
which is equivalent to 00
L
[ak+l(k
+ 1)(k + £ + V -
~ak(4k + 2£ + 3 - 2€)] pk
= o.
k=O
Each coefficient must vanish, so 4k + 2£ + 3 - 2€ ak+l = 4(k + 1)(k + £ + ~) ak
t!'
1
rv
k
+ 1 ak,
(4.17)
as k-+ 00. Hence ak and so in general v.e(p) eP , which is unacceptable. The only exceptions occur when the series Eq.(4.16) terminates, for then u.e(r) is square integrable. This occurs when I"V
I"V
€
= 2k+£+ ~2
for some k = 0,1,2, ... , for then ak+l = 0 and therefore also aj j>k+l. Accordingly,
E
= En =
(n
+ ~ )nw ,
o for
all
(4.18)
with n = 0,1,2, ... , where n = 2k + £. This agrees with Eq.(2.29), which we obtained by treating the isotropic oscillator in Cartesian coordinates. For a given principal quantum number, n, the angular momentum quantum number therefore satisfies £ = n - 2k,
i.e., £ = n, n - 2, n - 4, ... , 1 or O. Thus £ has the same parity (evenness or oddness) as does n. This is a property that we did not appreciate in the Cartesian treatment of Chapter 2. To calculate the degeneracy of each level, we recall that each value of £ is associated with 2£ + 1 values of the azimuthal quantum number m. Hence for £ = n - 2k there are 2n - 4k + 1 degenerate levels. This must be summed over k = 0,1,2, ... , [¥-], since the energy does not depend separately on £, but only on n. Now £ is non-negative and we have terminated the k-sequence with the integral part of ~. The total degeneracy of the n th energy level En is [~ ]
I)2n - 4k k=O
+ 1) =
Hn + 1)(n + 2) ,
71
Square- Well Potential
in agreement with Eq.(2.30). Further aspects of the isotropic oscillator will be brought to light in Exercise 4.10.
4.4
Square-Well Potential
Consider first of all a free particle, I.e., one for which V(r) Schrodinger equation,
-!{\12~(r) = E~(r), 2m
O.
The free
(4.19)
has a solution of the form Eq.( 4.2), with (4.20) where k 2
= 2mE /n?
Set p = kr, so that
d2 f(f + { dp2 +1p2
I)}
u£(r(p)) = O.
(4.21 )
A solution of Eq.(4.21) that vanishes at the origin is
where j£ (p) is called the spherical Bessel function. We shall shortly return to the general case, and to the properties of this function, but for the moment we concentrate on the S-wave, i.e., the case f = 0, for which Eq.(4.21) reduces to
d2 } { dp2 + 1 u£
= 0,
(4.22)
with the solution uo(r) = pjo(p) = sinp = sinkr, vanishing at the origin. Consider next the situation in which V (r) = - Vo (a constant) for r < a, and VCr) = 0 for r > a. If E > 0 and we set
k=
V2m (E + Vo) 1i
'
the radial Schrodinger equation has the form Eq.(4.20), but with k. Accordingly we can write, for r < a,
(4.23)
k in place of (4.24)
Central Potential
72
For r > a, we have the free Schrodinger equation Eq.(4.19)j but the more general solution
Ul(r)
= kr [Ai£(kr) + Bnl(kr)]
(4.25)
is possible, where the spherical Neumann /unctions, nl(p), will also- be defined shortly. For the S-wave, the corresponding solution is pno(p) = - cosp, and Eq.( 4.25) reduces to
uo(r)
= Asin(kr) -
Bcos(kr).
(4.26)
The coefficients A and B have to be determined by matching Ul (r ), and its first derivative, as given by Eq.(4.24) and Eq.(4.25), at the point r = a. A similar situation obtains if E is negative but larger than - Vo, for then E + Vo > 0, and so k is real and the solution Eq.(4.24) is still valid when r < a. However, k is now imaginary, and it is convenient to write Ii
= -ik =V -2mE ,
(4.27)
1i
so that the solution for r
> a can be written
For the S-wave this reads simply
Uo (r)
= A sineilir) -
B cos( ilir)
= iA sinh (lir)
- B cosh( lir) .
(4.28)
Again, A and B have to be determined by matching Ul (r) and its first derivative at the point r = aj but the problem now is that in general the solution will explode exponentially as r -t 00. Only for the special instances (if they exist) in which B = iA will (4.28) reduce to
uo(r)
= -B exp( -lir) ,
( 4.29)
which is an acceptable, square-integrable function. The special cases are boundstate or eigenvalue conditions. We must match the solutions, and their derivatives, at r = a. In the limit € -t 0, U~( a -
€) = u~ (a + €) ,
and these constraints can conveniently be combined into the eigenvalue equation,
(4.30)
73
Square- Well Potential
For the S-wave, this matching condition takes on the form d
-
da log sin ka
d
= da log e -Ita,
which leads to (4.31)
= -kcotka.
K,
However, K, and k are related, according to their definitions Eq.(4.23) and Eq.(4.27), by 2 K,
-22m +k = - 2 Vo,
(4.32)
n
which is a circle in the K, - k plane. In Figure 4.1 we show the first quadrant of this circle for a number of different values of the potential strength, Vo (with a = 1). Also plotted are the curves corresponding to Eq. (4.31). Intersections of the circles and the cotangent curves correspond to energy eigenvalues, i.e., to solutions of the bound-state problem. When the radius of the circle is less than ~, there is no bound state: the potential is in this case too weak to bind even one state. With the radius just equal to ~ (the smallest circle in the figure), there is a zero-energy bound state, and, as Vo is increased, the bound state becomes more deeply bound, until at a radius of a second bound state makes its appearance (this corresponds to the second circle), and so on.
10
3; ,
37T" Figure 4.1
Clearly, if the potential strength is such that
(2N - 1)7T" 2
a
12 Vr
1.70343}. Thus Va
= EB +
(klic) 2 mnc
2
= 125 MeV.
(4.38)
Further refinements are possible, for example the square-well potential can be replaced by a Yukawa potential,
V(r) = -~ e- cmrr / fi r
,
where m7r is the mass of a 7r-meson or pion (140 MeV), and 9 is the the attractive force engendered by the exchange of a pion between and the neutron. However, the square well gives a rough idea of the the interaction that just manages to bind the nucleons into a stable
4.5
strength of the proton strength of nucleus.
Spherical Bessel Functions
The spherical Bessel and Neumann functions were introduced as special solutions of Eq.(4.21); but so far we have only looked at f = O. To study the general case, consider the expression (4.39)
76
Central Potential
By making various choices for the function 10 (p), we will shortly be able to define not only the spherical Bessel and Neumann functions, but also the spherical Hankel functions, which are convenient for use in bound-state problems. Consider further the differential operators
d
be = dp
f
t d b = -e dp
+-p
f + -. p
(4.40)
It is easy to check that,. for any twice differentiable function, g(p),
f(f-l)} () { _~ 2 gp d 2+
P { _~ dp2
P
+
f(f+1)} () p2 9 P .
(4.41)
b~ and be are ladder operators for the functions fe(p) of Eq.( 4.39), in particular one finds
b~fe-l(p) = fe(p)·
(4.42)
It follows therefore that (4.43) and hence from Eq.(4.41) that (4.44) If the function fe-l(P) is such that
(4.45) then it follows from Eq.( 4.44) that (4.46) However, from Eq.(4.42) this means that (4.47)
77
Spherical Bessel Functions
Since Eq.(4.47) has the same form as Eq.(4.45), with l - 1 replaced by l, it follows by induction that, for any function fo(p) that satisfies
d2
(4.48)
- dp2fo(p) = fo(p) ,
the corresponding function h(p), defined by Eq.( 4.39) with this fo(p), necessarily satisfies d2 { dp2
+1-
l(l + 1) } p2
h(p)
= o.
(4.49)
Note that, whichever solution of Eq.(4.48) is chosen, the recurrence relation
(4.50) holds, this being simply the content of Eq.( 4.42). This recurrence relation may be used to generate higher from lower functions on the ladder. The choice fo (p) = sin p leads to h(p)
= pjf(p)
(4.51)
where the spherical Bessel function is defined by
d)
. ( ) _ ( )f (1 JfP--P -pdp
f
sin-p. p
(4.52)
The choice fo (p) = - cos p leads to
(4.53) w here the spherical Neumann function is defined by nf(p) =
f -( -p) (
1 d - - )
i
pdp
cosp
(4.54)
--. p
To analyze the threshold behavior of jf(p) and ni(p), it is convenient to introduce the variable a = p2, in terms of which (_2)faf / 2 (~ \) f sin Vada) Va(_2)faf/2
(~)\ f da
f
p=o
<Xl
(_2)ia i / 2 ' " (-1)P
L..J
p=i
(-a)P (2p + 1)! ,
p.
(p - l)!(2p + 1)!
a P- i .
Central Potential
78
The leading contribution comes from the term p the variable p, we find
=f
in the sum. Reverting to
(4.55) The spherical Neumann function has on the other hand non-integral powers of a in the expansion of (4.56) To get the leading power, it suffices to replace cos y'(i by unity, and since
= (-2)-£ (2f)! -(£+1)/2 ( ~)£_1_ da y'(i f! a , it follows that n£ (p )
rv -
(2f)! ( -p )-£-1 .
--
f!
(4.57)
In view of the threshold behaviors Eq.( 4.55)-( 4.57), and the discussion given in Section 4.1, it will be readily understood why only the spherical Bessel function is allowed as r -+ o. Thus
is an acceptable solution of the free Schrodinger equation. A much more general solution is obtained by superposing solutions: 00
1jJ(f) =
£
L L
A£m(k)j£(kr)Yim(O, ¢;),
£=0 m=-£ where A£m are arbitrary coefficients, constrained only by convergence requirements of the infinite f-series. They may depend on k, but not on r, 0 or ¢;, the essential point being that the Schrodinger equation Eq.(4.1) is linear and does not contain f or m. To consider asymptotic behaviors as p -+ 00, it is convenient to introduce the spherical Hankel function of the first kind, defined by
(4.58)
79
Spherical Bessel Functions
In particular, h~1 ) (p) = -i eip / p, corresponding to the choice 10 (p) Eq.(4.39). Accordingly, the explicit form for this Hankel function is
(1-pdp-d) -p . £
h (1) (p) =
£
. £ -l,(-p)
=
-ie ip in
.
etp
(4.59)
For large values of the argument, p, the leading contribution comes from repeatedly allowing the differential operator to act on e ip , ignoring its effect on 1/ p, since those differentiations produce terms that vanish more quickly than does 1/ p itself. Thus we find, for p -t 00, (1)
i
h£ (p)~-p(-I)
£(
d )
dp
i.
£ iP
i
£ i
(.
i7rf)
e =-p(-l,) eP=-pex p l,P-T
(4.60)
From Eq.( 4.58) we deduce (4.61 )
(4.62) Returning to the discussion of the square-well potential, we recall from Eq.( 4.27) that we had set k = iK for negative energies, where K is positive. In this case,
h (1) (iKT) £
~ - -
1 exp (
-KT -
KT
i7rf)
-
2
.
(4.63)
Thus the Hankel function of the first kind tends to zero exponentially at infinity, and it is the precise combination of spherical Bessel and Neumann functions that is needed for a bound state: for this reason, it is sometimes affectionately called the friendly Hankel function. The Hankel function of the second kind, h?)(p) = j£(p) - in.e(p), is on the contrary unfriendly, in the sense that it explodes exponentially at infinity. The eigenvalue condition Eq.(4.30) reduces to d da log [ j.e(ka - €) ]
=
d da log [Aj.e(ka + €)
+ Bn.e(ka + E)] .
(4.64)
As in the discussion of the S-wave, we have a bound-state condition (i.e., a square-integrable solution) only when B = iA, for only then do the spherical Bessel and Neumann functions combine to form the friendly Hankel function.
80
Central Potential
In this case, Eq.( 4.64) reduces to
! 4.6
log [jl(lca -
E)]
=
d~ log [h~l)(ka + E)]
(4.65)
Vibrational-Rotational Spectra
In this section we shall consider a simple diatomic molecule, for example carbon monoxide, CO, in which the constituent atoms have different masses, say ma and mb. Between the two constituents there is an attractive force, due to the presence of the valence electrons that are shared between the two nuclei. Such a force is described typically by a potential that is positive for small nuclear separations, due to the Coulomb repulsion between the nuclei, becomes negative for intermediate energies, with a well-defined minimum, and tends to zero for large separations, corresponding to the dissociation of the molecule. The Hamiltonian of the molecule can be written 2
H = ~ 2ma
2
+~ + V(I-->qa 2 mb
--> I) , qb
(4.66)
where the potential is a function of the modulus of the difference of the position coordinates of the atoms, ifa and if b. We need to make a canonical transformation to the center-of-mass and the relative coordinates,
Q=
maifa + mbifb ma+mb
P=Pa+ih mbPa - maPb P=-----ma+mb The quantization conditions
with all other commutators vanishing, lead to
with all other commutators vanishing. Thus the center-of-mass and relative coordinates, respectively (Q ,P) and (if,p), are independent canonical variables.
81
Vibrational-Rotational Spectra
The Hamiltonian Eq.(4.66) becomes p2
e2
p2
H=-+--21M 2m ItIl' where
M=ma+mb, and 1
1 ma
1 mb
-=-+-, m
( 4.67)
which shows that, in the treatment of the relative motion of the two atoms, the reduced mass m must be used. Note that the discussion preceding Eq.( 4.35) refers to the special case ma = mb. The motion of the molecule as a whole is free, governed as it is by the kinetic term p2 / (2M) only, while the internal motion of the atoms in the molecule is governed by the Hamiltonian
where we have now specialized to the configuration representation in the relative variables, in which the potential is central. The radial equation has the standard form Eq.(4.3), and we may expand the potential about the point, r = ro, at which it has its minimum:
where we have neglected terms of order (r - ro)3. Here V"(ro) > 0, since V(ro) is the minimum value of V(r). Accordingly, we may write Eq.(4.3) in the approximate form u£"() T
+ { -2m 2[ E -( V ro)] - .:\(T - TO) 2 - f (f + 2 1) } ~
h
U£ (r) =
0,
(4.68)
where A = mV"(To)/h 2 is the coupling parameter of a one-dimensional harmonic oscillator. The energy levels are therefore given by Enl =
V(TO)
1
+ hw(v + 2) +
h 2 f(f + 1) 2 2 mTo
'
where the vibrational angular frequency is determined by the oscillator strength, A. Here v is the oscillator quantum number. A further refinement would be to allow both ro and w to depend on £, since in fact the centrifugal term distorts
82
Central Potential
the form of the attractive potential; indeed, for large .e values there may well be no minimum in the effective potential. The electronic levels are separated by a few electron volts, and transitions between them give rise to photons that are typically in the visible spectrum, but each electronic level is split into equally-spaced vibrational levels, separated by a few millivolts. Transitions between these levels are in the infrared. The vibrational levels are themselves split into sublevels by the rotational degree of freedom, this splitting being not equally spaced, since the centrifugal term is proportional to .e(.e + 1). Transitions between levels at different f-values but the same v-value are in the far infrared. The rotational structure also shows up in the fine structure of the near infrared spectrum, corresponding to transitions say between vibrational level v and v-I, complicated by the fact that the initial state can be in one or another rotational state. Thus the energy difference between the vP and the (v - 1)8 level is slightly different from that between vP and (v - l)D, or between v8 and (v - l)P, and so on. The combination of vibration and rotation thus produces a complex spectrum, which can be used to elucidate the details of the molecular structure.
4.7
Exercises
Problem 1 The spherical Bessel and Neumann functions are respectively
d) pdp
. ( P) -_ ( -p )£ (1 sinp JR. -- £ -
nR.(p)
=
£ -(-p) (
1 d
- -) pdp
p
£
cos p --. p
(1) Prove the recurrence relations j£+l(p)
= -j~(p)
n£+l(p) =
f
+ -iR.(p) p
-n~(p) + ~nR.(p). p
(2) Work out j£(p) and n£(p) for f = 0,1,2,3, and sketch them graphically.
Exercises
83
Problem 2 Let 'ljJ(r) = uo(r)lr be an S-wave solution of the Schrodinger equation for a potential that has bound states, and which is strictly zero for r > a. Define the following function of the total energy:
feE) = u~(a) . u(a) •
Let E B be a bound-state energy. Evaluate f (E B)' Problem 3 Solve the bound state problem for a particle of mass m in the central potential
VCr)
o 00
where a
O b. Calculate the energies of the ground state and the first excited state, and obtain an approximate expression for the energy splitting of these levels if Vo is very large compared to these energy levels. Problem 6 Show that there cannot be more than one linearly independent bound state wave function for a given E and f, i.e., with the same energy and angular momentum. Problem 7 Given the radial potential energy for a diatomic molecule,
VCr) = V(ro)
+ ~mw2(r -
ro)2
h2
+ 2mr2f(f + 1),
find the radius, rl, at which this potential energy is minimum. Evaluate A and B in the expansion
84
Central Potential
Problem 8 A particle is in a cylindrical potential in three dimensions, defined by V (p) = 0 + x~ < and V(p) = 00 elsewhere. if p =
Jxi
a
(1) Determine the three lowest eigenvalues for states that have P3
L3
=
= o.
(2) Determine the three lowest eigenvalues for states that have P3 =
0 and
o.
Problem 9 A particle in a spherically symmetric potential is described by a wave packet tP = (XIX2 + X2X3 + X3Xl) e- ar2 • What is the probability that a measurement of the square of the orbital angular momentum yields zero? What is the probability that it yields 6h 2 ? If the orbital angular momentum is 2, what are the relative probabilities for m = - 2, -1, 0, 1, 2? Problem 10 With Ii = 1, consider the isotropic harmonic oscillator Hamiltonian in three dimensions,
and the operator
Let Enl.m be the energy level corresponding to principal quantum number n, angular momentum quantum number l, and azimuthal quantum number m,
Hlnlm) = Enl.mlnlm) . (1) Show that [H, U] = 0, (2) Show that (xIUlnll) oc (xln, l + 2, l + 2). Under what restrictions on n and l is this true? (3) Deduce from (1) and (2) that Enl.l. = E n ,I.+2,1.+2, under the same restrictions on nand l as in the previous question. (4) The isotropic oscillator Hamiltonian is invariant under rotations, i.e., orthogonal transformations of the coordinate axes. This leads to the fact that Enl.m is actually independent of m. By expressing Hand U in terms of creation and annihilation operators, discover the larger symmetry of the Hamiltonian that is responsible for the fact that Enl.m is also independent of l.
Chapter 5
Hydrogen Atom and Charmed Quark
In preceding chapters, we have encountered the potential r2 in connection with the harmonic oscillator, and this led to a study of the Hermite polynomials. We have also considered the repulsive, centrifugal potential r- 2 that required the introduction of the spherical Bessel functions. In this chapter we shall study the attractive potentials rand r- 1 , which will lead us to look at Airy functions and Laguerre polynomials. The case of the linear potential will be applied to the problem of the confinement of quarks in mesons, while the Coulomb potential, which we will first study, will be applied to the calculation of the energy levels of the hydrogen atom.
5.1
Radial Equation for Coulomb Potential
The potential energy of the electron in a hydrogen atom, or a (He+, Li++, etc.), is
hydrogen~like
ion
Ze 2 V(r) = - - , r where e is the elementary charge, and Z is the atomic number, i.e., the number of protons in the nucleus. With the notation ~ _ 2mZe 2 ':. -
2
n
and, for negative energies, 2 K
2mE
=---2->0, 11,
85
Hydrogen Atom and Charmed Quark
86
we may write the wave function
(5.1) where ul(r) satisfies
Ul" (r) + ['-:;. - £(£+1)] r2 Ul(r) = '" 2 Ul(r),
(5.2)
and we require solutions with the asymptotic behaviors r l + 1 as r -+ 0, and e- Kr as r -+ 00. Let us define p
=
2"'T
=
.j-BmE
n
)..=~=e2ZJ
n
2",
T
m
(5.3)
2E'
so that the radial Schrodinger equation Eq.(5.2) takes the form
d2ul dp2
+ [~. _ £(£ + 1)] Ul = ~Ul. P
Given the required asymptotics at p
= 0 and p =
(5.4)
4
p2
00,
it is sensible to write
Ul(r) = /+l e - p /2 L(p),
(5.5)
and we find from Eq.(5.4) that
pL"(p) + [2(£ + 1) - p] L'(p) + [A - £ - 1] L(p)
=
O.
(5.6)
Let us try a power series 00
L(p) =
L ajrJ,
(5.7)
j=O
with, say, ao = 1 (arbitrary normalization). From Eq.(5.6) we readily deduce the recurrence relation
j+£+l-A For very large j, this approaches
(5.B)
Radial Equation for Coulomb Potential
87
and so one intuitively expects
L ~J. = e
L(p) '"
P;
j
from Eq.(5.5) this means that
ul(r) '" e+ p / 2 = e KT , up to powers. This behavior is unacceptable, since it leads to a wave function that is not square-integrable, and hence one that is not subject to the probability interpretation. The above sloppy argument can be made rigorous by observing that, for i>A all the aj have the same sign, so no cancellation is possible; and, for any € > 0, it is possible to find a iE such that, for all i > ie ,
-.e,
a"+l _3_
> 1-
~ 2
i +1
aj
Hence aj >ie! - ( 1--€ )j-j. aj. -
i!
2
It is now easy to show that indeed ul(r) blows up exponentially as r -+ 00. Note that analogous reasoning establishes the correctness of the corresponding analysis of the infinite series solution for the isotropic harmonic oscillator (cf., Eq.(4.17) et seq.). The only way to prevent the above disaster is to require L(p) to be a polynomial, which will happen if the series Eq.(5.7) terminates. Then ul(r) behaves like e- KT , up to powers of r, so that it is indeed square-integrable. The series can terminate only if, for some i, say imax,
imax + .e + 1 -
A = 0,
(5.9)
for then, according to Eq.(5.8), ajInax+ 1 = 0, and hence aj = 0 for all i ~ imax+ 1. However, imax is necessarily a non-negative integer (it could be zero); and since A = imax + .e + 1, it follows that A must be an integer, not smaller than .e + 1. Let us write A = n,
(5.10)
which is called the principal quantum number. Writing E = En for the nth energy-level, we have from Eq.(5.3) and Eq.(5.10)
En = -
me 4 Z 2 1 2 2· 21i n
(5.11)
88
Hydrogen Atom and Charmed Quark
Note that Eq.(5.9)-(5.10) imply
f'
~~,
and now improve the approximation by setting ¢" =
~'
in Eq.(5.32), giving (5.33)
Hudroaen Atom and Charmed Ouark _
96
This equation integrates to
and hence we find (5.34) the missing lower limit of the r'-integration accounting only for a normalization factor. To see how this works out for the cc bound states, we recall that we had scaled the Schrodinger equation, with the linear potential, to yield Eq.(5.24). In our present notation, with e in place of r, and /'\,2 = e, we see from Eq.(5.34) that, for e > 0, (5.35) with arbitrary normalization. For negative e, we set e = lei e±i7r, corresponding to a rotation from positive to negative values of e, the upper over an anticlockwise, the lower over a clockwise arc in the complex e-plane, yielding (5.36) The required value of uo(r), which would be obtained by integrating the Schrodinger equation from positive to negative values of e, is the real part of Eq.(5.36), so (5.37) Expressions Eq.(5.35) and Eq.(5.37) give indeed the correct asymptotic expressions for the Airy function, respectively for e -+ 00 and e -+ -00, up to normalization. The first two zeros of the expression Eq.(5.37) are obtained by firstly setting the argument of the cosine equal to ~ and secondly to 3;. This yields the estimates
97r] 1 = 161 ~ [8
2
2.32
161 '"
[2~'r =
4.08,
which compare well with the zeros of the Airy function that were quoted in Eq.(5.28).
97
Exercises
5.4
Exercises
Problem 1 The associated Laguerre polynomial is defined by
LHp)
= Up)' eP
UJ
[pte-Pl·
(1) Prove the recurrence relation
L:+1(p) = pL:+l(p) (2) Work out Lt(p) for t them graphically.
+ (s + t + 1- p)L:Cp) - SL:-l(p).
= 0,1,2,3 and all integral values of s, and sketch
Problem 2 Is the electron in a hydrogen atom on the average further away from the proton when it is in the 2P orbital than when it is in the 28 orbital? Problem 3 Calculate the wavelength of the 2P -+ 18 transitions in deuterium. Use these masses for the deuteron and the electron: mdc2 = 1875.6 MeV, m e c2 = 0.51100 MeV, a = e2/(hc) = 1/137.036. Problem 4 Do the bound-state eigenfunctions corresponding to the following potentials form a complete set? Motivate your answers. (1) Attractive Coulomb: VCr) = -e 2 /r (2) The above plus a harmonic oscillator: VCr) = Ar2 - e2/r Problem 5 k)(¢>kl1/Jn(A))
.
k
Upon insertion of this expression in Eq.(7.6), we obtain
En(A) = En(0)
+ A(¢>nIHII¢>n) +
A """ (4)nl1/Jn(A)) ~ (4)nI H II4>k)(¢>kl1/Jn(A)).
(7.7)
We take a matrix element of Eq.(7.3) with a state other than (¢>nl, say (¢>kl:
(7.8) which implies
(¢>kl1/Jn(A)) = A (¢>kI H ll7jJn(A)) En(A) - EkO)
(7.9)
On substituting this into the last matrix element in Eq.(7.7), we find
En(A)
= E(O) + A(¢>nI H II4>n) + n
A2 ~'(¢>nIHII¢>k)(¢>kIHll1/Jn(A)). (4)nl1/Jn(A)) k En(A) - EkO) (7.10)
Note that Eq.(7.10) is exact. From Eq.(7.9) we see that the second term on the right-hand side of Eq.(7.4) is of first order in A, since En(A) - EkO) is 0(1), and hence. (7.11)
115
Perturbation Theory
From Eq.(7.6) or Eq.(7.7), (7.12) Inserting Eq.(7.11) and Eq.(7.12) into the right-hand side ofEq.(7.10), we obtain finally En()..) =
E~O) + )..(¢nIHll¢n) + )..2L:' 1(¢ZOI)Hll¢~~t + O()..3). k
En
(7.13)
- Ek
This is the required expression for a perturbed energy level, to second order, in the absence of degeneracy. The above treatment, based on the exact formula Eq.(7.10), is much more attractive than the expansions given in many textbooks. The second-order expression is often very difficult to calculate for interesting problems. An enormous simplification takes place if, for a given Ho and HI, one can find an operator, 0, such that (7.14) In this case
E~O) + )..(¢nIHll"HI in first-order perturbation theory.
Problem 2 Two spin-half particles, 1 en 2, have the unperturbed Hamiltonian
A perturbing Hamiltonian is brought to bear of the form HI
= B(SlxS2x
-
SlyS2y).
(1) Calculate the eigenvalues and the eigenfunctions of Ho. (2) Calculate the exact eigenvalues of Ho + HI. (3) By means of perturbation theory, calculate the first- and the secondorder shifts of the ground state energy of H 0, as a consequence of the perturbation HI. Compare these results with those of (2).
Problem 3 Suppose that a hydrogen atom is exposed to a uniform electric field, f, and a parallel, uniform magnetic induction, jj. Consider the first excited energy level, corresponding to n = 2, which is quadruply degenerate if f = 0 = jj . (1) Show that in general the level is split into four nondegenerate energy levels. (2) For what values of f and jj are there instead only three levels, and what are the degeneracies of these levels? (3) For what values of f and jj are there only two levels, and what are the degeneracies of these levels? In the answers, assume that the fields are strong compared with the fine structure effects, and ignore the complications caused by the electron spin.
Problem 4 An electron is in a superposition of the Coulomb and the 3D oscillator potential: e2
VCr) = -- + >..r2. r
(1) What is the shift of the ground state energy to first order in >... (2) How large must>.. be in order that the ground state have zero energy? (3) What are the first-order shifts of the n = 2 energies?
127
Exercises
Problem 5 Consider the two-dimensional oscillator Hamiltonian,
Ho
1
p2
= -+ -mw 2m 2
2 2
q
'
with the perturbation HI = Aqlq2 . (1) Calculate the energy shift of the ground state in first- and second-order perturbation theory. (2) Calculate the energy shift of the first excited state in first-order perturbation theory. (3) Calculate the first two energy levels exactly. Problem 6 Show that a function 'l/J(x) exists, such that, if V(x)
1
00
-00
dx'l/J*(x)
[
< 0 for all
n,2 d -2 d 2 + V(x) 1'l/J(x) m x 2
..) = L
2
2m
+ q2 + >..q4
and set m = ~ 11,2 , so that the ground state energy of the oscillator in the absence of the quartic perturbation is unity, i.e., Eo(O) = 1. (1) Calculate the ground state energy of the oscillator to second order in perturbation theory,
giving explicit expressions for the coefficients al and a2. (2) From this second-order perturbative result, calculate the [1,1] Pade approximant, Eb1 ,11. The [M, N] Pade approximant to a function that has . . a power senes expanSIOn,
is defined by the relation j[N,Ml (>..)
2:~=o bn>..n 1 + 2::=1 cn>..n M+N
2:: an>..n + O(>..M+N+1) ,
n=O
which defines the coefficients bn and Cn uniquely in terms of the first M + N + 1 coefficients an. (3) Make a table of EbO,l] (>..) ,EbO,2](>..) and Eb1,1] (>..), giving 6 significant figures, for each of the cases>.. = 0.1, 0.2, 1.0, and, if you have the computer facilities, write a program to calculate E o(>") to sufficient accuracy to guarantee the correctness of 7 significant figures.
Chapter 8
Scattering Theory
Up to this point we have mainly been interested in bound states, which are described by square-integrable wave functions. In this chapter we introduce the subject of scattering, in which the energies are positive and the scattering solutions are not square integrable.
8.1
Lippmann-Schwinger Equation
Consider solutions of the Schrodinger equation for positive energies, E =
n;.!2, (8.1)
To convert this partial differential equation into an integral equation we define the Green's function,
(8.2) where
E
-+ O. In polar coordinates this is
_1_1 (2 ) 7r
3
_1_
(27r)2 1 --
(27r)2
00
q
0
2d
roo dq
io
1
1 (27r)2ir
00
0
q
111' dB sm. B 1211' dA.. exp( iqrk2 cos .B) 0
If'
0
r i-I 1
q2
q2 - k 2 - if
2
q -
-
'If
dz eiqrz
q2 e iqr _ e- iqr dq-----------------q2 - k 2 - iE iqr
roo
i-oo dq (q 129
q eiqr
k - if')(q + k
+ if')
,
(8.3)
Scattering Theory
130
where E = 2kE' and k = +.vk2. This integral can be performed by closing the contour of integration in the upper half-plane of the complex variable q. The only pole of the integrand inside the contour is that at q = k +iE', and its residue yields 27ri k eikr -Gk(r) (27r)2ir 2k 1 eikr --- . (8.4)
47r r
We now claim that
(V'2
+ k 2 ) Gk(r) = (V'2 + k 2 )
ikr _e_
41Tr
= -8 3 (r).
(8.5)
We shall now give a direct proof of this equation, using distribution theory methods. Note that the special case k = 0 corresponds to Eq.(4.7)-(4.9), and so the following proof, in which no appeal will be made to the Fourier theorem, may be used in place of that to be found in Chapter 4.
Proof of Eq.{8.5) Let g( r) be an infinitely differentiable test function of finite support. Then
by two partial integrations. In spherical polars, this can be written
f /.00 dO
eikr 1 8 2
r 2dr-- 8 2rg(r).
orr r
(8.6)
We have neglected the angular parts of the Laplacian, since they give zero contribution. This follows from the fact that 7r /.27r /.o sin ede 0 d¢
/.00 dr e~kr . {sin1 e 8e8 sin e 8()8 + sin21 e 8¢2 82 } g(r) 0
can be written
/.= where
and
dre'kr
{[~ dA(i') + [
dOB(i')} ,
Lippmann-Schwinger Equation
131
Clearly A and B are zero because of the properties of the test-function space: it is enough that the first derivative of 9 with respect to () is bounded, and that the first derivative of 9 with respect to ¢ is continuous, so that its values at ¢ = 0 and ¢ = 271" are the same. By two partial integrations of Eq.(8.6) with respect to r, we see that
100 o
. 02 dr e'tkr -rg(r)
[ eikr ~rg(i)] 00 _ ik [00 dr eikr ~rg(i)
or2
or
[eikr
Jo
0
(1 + r! -
or
ikr) g(r)[ - k 2 fo~ dreikr rg(r).
Because of the continuity of the test functions at r = 0 and their vanishing as r ---+ 00, it follows that the integrated term above reduces to ~g(O). Thus we have shown that
!
d3 xg(r)V
In other words
2e~r = ! dO [-g(O) _ k21~ r 2dr e~r g(r)]
!
d3 xg(i) [\7 2
eikr
+ k 2 ] -r-
=
-471"g(0) ,
which is equivalent to Eq.(8.5). END OF PROOF
The general solution of Eq.(8.1) satisfies the Lippmann-Schwinger equation:
1/Jk(i) = 1/J2(i) -
~":
!
d3 r'Gk(1f - i'I)V(i')1/Jk(i') ,
(8.7)
where 1/J£(i) is the general solution of the free Schrodinger equation (8.8)
A special solution, corresponding to a plane wave transmitted along the z axis, is obtained by setting 1/Jf(i) = eik .r = eikz , so that in this case 1/Jk(i) = exp (if. i) -
2~ n~
!
d3 r'Gk(1f -
r'I)V(i')~k(i') .
(8.9)
This is the form of the equation that we use in a discussion of a typical experiment, in which a collimated beam of monochromatic particles is directed at a target, often in the form of a small sample of material.
132
Scattering Theory
Proof that Eq.(8.7) and Eq.(8.8) yield a solution of Eq.(8.1) Let 'l/Jk(i) be any solution of Eq.(8.7). Then
o
(V2
+ k2)1/J~(i)
(\7 2
+ k 2 ) {'I/I.(r) + :~
(V 2
+ k2)1/Jk(i) -
{ \7 2
1 ~": 1
d"r'G.(IT - r'I)V(r')'I/I.(r') }
d 3 r'b 3 (i - i')V(i')1/Jk(f")
+ k 2 - :~V(r) } 'I/I.(T).
(8.10)
END OF PROOF
Typically r = iii refers to distances of the order of meters or more, in the laboratory, whereas r' = Ii' I refers to atomic or nuclear dimensions. Hence it makes sense to consider an approximation that is valid when r > > r',
Ii - i'l ~ ry'l -
-,
....
2i· i' fr2 ~ r - ~, r
so that
(8.11) where k' = ki fr. Note that k' is a vector of the same length as k, but it is in the direction of i, whereas of course k itself is parallel to the z-axis. On condition that V(i') tends to zero sufficiently quickly as r' ---+ 00, we find asymptotically
(8.12) where, according to Eq.(8.7), the scattering amplitude is
fk((},4» = -
2: 1d 2
3 r'
e-ik'or'V(i')1/Jk(i').
(8.13)
Here () and 4> are the polar angles of i with respect to the incident wave. The form of Eq.(8.l2) is that of an outgoing spherical wave, superposed on the incident plane wave.
133
Scattering Gross-section
8.2
Scattering Cross-section
The wave function describing a scattered particle satisfies the time-dependent Schrodinger equation,
and for a real potential the Hermitian conjugate of this equation is
1
. a'!f;* * H='IjJ * [ - n2V +-2 +V . -1,n-='IjJ 2m
at
Therefore
(8.14) H
+-
-+
where V V - V . The rate of change in time of the probability of finding the particle within a given volume V is
(8.15) where A is the surface of the volume V and the suffix N indicates the normal component to this surface. The GauE theorem has been used here. Now consider the scattering solution (8.12), which can be rewritten (8.16) since
k'
= ki
and therefore
Ir.
For large
T,
Scattering Theory
134
where
In a moment we are going to average the flux over a small area, ~A = r2 ~O (see Eq.(8.18) below); and we introduce a normalized, infinitely differentiable smearing function, g(r), that has support ~A. We then replace Sk(r) by
and by repeated partial integrations in the z = cos 0 integral, we obtain at least three negative powers of ikr, on condition that fk (0, ¢) is at least thrice differentiable in the variable z. Under this condition, the smeared Bk(r) can be neglected, since it is o(r- 2 ). Accordingly, in leading orders (8.17) Suppose now that N particles per unit area per unit time fall upon a scattering center. The number that pass each second through a small area ~A is
[*
inN +-+] nN [.... - 2m ~A 'ljJk \1 'ljJk N = m ~A k
2] + k' r2Ifk (O, ¢)I N
(8.18)
If ~A is normal to k, the dominant term is the first one, the other being negligible for large r. The flux (the number of particles per unit area per unit time) is accordingly 1i~k in the forward direction. In any other direction, the
term k can be neglected. The reason is that the incident beam is never really plane, for that would imply an infinite lateral extent. Instead it is finite, although enormous compared with atomic scales, but at a macroscopic distance from the target, in a direction specified by the polar angles 0 and ¢, the detection apparatus lies outside the domain of the incident beam. Here only the second term in Eq.(8.18) contributes, and so there the flux is n~klfk(O,¢)12/r2. This flux tends to zero of course as r --+ 00, since the scattered energy is spread out in all directions. However, at radius r the area subtended by an increment of solid angle dO = sin OdOd¢ is r 2dO, this incremental area being normal to the vector k', so that finally the energy that is scattered into the solid angle dO, divided by the incident energy per unit area per unit time, is (8.19)
Born Approximation
135
This is called the differential cross-section. In practice one does not deal with just one scattering center, but with a target full of them. The measured flux of particles at angles ((), ¢) must then be divided by the number of scattering centers that are to be found per unit area of the incident beam. In most atomic, nuclear or high-energy scattering experiments there are three typical scales that are very different from one another: (1) Size of the target particle (perhaps a proton ~ la- 15 m). (2) Area of the target and incident beam (perhaps a square centimeter). (3) Distance of the target from the detection apparatus (often many meters). Under these conditions the approximations we have made are very good. The total cross-section,
a=
I
da dOdO'
(8.20)
is the scattered energy divided by the incident energy per unit area, both for unit time and per scatterer. The cross-section therefore has the dimensions of an area. It is a basic nexus at which the theoretician and the experimentalist meet: the former calculates the scattering amplitude from his theory and predicts the cross-section from Eq.(8.l9), while the latter measures the intensity of the scattered particles and deduces the cross-section. Such measurements serve to verify or falsify a theory and hence to support or undermine it.
8.3
Born Approximation
The Born approximation consists in iterating Eq.(8.7) once to yield (8.21) For an incident plane wave, ¢2(f) = eik .r , this leads to (8.22) cf., Eq.(8.l3). We see that the scattering amplitude in Born approximation is proportional to the Fourier transform of the potential. In the case that the potential is a central one, we use polar coordinates to perform the angular integrals in Eq.(8.22). It is advantageous to choose the z' axis to be parallel to the vector k - k " for then the ¢' integration is trivial, and
Scattering Theory
136
we obtain
_ ~ foo r,2 dr'V(r') (rr sin 8' d8' eilk -k'lr' cos 0'
n Jo
2m .... 2 Ik - k'in
-....
Since
Ik
- k' I =
k
kk - rr
1
00
Jo
.... r'dr'V(r') sin{lk.... - k' Ir'} .
(8.23)
0
= k· /2 - 2 cos 8 = 2k sin
v
~2'
(8.24)
it follows that the amplitude depends only on the scattering angle 8, and, in this approximation, it is real. We have then
fk(8)-1i~ m
e loo r'dr'V(r') sin {2kr'sin-28 } •
ICsm'2
10
. (8.25)
Consider the potential,
V( r ) -- g e -p.r/Ii , r
where J1, is the mass of the 7T'-meson or pion. This is the form guessed by Yukawa, as a model for the strong interaction between protons and neutrons. The integral Eq.(8.25) can be worked out straightforwardly, giving
The differential cross-section is accordingly
Note that n has dropped out of this expression, and that the sign of 9 is immaterial: it makes no difference to the cross-section whether the interaction is attractive or repulsive. The Coulomb potential is the special case J1, = 0, 9 = e2 • One then obtains dO' e4 dO. - 16E2 sin4 !!. ' 2
which is precisely the classical result that was applied to the a-Au scattering measurements by Rutherford. This led him to the postulation of the existence of a pointlike source of positive charge in the atom, the atomic nucleus.
137
Unitarity and Phase Shifts
8.4
U nitarity and Phase Shifts
When the potential is central, the wave function can be written
(8.26) where Dim are constants. If we choose the x3-axis to be parallel to i, we see that all terms with m -# 0 give zero in Eq.(8.13). Thus fk(f), ¢) A(f) does not depend on ¢, and we can write Eq.(8.12) as
(8.27) effectively setting Dim = 0 for m expanded in a Legendre series,
-# o.
The scattering amplitude can now be
00
= 2::(2£ + l)h(k)Pi(cosf)
fk(f)
(8.28)
,
£=0
where P£(z) is the Legendre polynomial, Eq.(3.56). Here h(k) are called the partial-wave amplitudes, and those corresponding to £ = 0,1,2,3, ... are called respectively the S, P, D, F, ... waves. The incident plane wave can be expanded as follows: 00
'ljJ~(i)
=
eikrcos(J
= 2::(2£ + l)i£j£ (kr)P£ (cos f)
,
(8.29)
£=0
and since r is large, we may use the asymptote Eq.(4.61) to obtain
(8.30) Proof of Eq.(8.29) Let us write 00
eipz
= 2:(2£ + l)A£(p)P£(z) ,
(8.31)
£=0
where we wish to evaluate the partial-wave amplitudes A£ of the plane wave. The tool we need is the orthogonality of the Legendre polynomials:
1/1
"2
-1
8kf dzPk(z)P£(z) = 2£ l'
+
(8.32)
Scattering Theory
138
which is the special case m = 0 of Eq.{3.62). From Eq.{8.31) it then follows that
(8.33) where £ partial integrations have been performed, the integrated terms all being zero. Consider
Bl(p)
=
i:
dZ(Z2 - I)le;pz.
Clearly i
r i-I 1
_i2
dz{z2 _ l)f-l ze i p z
!
1
-1
d (z2 - 1)f. ipz _ ~ B ( ) Z £ lpe - 2£ f P ,
(8.34)
where an integration by parts has been performed. Hence Bf
=
2£~..!!-. B f - 1 = 2f£! [~..!!-.]f Bo = 2f+l £! [~..!!-.]f sin p . pdp
pdp
pdp
p
It follows from Eq.(8.33) that ( _ip)f 2f+l£! Bf(p) .)f [1 d]f --=1,Jf sin p .f . ( p, ) ( -lP -pdp p
(8.35)
where the spherical Bessel function was given in Eq.( 4.52). END OF PROOF
We can also expand the wave function in a Legendre series: (8.36)
139
Unitarity and Phase Shifts
cf., Eq.(8.26) with Dim = (2£ + 1)8mo , and we know that ui(r) satisfies the radial Schrodinger equation (4.3), which can be written
"() + {k2 -
ui r
On combining Eq.(8.27) -
£(.e + 2 1) r
-
2m 2 V ()} r 11,
Ui () r -- 0 .
(8.37)
Eq.(8.36), we see that (8.38)
However, we know that the regular solution of Eq.(8.37) behaves like r i +1 for small r. Suppose we normalize ui(r) so that
for r ---t 0, which is always possible, since the Schrodinger equation is a homogeneous differential equation. Now Eq.(8.37) is a differential equation that can be integrated out from small to large r values. Because V(r) is a real potential, the solution ui{r) must remain real, and for r -+ 00 it has the asymptotic form (8.39) where Ci and 6i are necessarily real, since otherwise ui(r) could not be real. This can be rewritten
so that, on comparing with Eq.{8.38), we find (8.40) where the phase shift, 8i {k), is real and the scattering matrix, Si{k), is unitary. The scattering amplitude can be written (8.41) and so the partial-wave series Eq.{8.28) can be written in the standard form
100. fk(()) =
k 2:(2£ + 1) etOl(k) sin 8i(k)Pi(COS ()) . i=O
(8.42)
140
8.5
Scattering Theory
Phase Shifts and Resonances
If V(r) = 0 for r > a, the solution of the free radial Schrodinger equation there can be written as in Eq.( 4.25),
(8.43) where we stress that the constants Ae and Be depend on.e. For large r, we use the asymptotic formulae Eq.(4.61) and Eq.(4.62) to conclude that
ul(r)
~ Alsin (kr - ~l)
However, from Eq.(8.39) we have Ae
_BlCOS (kr _ ~l) .
(8.44)
= Cecos6e and Be = -Cesin6e, so Be Ae
tan6e = - - .
(8.45)
The continuity equation at r = a has the same form as Eq.(4.30), except that it applies here for any positive E, i.e., any real k. We write it in the form (8.46) where it is understood that the radial wave function on the left, ue(a), is to be obtained by integrating the Schrodinger equation from r = 0 out to r = a. Let us define, in terms of that function,
,eek) = ~ (u~(a) _ ~) k ue(a) a Then, with use of Eq.(8.45), we find
(k)
= j~(ka)
,e
- tan6en~(ka) je(ka) - tan6e ne(ka) ,
(8.47)
Il(k)jl(ka) - j~(ka) Il(k)nl(ka) - n~(ka)
(8.48)
from which it follows that tan6l(k)
=
From the small-k behaviors of the spherical Bessel and Neumann functions, Eq.(4.55) and Eq.(4.57), we find tanol(k)
rv
k 2l+1 ,
and if we stipulate that the phase shift begins at zero, OleO) = 0, this will also be the threshold behavior of the phase shift itself. It can happen that tan6l(k) rises through positive values and passes from plus to minus infinity, corresponding to
141
Neutron-Proton Scattering
a zero in the denominator of Eq.(8.48). If such a zero occurs at k = write, for k ~ ko,
~o,
we may
f/2 tan8e(k) ~ ko - k ' so
Se(k) -1 !£(k)
=
=
2ik
e 2i6t -1 2ik
~
1 f/2 ko ko - k - if/2
In the vicinity of this resonance, the scattering amplitude will be dominated by the contribution from this partial wave, so from Eq.(8.28), we read off
A(())
~
(2£ + l)!£(k)Pe(cos()).
Thus the differential cross-section is approximated by du
2
dO. ~ {( 2£ + 1) IIe (k ) IPe (cos ())} ,
(8.49)
and, from Eq.(8.20), the total cross-section is approximately (8.50) Note that u(k) has its maximum value at k = ko and is one half of its maximum at k = ko ± f /2, whence .r is called the width of the resonance, ko its mass. It is an unstable state, somewhat analogous to a bound state. The expression (8.50) is called the Breit-Wigner resonance formula.
8.6
Neutron-Proton Scattering
In the proton-neutron system, there is a spin-1 bound state, the deuteron, as we saw in Chapter 4. This is a triplet state, since the z-component of the spin has three possible values, namely -1, 0 or 1. However, when one combines two spin! states, there is also a spin-O component, as we saw in Chapter 6. There exists therefore both a triplet and a singlet scattering amplitude, and the effective potential is not necessarily the same in the two states. The singlet potential turns out to be too weak to produce a bound state, but it does give rise to a resonance in the S wave, as we shall see. Let us consider the square-well potential, of width a, as in Section 4.4. For r < a, the regular solution of the Schrodinger equation is
Uo (r)
= A sin kr ,
(8.51)
Scattering Theory
142
where k 2 = mn(E+ VO)/h,2, Vo being the depth of the potential well (recall that the reduced mass of the two-nucleon system is half of the nucleon mass, m n ). For r > a we write
uo(r)
= C sin(kr
+ 80) ,
(8.52)
where k 2 = mnE /h,2, this being equivalent to a superposition of sin kr and cos kr. The matching condition at r = a is logsin(ka + 80 ), dda logsinka = ~ da
(8.53)
+ 80 ) = k
(8.54)
which leads to k tan(ka
tan ka.
-2 2 2 We also have the relation k = k 2 + k5 , where we have defined ko = mn Vo/h, , and we recall from the discussion in Section 4 that the condition for a bound state at zero energy is k5 = 1r 2 /4. Eq.(8.54) can be recast in the form
ktanka - ktanka
£
tanuo =
,
(8.55)
k [tan koa - koa] . ko
(8.56)
k
+ k tanka tanka
and as k ---+ 0, we have k ---+ ko, and so tan 80
rv
-
The triplet S-wave scattering length is defined by Q:ot
. £ / = - hm uo k = a - -1 tan koa .
k~O
ko
(8.57)
From our study of the deuteron bound state in Chapter 4, we have K,
=
v'mnEB h,
-
-
= - k cot ka ~
- ko cot koa ,
(8.58)
so that Q:6 = a + 1/ K,. The triplet scattering length is known to be 5.4 F, and we recall from Eq.(4.37) that 1/K, = 4.4 F, so
a
= Q:ot - -1
= 5.4 - 4.4
= 1 F,
(8.59)
/'i,
in agreement with the value we used for the evaluation of the width of the potential well from the deuteron binding energy. A similar treatment will now be given for the singlet amplitude. We will assume the same width of 1 F; but it will turn out that the depth, ¥Os, will
143
Neutron-Proton Scattering
be less than the corresponding quantity for the triplet amplitude. The singlet S-wave scattering length is Q
s_ 1· s:s/k _ tan ksa - a k ' o - - 1m Uo k-tO
s
where k; = mn Vos Ih 2 • The singlet scattering length has been measured to be -23.7 F; and with a = 1 F we find tanks ks
-- = a -
Qo
= 24.7 .
The Mathematica command line FindRoot[Tan[k]/k
= 1.54, we find
returns {k -> 1.54459}. With ks
vt =
== 24.7, {k, 1.5}]
k;h2 = k 2 (hc)2 = 101 MeV. mn s m n c2
This potential strength is indeed just too small to produce a bound state; recall that the strength required to produce a zero-energy bound state is 105 MeV, and the strength in the triplet state that produces the deuteron (with a binding energy of 2.226 MeV) is 125 MeV. The singlet S-wave phase shift is positive and rises rapidly to ~. The resonance is the reason that the singlet scattering length is so much larger in absolute value than the triplet scattering length. The S-wave scattering length, either for the triplet or the singlet state, yields lim fk(O)
k-tO
= lim 60k(k) = -Qo . k-tO
Thus the threshold value of the total cross-section is
a
= k-tO lim
J
dOlfk(O)1 2
= 47rQ~.
For the triplet and the singlet cases, this leads to
= 366 F2
at
47r(5.4)2
as
47r(28.7)2 = 7058 F2 .
Since the triplet amplitude ~ spin one - has three states (83 = -1,0,1), while the singlet amplitude has only one, the total cross-section, averaged over all four states at zero energy, is
144
Scattering Theory
which agrees well with the measured value. The proton-proton scattering amplitudes are very similar to the neutronproton scattering amplitudes; but there is an important difference in the calculation of the cross-sections from the amplitudes. This is due to the fact that the incident and target protons are indistinguishable, so there can be no difference, in the center-of-mass frame, between scattering through an angle of () and 7r - (). Classically one would simply add the cross-sections for these two angles; but in quantum mechanics we must add the amplitudes. A general property of fermions (i.e., particles with half odd integral spins, like the proton, which has spin ~) is that their wave functions, and scattering amplitudes, are antisymmetric under interchange of two identical fermions. This property is part of what is called the spin-and-statistics theorem, which we will explain more fully in Volume 2. The spin part of the wave function of two protons in a triplet state is even under interchange of the particle labels, indeed from the table of Clebsch-Gordan coefficients for the case ~ ® ~ we have
11,1)
I ~)al
11,0)
~ (I ~)al_ ~)b+ 1_ ~)al ~)b)
11, -1)
~)b
I_~)al_~)b.
Since this spin part of the wave function is even under interchange of the particle labels (a and b), the space part of the wave function, which multiplies the spin part, must be antisymmetric, so that there is overall antisymmetry under interchange of the two protons. This is achieved by writing
Ifk(()) - fk(7r - ())1 2 Ifk(())1 2
+ Ifk(7r - ())1 2 - 2Re[fk*(())fk(7r - ())] .
The spin part of the singlet wave function is on the contrary odd,
so that the spin-and-statistics theorem decrees that the space part must be even. In this case we write
+ fk(7r - ())1 2 Ifk(())1 2 + Ifk(7r - ())1 2 + 2Re[fk* (())fk (7r - ())] . Ifk(())
145
Exercises
8.7
Exercises
Problem 1 The Legendre polynomial is defined by
e old 2 e ( ) Pe(Z) - Pe (Z) = 2et'! dz (Z - 1) . (1) Show that Pe+l(Z) = zPe(z) + ze2;11 Pl(z). (2) Demonstrate (t' + 1)Pe+1 (z) - (2t' + l)zPe(z) + t'Pe- 1 (z) = O. (3) Work out Pe(z) for t' = 0,1,2,3, and sketch them graphically.
Problem 2 Calculate the phase shifts, oe(k), for the hard core potential, V(r) = and V(r) = 0 otherwise.
00
if r < a
(1) Simplify the result for the S-wave. (2) Calculate the total cross-section in the limit k--+ O. (3) Calculate the total cross-section in the limit k --+ 00.
Problem 3 Prove the optical theorem for elastic scattering, that is aT
!
da dO. dO.
47r
=
T1mfk(0) .
l ..
Problem 4 .t1.f (~" ~t) When there is inelasticity, the partial wave scattering function, Se(k), can be written 1}e.9R{20e), where 0 < 1}e(k) < 1. Show that the optical theorem is still true, where now aT is the sum of the elastic and inelastic cross-sections. Problem 5 Calculate the scattering amplitude in Born approximation with the following spherically symmetric potentials, V(r):
(1) -g()(a - r).
(2) - ~:: exp ( - ~ ). Problem 6 Calculate the differential scattering cross-section in Born approximation for the scattering of two identical particles that move under a Yukawa interaction:
146
Scattering Theory
Problem 7 Consider two different particles of spin
~.
The interaction potential is
V(r) = W(r)81 ·82, where 8j = !1iiJj and W(r) is a continuous, scalar function. Calculate in Born approximation the ratio of the probability that, after scattering'one finds ~ 1 0'( ~l z _.L~ 1 SIz = 2h, ... s'joz\ given that, before scattering, Slz = 2h, = -S2z· f"'cc~~ I Problem 8 Let cSt(k) and 8t(k) be the f-wave phase shifts corresponding to the central potentials V(r) and V(r), respectively. Prove that sin(cS- t
-
cSt) =
- -2m 2-
h, k
1
00
dr[V(r) - V{r)]ut(r)ut{r) ,
0
where ut(r) and ut{r) are the radial wave functions. If V(r) = V(r) + AW(r), where W{r) is suitably well-behaved, sketch a perturbation method to be used for small A. Problem 9 By considering the differential operators £±
d
= tanh r ± dr '
solve the S-wave Schrodinger equation for the spherically symmetric potential V (r)
h,2
= - - cosh - 2 r . m
Calculate the S-wave phase shift. Are there any resonances and/or bound states? Problem 10 A nuclear physicist measures a scattering process, and finds on analyzing his data that the following phase shifts (in radians) describe his data well: cSO = 0.88779, cSl = -0.56640, c)2 = 0.23447, and cSt = 0 for f > 3. (1) Draw a graph of the differential cross-section against cos (). (2) A research student of the nuclear physicist analyses the same experimental results but she arrives at the following phase shifts: cSo = 0.10930, cS l .. -0.76993, cS2 = 0.23447, again with all the higher phase shifts vanishing. Draw a graph of the differential cross-section against cos (). Has the research student, or the nuclear physicist made a mistake, or are they both correct? Explain. (3) If there is an infinite number of non-zero phase shifts, are they uniquely determined by the differential cross-section?
Chapter 9
Atomic Physics
9.1
Two-Electron System
Consider a negative hydrogen ion, H-, a helium atom, He, or a positive lithium ion, Li+. In all cases, we have to do with two electrons, at positions, say, i 1 and i 2, and a nucleus of charge Ze, where Z = 1,2 or 3. The Hamiltonian operator in configuration space is
where
In the first instance, we concentrate our efforts on the calculation of the ground state .of this system; and the strategy will involve three steps: • Neglect of the electron-electron repulsion, V12 • • Treatment of V12 by first-order perturbation theory. • The variational method.
Neglect of repulsion term We choose a separable wave function, 'ifJ(i b i 2)'
'ifJ(i 1, i 2)
= 'ifJnlilml (i d'lfJn i m (i 2), 2
147
2
2
Atomic Physics
148
for which the energy is
Here En is the nth energy level of the hydrogen atom, except that e2 Z replaces e2 , i.e., eVZ replaces e. Thus
where a = ~: ::::::: 1~7 is the fine-structure constant. Hence the (nb n2) energy level in the present approximation is
122(1-+1) mc.2 n n
E=--Za 2 The ground-state energy (nl = n2
2 1
2 2
= 1) is therefore given by
.!£ = -Z 2 a? mc 2
(9.1)
The ground state of the hydrogen atom is -~a2mc2 = -13.6 eV; it follows that the above energy is 2Z 2 times that, i.e., -27.2Z2 eV. For the helium atom, Z = 2, so E = -108.8 eV, to be compared with the experimental value of -78.975 eV. We conclude that the electron-electron interaction term, which has so far been neglected, must have quite a large effect - at any rate if quantum mechanics is to be successful in describing the helium atom. First-order perturbation Treating V12 as a perturbation, we have in first order
Here 'l/JIOO is the normalized ground-state wave function of hydrogen, after the replacement of e by eVZ, so
'l/J100(r) =
~ ~ exp [-z~l ' V7r (~) ao ao
where 1i2 me2
ao = - - .
(9.2)
149
Two-Electron System
After the scale change ....
2Z ....
i
Pi = -ri, ao we find 6E =
e 2Z
327r2ao
II .
= 1,2,
d3pl d 3p2
IPl-P2I e
-(PI+P2)
.
In performing the P2-integration, at fixed P1, we use polars and choose the z-axis in the direction of Pl ' Hence Ip 1 - P21 = Pt + p~ - 2P1P2 cos 0 and the integrand is independent of the azimuthal angle, cp, which can accordingly be integrated out to give 27r. Moreover, since
J
d ./ 2 2 d oV Pl + P2 - 2P1P2 cos () =
PlP2 sin 0 2 0' P1P2 cos
J P12 + P22 -
it follows that
(7r
sinOdO
J0 J Pt + p~ -
_1_
PlP2
2P1 P2 cos 0
P1
[J
pi
+ p~ -
2PlP2 cos
0] 7r 0
+ P2 -IPl -
P21
P1P2 2 max(p1' P2) . Hence AE -_ e2 Z / d3 Ple - PI 87r ao
o
1°O 0
p~dp2 maX(Pl, P2)
e -P2 .
The angular integrals in the Pl-integration can now be done trivially, giving a factor 47r, and the remaining integrals are elementary: 6E
1 1
00
Z e 2 -2
ao
P12dP1 e -PI
0
{l 0
00 pidp1 e- P1 e 2 -Z ao 0 5 e4 m - Z 28 1i '
1.
00
PI
P22 -dP2 e P1
-P2 +
1.
00
PI
P22 -dP2 e P2
-P2}
P2dp2 e- P2
PI
In proceeding from the first to the second lines above, we reversed the order of the P1 and P2 integrations, in the first piece of the P2 integration only, and then we relabeled the integration variables (Pl f-t P2)' It is convenient to write 6E _ ~Z 2. me2 - 8 a,
(9.3)
Atomic Physics
150
and if we add this to Eq.(9.1) we find
E
+ .6.E =
TinC 2
-z (z - ~) a?
(9.4)
8
For He, Z (Z - ~) = 11 and so this energy is 121 times the H ground-state energy, i.e., -74.8 eV. This is much closer to the experimental value of -78.975 eV, being out by about 5%. To get better agreement, one could in principle go to the second order perturbation theory; but this involves a lot of numerical work. Variational method We recall from Eq.(7.20) that, given a Hamiltonian H and any vector l'if),
where E1 is the smallest eigenvalue of H, the ground state. We shall now apply the method to the helium atom. On comparing Eq.(9.1) and Eq.(9.4) we see that the effect of the electron-electron repulsion is to reduce the effective value of Z. This effect is called screening, and it is physically understandable at a classical level: each electron does not 'see' the bare nuclear charge all the time, since the other electron is sometimes between it and the nucleus, thus reducing its attraction. This classical picture must not be taken too literally, since really we have a quantum system of two indistinguishable particles; but it suggests that we try for I'l/!) the ground-state vector of a twoelectron atom, in which the electron-electron interaction has been omitted, but where the atomic number Z is replaced by a parameter Z, less than Z, to account for the screening. Z will be chosen at the end to minimize ('l/!IHI'l/!) and hence to optimize the estimate of the ground-state energy.
~et ~lOo(r) be the ground-state wave function Eq.(9.2), but with Z replaced by Z. Then 'l/!100 satisfies
- 2] 'l/!100(r) nt.2 [-2Tin - v 2 -Ze= r
1
-2
--Z a 2
2
2 TinC
'l/!100(r)
so that [-
;,,2
2
ze 2 ] -
2Tin V1 -~ 'l/!lOo(rd
e 2 - -Z 1 - 2Q 2TinC2][ (Z- - Z)'l/!lOo(rd
r1
2
and similarly for H2~lOo(r2)' Hence, if we choose
151
Two-Electron System
then
The only new term to be calculated is the first one, and this is elementary:
The last term has the same form as the first-order perturbation: . we can take over the result Eq.(9.3), of course changing Z there to Z. On gathering these results together, we find
(¢IHI¢) = Z [Z - 2Z + ~] a 2mc2 = {(Z -
z+
156 )2 -
(Z -
156
)2}a 2mc 2 ,
and this must be greater than the ground-state energy for all Z. The minimum of the right side evidently occurs when Z = Z - 156' and we obtain El ( 5)2 2 -< a. - - Z - 16 mc2
(9.5)
For Z = 2 we find El < - 77.5 eV, which is better than first-order perturbation theory, being about 2% too high. The variational result can be improved by fine-tuning the trial function. The following parametrization was used by Hylleraas in 1929, when quantum mechanics was still young:
1f(i' 1, i' ,)
= exp [_iT1 :
,,]
.L ajkmh + T2)i(T1 - T2)klf
1 -
i' ,1 m
.
},k,m
The parameters are the coefficients ajkm and Z. With just one term, aooo, the calculation is the one we have just completed. With three terms, Hylleraas obtained -78.98 eV. Notice that the result Eq.(9.5) has the same form as Eq.(9.1), but with Z replaced by Z - 156' This indicates that the effective charge 'seen' by each electron is reduced by 5e/16, as a result of the presence of the other
Atomic Physics
152
electron. For He the effective nuclear charge is thus 27e/16. Even better results are possible, for example with 13 terms in the Hylleraas series (see table below).
Method Oth perturbation 1st perturbation 1 term variation 3 term variation 13 term variation
Formula (Z = 2) -Z 2a 2mc2
-Z(Z - ~)a2mc2 -(Z - 156 )2a2mc2
Experiment
eV. -108.8 -74.8 -77.5 -78.98 -79.015 -78.975
Ground-state energy of the He atom For the hydrogen anion, H-, the effective charge is 11e/16. This gives a ground-state energy for H- of - (~i)2 a 2mc2 = -0.47a 2mc 2 ~ -12.86eV, whereas the ground-state energy of a hydrogen atom, -~a2mc2 ~ -13.60eV, is slightly lower. If this were the last word, it would mean that H- should be unstable into decay into a neutral H-atom and a free electron. However a treatment a la Hylleraas gives -14.34 eV, in good agreement with experiment. In the following table, we summarize the results of a 13-term variation calculation, with the experimental results, for the cases Z = 1,2,3,4. The agreement is truly impressive, the differences being always less than one per cent, attributable in all cases to relativistic effects that have been neglected in our treatment.
HHe Li+ Be++
Variation -14.34 -79.02 -198.1 -371.6
Experiment -14.35 -78.98 -197.1 -370.0
Ground-state energy in e V of H-, He, Li+, Be++
On the next page, output from a FORTRAN program (Schmid, 1987) for His displayed.
153
Two-Electron System
THE GROUND STATE OF THE HYDROGEN ANION BY THE HYLLERAAS METHOD Nuclear charge Z = 1 Energy of H-atom in ground state:
E = -13.60 eV
Energy of H-anion E = -14.294950 eV Expectation values: Kinetic energy Electron-nucleus interaction <W> Electron-electron interaction Hamilton operator
with 3 terms and
z- = 0.769
= 14.303991 eV = -37.522710 eV 8.923769 eV = = -14.294950 eV
Basis states used and corresponding components of the eigenvector:
State no. State no. State no.
1 2 3
J
K
M
0 0 0
0 0 2
0 1 0
Components of eigenvector .408970 .246321 .182189
*************************************************************************** Energy of H-anion E = -14.338050 eV Expectation values: = Kinetic energy = Electron-nucleus interaction <W> = Electron-electron interaction = Hamilton operator
with 13 terms and Z- = 0.680 14.338264 -37.384232 8.707918 -14.338050
eV eV eV eV
Basis states used and corresponding components of the eigenvector:
State no. State no. State no.
1 2 3
J
K
M
0 0 0
0 0 0
0 1 2
Components of eigenvector .499839 .312642 -.029391
Atomic Physics
154
State State State State State State State State State State
9.2
no. no. no. no. no. no. no. no. no. no.
4 5 6 7 8 9 10 11 12 13
0 1 1 1 2 2 3 0 0 1
0 0 0 0 0 0 0 2 2 2
3 0 1 2 0 1 0 0 1 0
-.000122 -.203656 -.038433 .005354 .040781 -.000042 -.002835 .104698 .013612 -.007983
Exchange Term
So far we have only considered the ground state of He, which may be symbolized (ls)2, indicating that each electron is in an n = 1, f = 0 state. We have completely neglected spin; but we will now consider this complication. Let the two electrons be designated A and B. Electron A can have spin 'up' or 'down' with respect to an arbitrary z-axis. Introduce the eigenvectors x~, such that
stxt = ±~nxt and similarly for electron B. The total spin of the two-electron system S is 1 or O. The eigenvector of the total spin operator belonging to S = 1 is
while the eigenvector belonging to S = 0 is Xsinglet =
(X~x~ - X~x!)/V2.
Suppose that electron A is in the orbital state given by (nA,fA,m A ), so that the spatial part of the wave function is
..'f'I,A(-) r1
)pmA( ()) im A 4>l = CnAlAmAPlA e -PLUA+1(2 n A +1 P lA cos 1 e ,
where P = nZ;~o. Similar considerations apply to 'ljJB (r2), describing electron B. If the electrons were distinguishable - for example if one were replaced by a muon - one would write 'ljJA (r1 )'ljJB (r2) and that would be the end of the story. However, the two electrons are indistinguishable; and it does not make sense to say that the one at r1 is specified by the quantum numbers (n A , fA, m A ) and the other at r2 is specified by (n B , fB , m B). The space part of the two-electron
155
Exchange Term
wave function can contain 'ljJA (rl )'ljJB (r2), but also 'ljJA (r2 )'ljJB (rl), in which the two electrons are exchanged; and a priori we could have any linear combination of these two products. There is a fundamental law according to which the wave function describing two or more identical bosons is symmetric under exchange of any two particles, while the wave function for identical fermions is antisymmetric under such an interchange. A boson is a particle of integral spin (in units of h), e.g., a pion (8 = 0), a photon (8 = 1) or a graviton (8 = 2). A fermion is a particle whose spin is an integer plus one half, e.g., an electron (8 = ~) or an n- particle (8 = ~). The law can in fact be proved within the framework of relativistic quantum field theory, where it is called the spin-and-statistics theorem (cf., the discussion of proton-proton scattering in the previous chapter). Since the triplet spin vector Xtriplet is even, and the singlet vector Xsinglet is odd under exchange of A and B, it follows from the spin-and-statistics theorem that Xtriplet must be combined with an odd, and Xsinglet with an even spatial wave function, in order that the total wave function be odd under A f-t B exchange. In other words, the two possible total wave functions for the twoelectron system, including spin, are
Wtriplet =
~ {'ljJA(rl )'ljJB (r2) -
'ljJA(r2)'ljJB (rl) } Xtriplet .
(9.7)
In the case that 'ljJA (r) = 'ljJB (r), i.e., the two spatial wave functions are the same, the triplet combination vanishes: this fact is sometimes expressed by saying that no two electrons can have the same quantum numbers, n, f, m, s, where s = ± ~ is the value of the spin, in units of h, along the quantization axis. This is called the Pauli exclusion principle. More accurately - since after all Xtriplet also contains a state in which 8 3 = 0 - we should say that the spin triplet combination is excluded if the spatial wave functions are identical. In our analysis of the ground state of the He atom, both wave functions are 'ljJ100, so we have to do with the singlet state W = 'ljJlOo(rI)'ljJlOo(r2)Xsinglet, which is Eq.(9.6) with A = B, except for the normalization. The spectroscopic notation for this state is l80 , where the superscript is the number of spin-states, namely 28 + 1, the letter gives the total orbital angular momentum (8 for L = 0, P for L = 1, D for L = 2, ... ), and the subscript is the total angular momentum, J, which is a strict constant of motion. The principal quantum numbers are not
Atomic Physics
156
specified in this notation. The notational alternative, (ls)2, indicates that both electrons are in n = 1, f = 0 states. Let us now consider the case that the two electrons are in different spatial states. Treating V12 in first-order perturbation theory, we find the first-order energy shift to be
The perturbation does not involve the spin variables, so these drop out, since X+ X = 1 for either the singlet or the triplet cases. Accordingly 6.E
=
~
JJ~;~~I
{7jJA(f1}t/JB (f2)
V±£
{t/lA(Tl)t/lB(i'i) ±.pA(i'i)t/lB(Tl)}*
± 7jJA(f2)7jJB (f1)} (9.8)
in which the direct term is
and the exchange term is
where the plus sign applies to the singlet case and the minus sign to the triplet case. The direct term V can be written
where
which exhibits its semiclassical form. The exchange term £ does not have such a classical analogue: it is a pure quantum effect arising from the coherence of superpositions of states. Both V and £ are positive - the former manifestly so and the latter in cases of interest. As a consequence, the electron-electron repulsion raises the (negative) energy-level that one would calculate without V12 - an effect that we already found for the (ls)2 ground state. The increase is
157
H artree and H artree-Fock Methods
greater for the singlet, 1) + E, than for the triplet, 1) - E. Although the electronelectron interaction does not depend explicitly on the spin of the electrons, the fact that the energy levels are different for the singlet and the triplet spin combinations means that there is indeed an implicit dependence. We can make this effect explicit by considering ---
---
.... A
S .S = S
---A
.S
---A
+ 25
___ B
.S
___ B
+8
2 {S(S + 1) - ~ 2S(8
___ B
.S
0 + 1) - ~ 0 + I)}
+ 1) - 3 = {
1
-3
for for
8=1
5=0
Hence we can write the first-order perturbation Eq.(9.8) .6.E =
9.3
~ \ 1 + if A.if B) E .
1) -
Hartree and Hartree-Fock Methods
The Schrodinger equation for an atom with Z electrons is
which is impossible to solve analytically, and very difficult to solve numerically, even when Z is quite small. Nevertheless, this 3Z-dimensional partial differential equation, supplemented by the spin degree of freedom, relativistic corrections such as the spin-orbit coupling, and the antisymmetrization condition that is required by the spin-and-statistics theorem applied to fermions, contains the essential physics of atoms and hence also the essence of chemistry. The Hartree approximation consists in the approximation of separability:
1/J (r 1, r 2, •.. , r z)
=
1/J1(r 1) 1/J2 (r 2) ... 1/J z (r z) .
Application of the variational method leads to the following system of equations:
(-:~ \]~ -~:2
+ Vi (Ti) )
.p,(Ti) = .,.p,(T.),
(9.9)
where
(9.10)
158
Atomic Physics
and (9.11) i
This equation has a clear intuitive meaning: the ith electron satisfies a oneparticle Schrodinger equation with a potential energy that is supplied by the nucleus (attractive) and by all the other electrons (repulsive). The effective charge-density of the jth electron, say Pjerj), is evidently el'l/Jj(rj) 12 , which may be interpreted as the average density of electric charge at the point j, due to the presence of the jth electron. Despite the intuitively appealing nature of the Hartree equation (9.9), it must be stressed that it is an approximation. With the additional (Pauli) requirement that no two electrons may be in the same quantum state (including the spin degree of freedom), this equation yields results for energy levels that generally agree with experiment at the level of 10% to 20% only. Solution is effected by the method of successive approximations: first the electron-electron repulsion terms are ignored, so the zeroth-order solution is a product of hydrogen-like wave functions, as in Section 9.1 for the case Z = 2. The repulsive potential terms Vi (r i) can be calculated in first order, and these are then included in the Schrodinger equation, which is solved to give improved values of the 'l/Ji(ri). The procedure is iterated to convergence, yielding the so-called self-consistent solution of the Z-electron Schrodinger equation in the Hartree approximation. This method is clearly better than the perturbation theory that we explained, since the electron repulsion term is not simply treated as a first-order perturbation. In practice a further central field approximation is often made by averaging over directions:
r
This approximation is less important now that we have high-speed computers with large memories than it was in the pioneering days of quantum atomic physics. The Hartree equation (9.9) can be written (-
:~ '\7 2 J;2) ,p,(r, s)
+e 2
L L Id r' 'l/J; (r', s') Ir. . ~ . . '1 'l/Jj (r', s')'l/Ji(r, s)
(9.12)
3
"-I-"
3.,-'t
S
I
r
=
€(IPi(r, s) ,
where the spin quantum number has been included explicitly. According to the spin-and-statistics theorem, interchanging two identical fermions results in
159
Periodic Table
a change in sign of the total wave function. This exchange effect, including the sign-change, is effected by subtracting from the above a term in which 'l/Ji (i, s) and 'l/Jj(if, Sf) are replaced by 'l/Ji(i', Sf) and 'l/Jj(i, s) respectively:
+ e2 ~ -
e2
~L J=I=1.
f
rl='t
L f d3rf'l/J; (if, Sf) Ii ~ ifl 'l/Jj (if, Sf)'l/Ji (i, s) 8'
d3r f'l/Jj(i f , Sf) Ii
~ if I'l/Jj(i, s)'l/Ji(i f , Sf) =
fi'l/Ji(i, s).
8'
This is called the Hartree-Fock equation; it is much more laborious to solve than the Hartree equation, since it is truly an integro-differential equation. The last term gives rise to exchange contributions to the energy levels, analogous to those discussed in the last section for the helium atom. The Hartree-Fock equations give results that are significantly closer to the experimental measurements than are those of the Hartree equation (9.9).
9.4
Periodic Table
We shall now discuss very briefly the periodic structure of the elements, based on the separable Hartree approximation, according to which there are Zone-particle states, labeled by Z different quantum numbers ni, fi' mi and Si, i = 1,2, ... , Z. It is understood that these one-particle states are modified by the electronelectron repulsion terms, and, if we include the Fock improvement of the Hartree method, also by the exchange effects. Nevertheless, the above quantum numbers can still be used as labels. Let n be the largest of the ni for a given atom in its ground state. For a given value of n, the angular momentum quantum number f can take on the values 0,1,2, ... , n - 1. For a given nand f, there are 2f + 1 allowed values of the azimuthal quantum number m, namely -f, -f + 1, ... , f, and for each of these values there are two possible spin values.
f 2(2f + 1)
s
P
0 2
1 6
D 2 10
F 3 14
Orbital Angular Momentum The seven periods in the periodic table of the elements each begin with the S states, containing 2 elements, and then the P states, containing an additional 6 elements, except of course for the first period, for if n = 1 then f = 0 necessarily. As one adds successively more and more positively charged protons to the
160
Atomic Physics
nucleus, and correspondingly more and more electrons outside it, so more and more energy levels are filled up. Period 1 2 3 4
5 6 7
# 2 8 8 18 18 32
S P D 2 2+6 2+6 2 + 6 + 10 2 + 6 + 10 2 + 6 + 10 Incomplete
F
n 1 2 3 4
5 6 7
Shell K L M N 0 P Q
Periodic Table The D states corresponding to n > 3 do not belong to the chemists' period n, but rather to period n + 1. The reason for this displacement is that, for R = 2, the electrons in question have a greater probability to be situated closer to the lower-lying electrons than are the R = 0 electrons - with the same n and they experience more inter-electron repulsion and so are less tightly bound. The result is that the n = 3, R = 2 electrons are more lightly bound than the n = 4, R = 0 and R = 1 electrons: this gives rise to the first transition series of elements, containing useful elements like iron, chromium and copper. A similar displacement occurs for the second transition series, containing silver, and for the third, containing gold. The n = 4, R = 3 electrons are even better shielded than are the R = 2 electrons, and the corresponding elements are displaced by two periods, giving rise to the 14 rare earths in the sixth period. The above language of repulsion and displacement is only a picturesque way of gaining some intuition about what goes on as one successively increases the nuclear charge and adds electrons: it is not strictly true that one particular electron is in a particular shell-rather the wave function for the electrons is an antisymmetric combination corresponding to all the (n, R, m, s) states that are occupied. Nevertheless, it is meaningful to say, for example, that as one increases the nuclear charge of a Co nucleus by one unit and adds an electron to keep the resulting Ni atom neutral, one more 3d state is occupied, and not a 4p state, since this would cost more energy. The latter option exists, but it is an excited state of the nickel atom. The seventh period, containing in particular radium, uranium and plutonium, is incomplete because of nuclear instability: a-decay occurs in some cases, for example Ra-'----tRn+He, which is due to a quantum-mechanical tunneling effect, ;3-decay in others, and more radical fission (in the presence of thermal neutrons) in yet others, for example n+Pu~fragments.
Periodic Table
161
Two empirical observations, the Hund Rules, are helpful in deciding how the shells are successively filled: a: Other things being equal, the state of highest spin has the lowest energy. b: If an incomplete shell is more than half-filled, the lowest energy corresponds to J = IL + 81, otherwise to J = IL - 81. The two most important corrections to the Hartree-Fock calculations are:
(1) Deviations from the Hartree separability Ansatz and the central-field approximation. We shall assign all these electrostatic corrections to the potential VES ' (2) Spin-orbit coupling, a relativistic (Dirac equation) effect. The corre:.. sponding correction will be designated VLs.
When the first term is the more important, and the spin-orbit coupling can be ignored, or treated as an extremely small additional effect, we speak of the LS or Russell-Saunders coupling scheme. Because the spin-orbit coupling is neglected, not only J, but also the total orbital angular momentum, L, and the total spin, 8, are constants of motion. A Russell-Saunders state corresponding to L = 2, J = ~, and 8 = ~, for example, would be written 4D!., where the superscript 4 2 is the multiplicity, 28 + 1. When the spin-orbit term is more important than the electrostatic correction, each electron can be characterized by the quantum numbers ni, li, ji, mji' with a spin-orbit coupling of the form
(9.13)
This is called the jj coupling scheme, and is important for very heavy atoms, where the large value of the central potential makes Eq.(9.13) the more important contribution. In practice, often both VLS and YES are included in a serious calculation: for all but very heavy atoms the initial classification is on the basis of the LS scheme, with spin-orbit splitting being treated as a small additional perturbation, whereas from about Pb onwards in the periodic table, the jj classification is useful, in which the orbital and spin angular momenta for each electron are combined into a total angular momentum.
162
Atomic Physics
LS:
VES»
l = ~ili ....
J
VLS
§ = Li Si ....
VLS»
JJ:
....
....
....
VES ....
J i =Li+ 8 i
J = LiJi
=L +8
LS and
ii Coupling
Finally, in order to impose the correct antisymmetrization of the Z-electron wave function, one works not with a simple product of one-electron functions, but rather with their Slater determinant:
'l/J1 (r 1,81) 'l/J1 (r 2,82)
'l/J2 (r 1, 81) 'l/J2(r 2,82)
'l/Jz(r 1,81) 'l/Jz(r 2,82)
(9.14)
which guarantees the Pauli exclusion principle, since if any two columns are identical, the determinant is zero.
9.5
Hydrogenic Molecular Ion
The simplest 'molecule' is H2 +, i.e., a hydrogen molecule in which one of the two electrons has been removed. We will treat this system first, before considering the neutral hydrogen molecule itself. Let the two protons, of mass m p , be at positions R 1 and R2, and the electron, of mass m, be at position r. The Schrodinger equation in configuration space is
(9.15) where
PI =
-itiVRl
163
Hydrogenic Molecular Ion
We shall not try to solve this 9-dimensional partial differential equation exactly, rather we shall make some approximations, and finally use the variational method to estimate the binding energy of this molecular ion. Firstly, instead of transforming to the centre-of-mass system of the three particles, we will transform to that of the two protons only. The mass of the twoproton system is nearly 4000 times that of the electron, and relativistic effects, which are being neglected, are relatively more important in any case. Such additional small effects can be included afterwards as first-order perturbations. Let us in the first instance remove the electron completely. The two-proton system is described by
[Pr
2mp
Pi· + 2mp + V12
-
1 - -
E 'if;(R 1 ,R 2 )
= o.
(9.16)
We transform from the canonical pairs {R 1, j5 I} and {R 2, j5 2} to the centreof-mass coordinates, as in the discussion of the deuteron in Chapter 4:
and the difference coordinates
In terms of these variables, the Schrodinger equation (9.16) becomes p2 [ 2~~m
p2
+ 2M + V12 -
1
I
-
-
E 'if; (Rcm,R) = 0,
(9.17)
which is called the Born-Oppenheimer approximation. Here
the total mass of the two protons, while 1\([ --
!.m 2 P
is the reduced mass. Here we use M for this reduced mass of the protons, since we reserve m for the electron mass. As expected, the centre-of-mass motion can be factored out, and it is in fact convenient to work with coordinates such that Rcm is identically zero at all times, i.e., the origin is chosen to be the point
164
Atomic Physics
midway between the two protons. The above system, with the centre-of-mass motion factored out,
p2 [ 2M
l ..
+ V12 -
E 'lj;(R)
= 0,
(9.18)
describes the scattering of two protons. For example, a beam of protons from a synchrotron might be incident on a bubble chamber filled with liquid hydrogen. There are of course no bound states, since the potential, e2
V12
=
R'
is purely repulsive. The situation changes dramatically when we add an electron. We have now (9.19) Since
Rem = 0, we have
By factoring out the centre-of-mass coordinates, we have reduced the dimensionality of the system to 6. We make now the reasonable assumption that, if the ion is in a configuration in which the two protons are close to equilibrium - in which their mutual repulsion is balanced by the attraction to the electron - 'lj; will be a much more slowly varying function of Ii than it is of Hence we neglect the proton kinetic term, :;'lj;(R, r) = - 2~ V'~'lj;(R, r), compared with the electron kinetic term. The Schrodinger equation (9.19) becomes
r.
Ii 2 2 [ - 2m V'r -
e2
e2
IT _ i R I - IT + i R I +
] ........ e2 R - E 'lj; (R , r )
= 0.
(9.20)
Here R can be regarded as a parameter: indeed we could have written Eq.(9.20) down immediately on intuitive grounds, since it describes the situation in which the two protons have no dynamics, but are nailed down at a mutual distance R from one another. The trick now is to use the variational method to estimate the ground-state of the ion. If the electron is very close to proton #1, with proton #2 relatively
165
Hydrogenic Molecular Ion
. far away, a reasonable guess for the eigenfunction would be the ground state wave function of the hydrogen atom, in which proton #2 is a distant spectator. This normalized function is
(9.21)
:;;2.
For cleanliness we have set the unit of length to be the Bohr radius, ao = If on the other hand the electron is close to proton #2, the guess would be
(9.22) with proton #1 as the spectator. However, this way of talking is all wrong: we must think quantum mechanically! The two protons are identical, and it makes no sense to talk about being close to proton #1 and not to proton #2. Rather, we must consider symmetric and antisymmetric superpositions of the above states:
Here C± are normalization constants; they are given by
The overlap integral is
I 11
~71"
d3 r exp[ -
IT - 12 R I - IT + 12 R I]
71"
d3 r exp [-
IT -
R- I - r]
roo drr 2 e-
/1
dcos (j exp[-Vr2
2
Jo
T
(9.23)
+ R2 -
2r Rcos (j]
.
-1
In going to the second line, we shifted the integration variable from if to if + ~ R . The cos (j integral can be performed by substituting u = J r2 + R2 - 2r R cos (j, so d cos (j = - ~~. The integral (9.23) becomes
21
R
00
o
drr e- T
l
R
T
+ duu e- U
•
IR-rl
These integrals are elementary and we find
(9.24)
166
Atomic Physics
Let us now consider the expectation values of the Hamiltonian in the states
'l/J± : ('l/Jl ± 'l/J2!H!'l/Jl ± 'l/J2) 2 [1 ± ('l/Jl!'l/J2)] ('l/JIIH!'l/Jl) ± ('l/J2!H!'l/Jl) 1 ± ('l/Jl! 'l/J2)
(9.25)
where the symmetry between 'l/Jl and 'l/J2 has been used. We have already evaluated the overlap integral in the denominator. Furthermore, since
!
--+ --+ [ n? 2 d3 r'l/Jl(R,r) --2 'V'r -
R
2 1
-:2 0
2!
2
2
me
-
e2
+R
d3 r
e2 - -;
!
1
e2
--+
if + ~R! e-
3
d r
--+
!r - :2R!
m
1 2 - -0 me2 + -e
e2
--+
-
--+
!r
2
e2
1 ....
+ :2R !
--+--+ + eR ] 1Pl(R,r)
2 [ 'l/Jl( R--+ r--+)] '
2r
(9.26)
IT + R! .
In proceeding to the last line, we shifted the integration variable from T to .... r--+ - 2"1 R • After switching the sign of cos fJ, we handle the integral with the same substitution as before, finding
~ 7r
!
d3r
e-2~ if + R!
=
~ roo rdr e-2r rR+r da, R
Jo
J1R-rl
the integrals now being elementary. We obtain
('l/Jl!H!'l/Jl) = - ~02me2
e2
+ R (1 + R) e- 2R .
(9.27)
The other matrix element in Eq.(9.25) is
I!
-
7r
!J
d3 r
e - r --+ exp [-
if+R!
if + R ! ]
roo rdre- r rR+r dae- u o J1R-rl
,
(9.28)
167
Hydrogenic Molecular Ion
where
(1
is as before. We obtain
(1/12IHI1/1,) =
[-!,,2 mc2+ ~] (1/1,11/12) - .2(1 + R) e-
R .
(9.29)
!:
Since we have set the Bohr radius to unity, we have e 2 = = o:2mc2 . Inserting the results Eq.(9.24), Eq.(9.27) and Eq.(9.29) into Eq.(9.25), we find finally (9.30) where
+ R) e- R 1± (1 + R + ~R2) e- R
1 - ~R2 ± (1
2
f±(R) = =f R e-
R
.
(9.31)
By the variation principle, ('lfJ±IHI'lfJ±) is, for all values of R, an upper bound on the ground state of the ion. Accordingly, we vary R to obtain the smallest possible value for the right side of Eq.(9.30). The following table gives values of f.... (R) as a function of R. obtained from the Mathematica function
f[R_, p_]:
=
-2
* p * Exp[-R] * (1 - (2/3)R * R + p * (1 + R) * Exp[-R])/ (R * (1 + p * (1 + R + R * R/3) * Exp[-R])); R
f+(R)
f_(R)
0.10 1.00 1.50 2.00 2.20 2.40 2.45 2.50 2.55 2.60 3.00 4.00
-18.009060 -.423267 -.009972 .107543 .122985 .129083 .129543 .129659 .129465 .128992 .118165 .073732
-21.634880 -2.090802 -1.145122 -.678293 -.557160 -.460042 -.438838 -.418716 -.399612 -.381467 -.264881 -.110226
Here p = ±1 corresponds to ± in Eq.(9.31). The command line
FindMinimum[-f[R, 1], {R, 2}] returns {-0.129662, {R- > 2.49283}}. The first number is the minimum of - f+(R), i.e., minus the maximum of f+(R). From the table we see that f+(R)
168
Atomic Physics
has a maximum of about 0.13 at R ~ 2.5. This maximum of f+(R) corresponds to a minimum of the energy, which indicates that this R value is an estimate of the equilibrium separation of the protons in the hydrogen molecular ion. Since our length unit is the Bohr radius, the calculated equilibrium separation is 1.3 A, compared with the experimental value of 1.06 A. The calculated binding energy is about 0.13 times the ground-state energy of the hydrogen atom (13.6 eV), i.e., 1.77 eV, as compared with 2.8 eV from experiment. As can be seen from the above table, the antisymmetric combination of wave functions does not yield a finite maximum, and this can be understood readily. In this case the wave function at the mid-point between the protons (our origin of coordinates) is zero, and the probability that the electron is between the two protons is much smaller than it is in the symmetric case. Accordingly, the mutual repulsion of the protons is insufficiently compensated by the attraction of the electron, and the minimum energy (the maximum of f_(R)) corresponds to R -+ 00. Notice that the stability of the hydrogenic molecular ion, that we have just explained, is a quintessentially quantum phenomenon. If we had only used 1/11, obtaining then Eq.(9.27) as our upper bound on the bound-state energy, we would have found, as in the antisymmetric case, a minimum value in the limit of infinite R, corresponding to the dissociation of the system into a neutral hydrogen atom and a free proton at infinity. This is what one might intuitively expect, and it is of course a possibility; however, by bringing the second proton up to about an Angstrom of the first, energy is gained. 2 If we could ignore the proton-proton repulsion term ~ in Eq.(9.20), we could imagine bringing the protons together, so that R = 0, and then we would have a helium cation, which is a hydrogen-like ion with nuclear charge Z = 2. This is not so whimsical as it seems, for although a diproton is not stable, a dideuteron is stable. If one brings two deuterons closer and closer together, at first there is a growing Coulomb repulsion, but at very small separation the attractive strong attraction takes over. Thisforce arises from the exchange of 7r-mesons, or pions, between the nucleons, and it is described by the Yukawa potential:
Vy
=-
[m7rIicR]
g R exp -
(9.32)
'
m7r
is the mass of a where g is the strong interaction coupling constant, and pion, which is about 140 MeV / c2 • This nuclear force is attractive and is much stronger at short distances than the repulsive electrostatic force, but it is of short
169
Hydrogenic Molecular Ion
range, because of the exponential term. In fact, we can rewrite Eq.(9.32) as (9.33) where 197 MeV F 4 R 7 -1 " -hc --'" ",1. F m7l"c2 140 MeV r-..J
r-..J
Here F stands for fermi, which is 10- 15 m, the typical nuclear scale of distance (cf., Sect. 4.4). Now since 1 A = 105 F, it is clear that the nuclear attraction wins over the Coulomb repulsion only at separations that are tiny, as compared with typical atomic distances like the Bohr radius of the hydrogen atom. Nevertheless, it is energetically well worth the effort, since the difference in mass between two deuterons and one alpha particle (a nucleus of He 4 ) (x c2 !) is released as energy. This fusion reaction occurs, directly of indirectly, in stars, hydrogen bombs, and in experimental fusion reactors. The ground-state energy of a hydrogen-like atom with nuclear charge Z is _~Z2Q:2mc2, and the corresponding normalized wave function is
-
'!/J(r)
=
Vb--;- exp [-] -Zr
.
(9.34)
After fusion of two deuterons the nuclear charge is 2e, and in that case Eq.(9.34), with Z = 2, should be used to estimate the ground-state energy, not Z = 1, which is what we had in Eq.(9.21) et seq. With the deuterons (or protons) separated by an Angstrom instead of a fermi, which is the actual situation in our hydrogenic molecular ion, it is reasonable to expect that the best effective Z to choose for the trial function Eq.(9.34) should lie between 1 and 2. At distances of many Angstrom, Z = 2 should give the tail of the wave function well, but at distances of less than an Angstrom one would prefer Z = 1. Evidently we should be able to improve the variational calculation by allowing Z to be a parameter that will be adjusted: we expect the optimal value to be closer to 1 than to 2. We replace Eq.(9.21) and Eq.(9.22) by
and insert these functions into Eq.(9.25). It is then easy to see that the overlap integral Eq.(9.24) becomes (9.35)
Atomic Physics
170
where
-
-
R=ZR. The Hamiltonian satisfies
[::, + ZV. + (1 - Z)V. + V2 + V'2] 1"'+) [-~Z-2 a 2me2 + (1- Z)V1 + V2 + V12 ] ItP+) ,
(9.36)
and by symmetry it is easy to see that (tP2IV1ItP1) = (tP2IV2ItP1). Hence
(nt. IHlnt. ) = _!z-2 2 2+v, +(1- Z)(1hIVl l1P1) + (tP1I V2ItP1) + (2 - Z)(tP2I V2ItP1) . Cf'+ Cf'+ 2 a me 12 1 + (tP2ItP1) By scaling arguments, we can see that, to obtain (1P1IV2ItP1) or (tP2IV21.~h) from the corresponding expressions without tildes, we simply multiply by Z and replace R everywhere by R. We find
(tP1IV1ItP1) (tP11lt2ltP1) (tP2IV2ItP1) We conclude that
(tP+IHltP+) {
-~a2me2 x
=
Z2 _ 2Z{(Z - l)R + [1 +_ (Z - l)R ~ (Z -_ ~)R2]_e-R + (1 + R) e- 2R }} . R{l + (1 + R + ~R2) e- R }
It is left as an exercise to the student to write a short program, in Mathematica or another language, in order to minimize numerically the above matrix element with respect to Rand Z. We find the optimal value of Z to be about 1.25, which fits our expectations well enough. The optimal R is 3.125, but this corresponds to R = RI Z ~ 2.5, which is the same value for the separation, in units of the Bohr radius, that we obtained before. The binding energy is now improved to 0.173 times the hydrogen atom ground-state energy. The results in more useful units are as follows:
-
Z 1 1.25 Experiment
-
-
R( Angstrom) 1.3 1.3 1.06
Binding energy (eV) 1.77 2.2 2.8
171
Hydrogen Molecule
The results are satisfactory, and the accuracy can be improved by choosing more complicated trial functions, a la Hylleraas.
9.6
Hydrogen Molecule
The hydrogen molecule, with its two protons and two electrons, is pictured below. The total Hamiltonian for the electrons is now
pi + p~ H =
2m
e2
-
e2
e2
e2
iiI - ~ R I - iiI + ~ R I -- Ir2 - ~ R I - 1T2 +
e2 ~ RI+ R
e2
+ ITI - r21 '
the last term being of course the electron-electron mutual repulsion term. This can be rewritten (9.37) where, for i
= 1,2, (9.38)
HI is the Hamiltonian for an H-
-e anion composed of electron 1 and -e the two protons, while H2 is that composed of electron 2 and the two protons. In each term, each electron is associated with both protons: we thus speak of the e molecular orbital method, and 2 it has much application also to complicated molecules. Notice Hydrogen Molecule 2 that the term e / R in Eq.(9.37) has a negative sign; this is right, since it occurs in HI + H2 twice with a positive sign. A reasonable first approximation is to drop the last two terms in Eq.(9.37): after all, one is attractive and one is repulsive, and the average separation of the electrons from one another might be expected to be very roughly the same as the separation of the two protons from one another, so the contribution of one term to the ground-state energy should largely be canceled by that of the other.
!R
e
172
Atomic Physics
With the approximate Hamiltonian
the ground state wave function is simply the spin-singlet combination
'I/J+(R, Ti, i 2) = 'I/J+(R, id'I/J+(R, i2)Xsinglet, where 'I/J+(R, i) is the H- ground-state wave function that we calculated in the last section. The ground-state energy, in this approximation, is just twice that of the H- ion, and hence the binding energy of the hydrogen molecule, being the difference between the energies of two H- ions and two H atoms, is thus twice the binding energy of the H- ion. We calculated 2.2 eV for the latter, so we get 4.4 eV for the binding energy of H2. The experimental value is 4.75 eV, so we seem to have booked a big success with little extra work. The goodness of the agreement is actually fortuitous, because if we use the experimental value of the H- binding energy instead of what we calculated in the previous section, we get 5.6 eV. Moreover, 'improving' the calculation by including the term - eR2 (easy) and the term + ,_TI-T2 e 2 _ , (hard!), does not improve agreement with experiment, rather it makes it significantly worse. This does not mean that quantum mechanics has failed; but it does mean that we have gone about as far as we can with hand calculations and simple programs on the personal computer. To proceed further one needs sophisticated programming, involving complicated trial functions and numerical integration routines. That way lies quantum chemistry, littered with messy calculations but hugely crowned with success.
9.7
Exercises
Problem 1 Consider the following terms: IS, 2p, 3p, 3D, 2D, ID, 4D. (1) What levels may arise from these terms (i.e., which values of spin, orbital angular momentum, and total angular momentum can occur)? Express your findings in the standard form, 2S+1L J . (2) Arrange in order of increasing energy the levels that may arise from the following configurations: ls2p, 2p3p, 3p3d.
Problem 2 Calculate the probability density, p(i) = \'I/J+(R, i)\2, for the electron in the Hi molecular ion, and plot it on the line joining the two protons. Evaluate
II ).
p( R ) / p( kR) and p( 0) / p( k
173
Exercises
Problem 3 Calculate the first and second ionization energies for helium (i.e., the energy required to strip first one, and then the other of the electrons from the atom). Compare your results with the experimental values of 24.85 eV and 54.40 eV. Problem 4 The first few S terms of helium have the following energies, relative to 182 IS, the ground state: 1828 IS : 20.89436; 1828 3S : 20.08735; 1838 IS : 23.23007; 1838 3S : 23.02549 (all in units of the electron volt) . Calculate the contributions to the energy coming from the overlap integrals in the 1828 and the 1838 configurations. Problem 5 Calculate the energy of the ground state of the lithium atom. With the Bohr radius set equal to unity, use the trial functions 2Z;/2 exp( - ZI r) for the Is electrons, and cZ;/2 exp( - Z2r /2)(1 - dZ 2r) for the 2s electron. Here ZI and Z2 are variational parameters, while c and d are determined by considerations of normalization and of orthogonality. Problem 6 Consider the beryllium atom. (1) Write down the Slater determinant for the ground term of this atom, and find an expression for its energy in terms of Coulomb and exchange integrals. (2) Find expressions for the energy in terms of the Hartree-Fock expression for the configuration 18 2 28 2 . (3) Evaluate the expectation value < wI H I w>
Problem 7 Write down the part of the ground-state wave function that represents the pstate electrons in the atom of nitrogen in terms of '¢21m, m = -1,0, 1. Assume that the Hund rules apply. Problem 8 Consider a 3 - d electron in an atom that is in a crystal lattice. The energy level of this electron is fivefold degenerate. The corresponding wave functions are
(ill)
=
xzJ(r)
(iI2) = yzf(r)
(iI3) = xyJ(r)
(1) Show that these are indeed 3d wave functions.
174
Atomic Physics
The influence of the crystal environment on the electron is described by the perturbing (rhombic) potential
V(r) = Ax2 + By2
+ ez 2 .
(2) Calculate the energy shifts induced by V in first order perturbation theory. (3) What are these shifts for a tetragonal field (A = B)? (4) Show that the expectation value of the z-component of the angular momentum always vanishes in the presence of this field (in first order). (5) Is the same true for the x- and y-components?
Problem 9 Carbon dioxide is a linear molecule (OCO) that can combine with an electron to form an ion. Suppose that the electron has energy Eo when it is attached either to one of the oxygen atoms or to the carbon atom. The state of the molecule when the electron is attached to the carbon atom is Ie), while the states corresponding to its being attached to one or other of the oxygen atoms are respectively 101) and 102 ). These three states are not eigenvectors of the Hamiltonian, because there is a small transition probability that the electron jumps from the central carbon atom to one of the oxygen atoms, or vice versa. The probability that a direct transition takes place between one oxygen atom and the other is very small, and may be neglected. (1) Write down a Hamiltonian for this ion. (2) Determine its energy levels and its eigenfunctions. (3) An external electric field is applied, so that the electron's energy when it is attached to one oxygen atom is Eo + 1) N
L anzn ,
P(z) =
n=O
where it is supposed that aN =J. 0, so this is truly of Nth order, but where some or all of the other coefficients may vanish. For Izl sufficiently large the term aN zN will have a larger modulus than the sum of the rest: indeed we will be able to find a real positive R so great that, on the circle Izl = R, we have IP(z) I > ~ laNIRN. Inside this circle of radius R in the complex z-plane, IP(z)1 must reach its minimum, non-negative value at least at one point, say zo0 We shall suppose that this minimum value, IP(zo)l, is not zero, but is strictly positive. This will lead to a contradiction, allowing us to conclude on the contrary that IP(z)l, and hence P(z) itself, has at least one zero in the circle of radius R. The polynomial can be rewritten in the form N
P(z) =
L cn(z -
(A.12)
zo)n ,
n=O
where the coefficients Cn can easily be computed from the coefficients an, given zo° It can occur that some of the Cn vanish, but certainly CN =J. 0, since P(z) is an Nth order polynomial, and Co = P(zo) =I 0, since IP(zo)1 > by assumption. We shall write z - Zo = r eiB and
°
N
P(z) = Co + Ckrk eikfJ
+
L
cnrn einfJ
n=k+l
Here k could be 1, but it could exceptionally be N, in which case the sum would be absent. Let us write the known complex constants
and choose the phase () of z - Zo, which is at our disposition, to satisfy () _
-
7r
+ cPo - cPl k
.
Fundamental Theorem of Algebra
183
Hence, with z satisfying this constraint, N
P{z) = (Icol - ICklrk) eir/>o
+
L
cnrn einlJ
n=k+l Now for all r such that
< ro
l/k
I~ I
, we can certainly find a real, positive constant,
0,
N
L
cnrn einlJ < or k + 1 ,
n=k+l so in this domain (A.13) For r small but non-zero the right-hand side of the above inequality is less than Icol = IP{zo)l, and this contradicts the assumption that IP{zo)1 > 0 is a minimum of IP{z)1 in the circle Izl < R. The theorem has thus been proved by reductio ad absurdum. The basic trick was to find a direction in the complex plane along which IP{zo)1 could be made smaller, so that a local minimum other than zero is excluded. Since we now know that a minimum occurs at a point where Co = P{zo) = 0, it follows that Eq.{A.12) can be written N-l
P(z) = (z - zo)
L
Cn+l{Z - zo)n ,
n=O
i.e., z - Zo times a polynomial of order N - 1. However, this latter polynomial can be subjected to the same reductio, showing that it too has a zero that can be factored out. The procedure stops when a zeroth order polynomial, i.e., a constant, is left. Thus Eq.(A.12) can finally be written N
P{z)
= CN
II (z -
zn),
n=l
where the zn,.which may.not all be, distinct, are called the.roots mial, being the solutions of the equation
P{z) =
o.
oLth~
polyno
N
184
A.4
Completeness of Eigenvectors
Diagonalization of a Hermitian Matrix
We shall show that any Hermitian matrix can be reduced to diagonal form by a unitary transformation. That is, if
= a,
at
then a matrix
u
exists such that
u tu
=
1 and
a diag =
(A.I4)
utau
is a diagonal matrix. Note that, in a finite number of dimensions, u t u = 1 implies Idet ul = 1, which in turn implies the existence of an inverse of u, which is of course u t. This means that also uu t = 1. A matrix that satisfies u t u = 1 = uu t is said to be unitary. Now Eq.(A.I4) is equivalent to ua diag
= au,
or, written out fully, Uu
Ul2
ulN
Al
0
u21
U22
u2N
0
A2
0 0
uNI
uN2
uNN
0
0
AN
au
al2
alN
Uu
ul2
UIN
a21
a22
a2N
u21
u22
u2N
aNI
aN2
aNN
UNI
UN2
uNN
where the Ai are not necessarily all distinct. For a constructive proof of this statement, consider the matrix equation au a21
A
al2 a22 -
aN!
aN2
A
alN
Uu
a2N
U21
aNN-A
UNI
=
o.
(A.I5)
We know from Sect. A.2 that a necessary and sufficient condition for Eq.(A.15) to have a nontrivial solution is the vanishing of the determinant of the matrix 0" tne left \.e. ditt t~ AI1 ;; 0 where 1 ii l he 'Ioit m,t"llC This 'iletefmiu"l\llt is however an Nth order polY.b.Omla\ ia A, and we know trorn the Sect. A.a ihai 1'1; has N roots, which may however not all be distinct. We call one of these roots Al (we know there is at least one), and we construct by the method of the Sect.
-
185
Diagonalization of a Hermitian Matrix
A.2 a nontrivial, normalized solution for the Un, the first column of the matrix u. Let us now require the remaining Uij, j = 2, 3, ... ,N to be normalized, and to be orthogonal to Uil, and also to one another, i.e., 2::1 UijUik = bjk , for all j, k E {I, N}. This is always possible, and of course there is much freedom in the choice, subject to the constraints. With this construction, U is unitary and U t au has the form .A1
0
0
0
0 0
b22 b32
b23 b33
b2N b3N
0
bN2
bN3
bNN
(A.16)
The zeros in the first column follow from the fact that 2: j .A1 2: j uljuj1
=
2:k uljajkuk1
=
.A1bil; and the zeros in the first row follow from this, and the
fact that Ut au is Hermitian: [u tau] t = Ut au. Hence the first step of the diagonalization has been performed. Let us write the above in shorthand notation as U
t
( .A1
au = (
(A.17)
\. 0
where b is a matrix of order N - 1. By using the property det ac at the end of Sect. A.2, we have
= det ca , proved
det[a - .AI] = det[(a - .AI)uut ] = det[ut(a - .AI)u]. From Eq.(A.17),
det[a - .AI]
= det (
.A1 - .A 0
b _0>.I )
= (>.. -
>')det[b -
>.II ;
where I is now the unit matrix in N -1 dimensions. The right side of Eq.(A.17) is a Hermitian matrix and so it follows that .A1 is real and b is a Hermitian matrix . .A1 is called an eigenvalue of a. By the above procedure we have reduced the eigenvalue problem for a Hermitian matrix in N -dimensions to one in N - 1 dimensions, in the process having 'split off' one eigenvalue. By iterating the procedure we can proceed to split off the next eigenvalue, thereby reducing the dimension again. It is important to stress that the next eigenvalue may possibly not be different from .A1. The method nowhere requires the successive eigenvalues to be different: it is enough that the matrix b be Hermitian to ensure that it has at least one eigenvalue.
186
Completeness of Eigenvectors
At each step of the procedure a unitary transformation of the matrix is made, so that finally, after the original matrix a has been reduced to diagonal form, N - 1 successive unitary transformations have been made, which of course can also be written in retrospect as one unitary transformation, as in Eq.(A.I4). The columns of U are just the eigenvectors of a: U is not unique, for the order of the eigenvalues in a diag can be changed, and in the case of degeneracy the choice of basis in the eigenspace is not unique, but the important point is that there are always N orthonormal eigenvectors. Since the N eigenvectors are mutually orthogonal, they are necessarily independent of one another, and so they span the entire N-dimensional space. This is really the end of the proof; but to do the job a little more formally, consider a basis of orthonormal vectors that obviously spans the space, namely {ej}, where 1 0 0
el =
e2
0 1 0
=
o
eN
=
o
0 0 0
1
Such a set of vectors is called an orthonormal basis. An arbitrary vector x can be expanded in terms of them: (A.I8)
x = LXkek. k
The N eigenvectors of the matrix a that we have constructed can be written
Uj = L Uijei· i
Using the unitarity of the matrix u, we find
ek
= L 6ki ei = L L UkjUijei = L UkjUj . i
i
j
j
Hence
x
= LXk LUkjUj = LCjUj, k
where
j
j
Commuting Hermitian Matrices
187
This concludes the explicit demonstration that an arbitrary vector x can be written as a linear superposition of the N orthonormalized eigenvectors Uj of an arbitrary Hermitian matrix a.
A.5
Commuting Hermitian Matrices
In this section we shall show that if two Hermitian matrices, say a and b, commute with one another, then it is always possible to find simultaneous eigenvectors. Let us first consider the case that .A is a simple (or nondegenerate) eigenvalue of a, i.e., ax = .Ax,
and there is only one independent vector x that satisfies this equation for the given .A. Since ab = ba, it follows that abx
= .Abx,
and hence bx is also an eigenvector of a belonging to the same eigenvalue .A. Since there is only one such eigenvector, it follows that bx must be proportional to x, i.e., x is also an eigenvector of b. When there is degeneracy the situation is a little more complicated. Suppose now that, for the given eigenvalue .A,
for q = 1,2, ... ,m, with m < N, where we can suppose without loss of generality that the m independent eigenvectors x(q) have been orthonormalized: X (p)tx(q) --
..
U pq
.
Again we have abx(q) = .Abx(q) ,
so bx(q) is an eigenvector of a belonging to the eigenvalue.A. However, since there are now m such independent eigenvalues, we can only assert that bx(q) IS some linear combination of them: m
bx(q)
=
L r=l
Bqrx(r) .
188
Completeness of Eigenvectors
The orthonormality of the eigenvectors implies that x(p)tbx(q) -
B
qp'
and from the hermiticity of b we find easily that B;p
= B pq ,
so that Bpq are the elements of an m x m Hermitian matrix. We know that a unitary matrix U can be constructed that diagonalizes B. To be precise, there is an array U pq such that
"U ~ pq B qr
"~ Bdiag pq Uqr
-
q
-
Bdiag Upr . pp
q
It follows then that b" Upq x(q) -- " " U pq B qr x(r) -~ ~~ q
r
q
Bdiag" U pr x(r) . pp ~ r
For each of the m values of p, the linear superposition :L: q Upqx(q) is an eigenvector of the matrix b. If all of the elements B~;ag are distinct, then the degeneracy of the eigenvalue ). of a has been resolved, in the sense that each of the vectors :L: q Upqx(q) can be labeled by a different pair of eigenvalues. In the case however that some of the B~~ag are the same, some degeneracy remains. One can look for a third matrix, c, that commutes with a and b, in order to lift the degeneracy completely, and if that does not work, then a fourth and a fifth matrix, until one eventually has a complete set of commuting matrices such that each simultaneous eigenvector has a unique assignment of eigenvalues. A.6
Infinite Dimensional Spaces
In this section we shall illustrate some of the differences between finite and infinite dimensional spaces, and the operators on them. It would be helpful to work through Chapter 1 before reading this section. The configuration-space representation of the kinetic energy of a particle is p2 h2 2 -=--V, 2m 2m
(A.19)
and it is an important operator in quantum mechanics (see Sect. 1. 7 et seq.). In this section we shall for convenience consider one spatial dimension only, in which V 2 reduces to D2 = d2 / dx 2 , but the general conclusions apply to the three-dimensional case as well.
189
Infinite Dimensional Spaces
Consider a twice-differentiable function, ¢>( x), of the one-dimensional variable, x, that is restricted to the interval -L/2 < x < L/2, where we shall consider separately the cases that L is finite, or that it is infinite. Let D2 be the operation of taking the second derivative, i.e.,
Is D2 Hermitian and does it possess a complete set of eigenvectors? The answers depend on how the infinite-dimensional space on which D2 acts is defined. We shall restrict the space to consist of twice-differentiable functions; and we define a scalar product between two functions, ¢> and 'ljJ, to be S('ljJ,¢» = /L/2 dx'ljJ*(x)¢>(x) , -L/2
see Sect. 1.4. The Hermitian conjugate of D2 is defined by S( 'ljJ, D2t ¢»
=
{S( ¢>, D2'ljJ)} * ,
for all ¢> and 'ljJ in the space. Thus D2 is Hermitian if
However, by two partial integrations we see readily that /
L/2 dx'ljJ*(X)¢>"(X) -L/2
= /L/2
dx'ljJ"*(x)¢>(x)
+ B,
(A.20)
-L/2
in which the boundary term is
The operator D2 is Hermitian only if B vanishes, which will not be the case unless we impose further restrictions on our space. With L finite and with a space of bounded functions that satisfy periodic boundary conditions, i.e., ¢>(L/2) = ¢>( -L/2), D2 is indeed Hermitian. Moreover, the two independent solutions of
(A.21) are ¢>(x) = exp(±vlAx). These solutions satisfy the condition of periodicity if
where n is an integer. Thus the eigenvalues form a discrete (but infinite) set, and the eigenvectors are bounded in modulus. Moreover, it is known from the theory
190
Completeness of Eigenvectors
of Fourier series that any bounded, twice-differentiable, periodic function can be represented by a convergent Fourier series, i.e., the eigenvectors form indeed a complete set: they span the space. Since the number of Fourier coefficients is in general infinite, the vector space describing the system has an infinite number of dimensions. To obtain the above desired result, however, we had to restrict the space by the artificial constraint of periodicity. If we relax this constraint, n2 is not Hermitian. If we let L = 00, which is more natural, and require the functions in the space to be bounded, as well as twice differentiable, then the eigenvalues are continuous, comprising all the negative real numbers, and the eigenvectors are e±iJ.Lx, where J.1 = J A. By the theory of Fourier analysis, we know that the functions e±iJ.Lx are complete, in the sense that any bounded, twice-differentiable function can be written as a Fourier integral, but n 2 is not Hermitian on this space, since the boundary terms in Eq.(A.20) are ill-defined in this case. True, the boundary terms oscillate themselves to death, so we might hope that n2 is 'almost' Hermitian, and shares the good features of such operators. Indeed, the kinetic energy operator, Eq.(A.19), has the desirable property that its eigenvalues are the positive, real numbers, and moreover its eigenvectors span the infinite-dimensional vector space. More can be said about the peculiarities of infinite-dimensional vector spaces; and we will return to the matter in Volume 2. The option chosen in this book is to drop the requirement that all physical observables be strictly represented by Hermitian operators on Hilbert space. The essential requirements are that they have real eigenvalues and that their eigenvectors are complete, and this is certainly true of the kinetic energy operator if it is defined on a suitable vector space. Dirac himself was careful never to limit his observables to those that are Hermitian on a Hilbert space, indeed he wrote in his book explicitly that the bra and ket vectors form a more general space than a Hilbert space. He requires an observable to have real eigenvalues and to have a complete set of eigenvectors; but he adds that this is a physical assumption and that it is often very difficult to decide mathematically whether a particular dynamical quantity is an observable. If one can prove that the corresponding operator is Hermitian, the job is done, of course; but if it is not one can sometimes still check the conditions of reality and completeness, as we did for the kinetic energy. Other options, which we will consider in Volume 2, are (1) Extending the space, so that the kinetic energy, and scattering operators, are truly Hermitian. This involves a 'rigged' Hilbert space. (2) Discarding Hilbert space and working in an algebraic approach with projection operators. Von Neumann handled the difficulty in this way.
191
Exercises
A.7
Exercises
In these exercises, whenever the word 'matrix' or 'operator' is used, this means a square matrix of finite dimension. Problem 1 Prove that the matrix (1 - iA) has an inverse if A is Hermitian. Problem 2 Let I be the unit matrix, and A an arbitrary square matrix. Show (1) det [1 + fA] = 1 + fTr (A) + O(f2), where the trace of the matrix is the sum of the matrix elements on the diagonal, i.e., Tr (A) = I.: j Ajj , (2) det[exp(A)] = exp[Tr (A)] .
Problem 3 A Hermitian operator P is called a projection operator if p2 = P. (1) Prove that the only projection operator with an inverse is the identity operator. (2) Let PI and P 2 be projection operators. Prove that if PI + P 2 is a projection operator, then P I P 2 = P 2 P I = o. Operators satisfying this condition are called orthogonal operators.
Problem 4 Consider the three orthonormal vectors, {I
ei
> H=ll given by
(1) Find the projection operators Pi associated with each of these vectors. (2) Prove that Pi projects onto the line along I ei > . (3) Verify that L~=I Pi = 13 •
Problem 5 Show that the symmetrizing operator, S, defined such that, for any matrix, Yes), depending on a variable, s, SV(s) = ![V(s) + V( -s)], and the antisymmetrizing operator, A, defined such that AV(s) = HV(s) - V( -s)], are projection operators. Problem 6 Prove that two Hermitian matrices commute if there exists a complete orthonormal set of common eigenvectors.
192
Completeness of Eigenvectors
Problem 7 Let s be a complex variable and A a constant square matrix of finite dimension. Define Yes) = exp(sA). Show (1) d~is) = A V = V A (2) {V(s)V( -s)} = 0 and hence that the inverse of the matrix Yes) is
:8
given by V-1(s) = V( -s) (3) Using the above results, show that if A is a Hermitian matrix, then eiA is a unitary matrix.
Problem 8 If A and B are square matrices and A commutes with [A, B], show that, for any
n=1,2, ... (1) [An, [A, Bl] = 0 (2) [An, B] = nAn-l [A, B]] (3) t@tA, B] etA [A, B] Problem 9
t ~f\ . 8 'J ~ ,t t A. t1 .
Let t be a complex variable, and define
J(t) = etA.etB.e-t(A+B) where A and B are square matrices that both commute with [A, BJ. Show
(1) f'(t) = 2tJ(t)[A, B] (2) J(t) = exp {t 2 [A, Bn (3) eA.eB=exp{A+B+[A,B]} Problem 10 Let t be a complex variable, and define
get) = etA B e- tA where A and B are square matrices that do not commute with [A, B]. Show
(1) g'(t) = [A,g(t)] (2) get) = exp[tOA]B, where OA is the linear operator, called the commutator operator, that is defined by nAG = [A, GJ, where G is any square matrix.
(3) eAB e- A = B + [A, B] + HA, [A, B]] + ~ [A, [A, [A, BJ]] + ... Here the (n + l)th term has a factor (n!)-l and contains n nested commutators.
Bibliography
Atkins, P.W., (1983) Molecular Quantum Mechanics, second edition, Clarendon Press, Oxford. Atkins, P.W., (1983) Solutions Manual for Molecular Quantum Mechanics, Clarendon Press, Oxford. Basdevant, J-L., (1986) Mecanique Quantique, Ecole Poly technique, Paris. "'Ballentine, L.E., (1989) Quantum Mechanics, Prentice-Hall, New Jersey. ..,.Dirac, P.A.M., (1958) Quantum Mechanics, fourth edition, Oxford University Press, Oxford . .J Gasiorowicz, S., (1974) Quantum Physics, second edition, John Wiley and Sons Inc., New York. Gol'dman, I.I. and Krivchenkov, V.D., Problems in Quantum Mechanics, Pergamon, London. JGoswami, A., (1992) Quantum Mechanics, Wm. C. Brown, Dubuque. JGreiner, W., (1989) Quantum Mechanics, An Introduction, Springer-Verlag, New York. Haar, D. ter, (1975) Problems in Quantum Mechanics, third edition, Pion Ltd., London. JItzykson, C. and Zuber, J.B., (1980) Quantum Field Theory, McGraw-Hill Inc., New York. Kok, L.P. and Visser, J., (1996) Quantum Mechanics, Problems and Solutions, second edition, Coulomb Press Leyden, Leiden. j Liboff, R.L., (1992) Quantum Mechanics, Addison-Wesley Publishing Company Inc., Reading. Mandl, F., (1992) Quantum Mechanics, John Wiley and Sons, Chichester. McMurry, S.M., (1993) Quantum Mechanics, Addison-Wesley Publishing Company Inc., Reading. ~Merzbacher, E., (1998) Quantum Mechanics, third edition, John Wiley and Sons, New York. JSchiff, L.L, (1968) Quantum Mechanics, third edition, McGraw-Hill, New York. Schmid, E.W., Spitz, G. and Losch, W. (1987) Theoretische Physik mit dem Personal Computer, Springer-Verlag, Berlin . .1 Townsend, J.S., (1992) A Modern Approach to Quantum Mechanics, McGraw-Hill, New York.
193
Index
Action, 3 Addition of angular momenta, 101 Clebsch-Gordan coefficients, 101, 112 spin and orbital parts, 105 two spins, 104 Airy function, 93 Alkali atom, 98 Almost degenerate levels, 123 Ammonia molecule, 124 maser, 125 Amplitude, 132 Angular momentum, 44 eigenvalues, 47 eigenvectors, 57 matrices, 99 operators, 43 in four dimensions, 63 Anharmonic oscillator, 118, 128 Anisotropic oscillator, 32 Annihilation operator, 32, 45, 76 Antisymmetric wave function, 155 Slater determinant, 162 Associated Legendre functions,56 Asymptotic behavior r -+ 0, 65,78 r -+ 00, 68, 79 Asymptotic expansion for Airy function, 96 for Bessel functions, 79 for plane wave, 137 Atomic structure, 147 Axioms of vector spaces, 9, 27
Bell inequality, 29, 110 Beryllium, 173 Bessel functions, 75, 82 Born approximation, 135, 145 Bound states, deuterium, 75 harmonic oscillator, 37 linear potential, 93 square well, 73 Bra vectors, 11 Breit-Wigner resonance, 140 Canonical momentum, 4 Canonical transformation, 2 Carbon dioxide, 174 Carbon monoxide, 80 Cauchy sequence, 12 Center-of-mass coordinates, 74, 81, 90 Central potential, 65 Charmonium, 92 Classical mechanics, 1 Clebsch-Gordan coefficients, 101, 112 Commutation relations, for q and p, 13 for components of I, 43 Compact support, 15 Complete space, 12 Completeness of Eigenvectors, 175, 188 Complex vector space, 9 Configuration representation, 20 Conservative field, 1 Coulomb potential, 85 194
195
Index
Creation operator, 32, 45, 76 Cross-section, 133 optical theorem, 145 identical particles, 144 Cuprate, high T c , 42 Degenerac~
124, 188 for harmonic oscillator, 39 for hydrogen atom, 88 in perturbation theory, 120 for simultaneous eigenvectors, 188 Determinant, 176 Deuterium, 97 Deuteron, 74, 110 Diagonalizaton of matrices, 184 Diatomic molecule, 83 carbon monoxide, 80 hydrogen, 171 Differential cross section, 133 Differential equation for Airy function, 93 for Hermite polynomial, 41 for associated· Laguerre function, 89 for associated Legendre function, 57 Differential operator, 21 Dirac delta function, 14 Dirac notation, 80 Distribution, 14 Domain of operator, 12 Eigenfunctions, 37, 50 Eigenvalues, 13, 186 Eigenvectors, 13, 188 Exchange term, 154 Exclusion principle, 155 Expansion postulate, 175, 190 Evolution operator, 22 Euler-Lagrange equation, 3 Fine structure constant, 148 Flux conservation, 29 Fourier theorem, 67 Functional, 10, 15 Fundamental theorem of linear equations, 176 of algebra, 182
Gaussian reduction, 178 Generator of time translation, 22. epees Horne-Zeilinger state, 111 Green's function, 66, 129 Ground state, 90, 118 of harmonic oscillator, 37 of He atom, 152 ofH atom, 90 of square well, 73 Hamilton equations, 5 Hamiltonian, 4, 22 Hamilton's variation principle, 3 Harmonic oscillator, 32, 69 Hartree-Fock method, 157, 173 Heaviside step function, 18 Heisenberg picture, 23 Heisenberg uncertainty relation, 41, 62 Helium atom, 147 electron-electron repulsion, 147 effective spin-dependent term, 157 Hermite polynomials, 36,41 Hermitian operators, 175 Hermitian conjugate, 12 diagonalization, 184 commuting matrices, 187 Hilbert space, 9 Hund rules, 161, 173 Hydrogen atom, 85 degeneracy of eigenvalues, 88 Hydrogen ion, 162 Hydrogen molecule, 171 Hyperons, 91 Identical particles in He atom, 155 in Ht ion, 165 proton-proton scattering, 144 Ions, 85, 152 helium ionization energies, 173 hydrogen ion, 162 Infinite dimensional spaces, 188 Jacobi identity, 6 proof for commutators, 7 proof for Poisson brackets, 7
196
Kronecker delta, 6 Ket vectors, 11 Ladder operator, 33, 45, 76 Lagrangian, 3 Laguerre polynomials, 88, 97 Laplacian, 25, 52 Legendre functions, 55 Legendre polynomials, 56, 137, 145 Levi-Civita symbol, 43 Linear potential, 91 Linear vector space, 9 linear operators, 12 Lippmann-Schwinger equation, 129 Lowering operator, 33, 45, 76 Maser, 125 Matrix, 176 commuting Hermitian matrices, 187 determinant, 176 determinant of matrix product, 181 diagonalization of matrix, 184 GauE reduction, 178 inverse of matrix, 180 orthonormal basis, 186 permutation, 176 Mean, 41, 63 Molecule, 80, 83, 174 H2 atom, 171 Ht ion, 165 vibrational-rotational states, 80 Momentum operator, 19 Neutron-proton scattering, 141 Nitrogen, 173 Norm, 12 N umber operator, 33 Null vector, 9 Observables, 175, 190 Operators, 12, 22, 23, 32 annihilation, 32 45, 76 creation, 32, 45, 76 hermitian, 13 ladder, 33, 45, 76 linear, 12
Index
lowering, 33, 45, 76 projection, 191 raising, 32, 45, 76 symmetry, 91 unitary, 22 Optical theorem, 145 Orbital angular momentum, 44 Orthonormal basis, 10, 27, 186 Oscillator, 32, 69 Overlap integral, 165 Pauli principle, 155 Pade approximants, 128 Pauli exclusion principle, 155 Pauli matrices, 101, 109, 126 Periodic table, 159 Perturbation theory, 113 first order, 115, 121, 126 second order, 116, 128 Phase shifts, 139, 145 neutron-proton S wave, 141 proton-proton S wave, 144 resonant, 140 threshold behavior, 140 Poisson brackets, 6 Polar coordinates, 51, 64 Position operator, 13 Positronium, 107, 110 Potential Coulomb, 85 simple harmonic, 32, 69 linear, 91 square-well, 71 Yukawa, 75, 146, 168 Potential energy, 2 Projection operator, 191 Proton-neutron scattering, 141 Probability density, 29 Proton-proton scattering, 144 Projection operator, 191 Quantum chromo dynamics, 92 Quadratic potential, 32, 69 Quantization of angular momentum, 44 Quarks, 91
197
Index
Radial equation, 65 Raising operator, 32, 45, 76 Range of operator, 12 Representation of 8-function, 19, 28 Relative motion, 74, 81, 90 Resonance, 140 singlet neutron-proton state, 143 Riesz theorem, 10 Ritz variational principle, 117 Runge-Lenz vector, 98 Scalar product, 9 Scattering, 129 amplitude, 132 Born approximation, 135, 145 cross-section, 133 phase shifts, 139 resonant, 140 scattering length, 142 square well, 74, 141 Schrodinger's cat, 125 Schrodinger equation, 24 asymptotics, 65, 79 center-of-mass coordinates, 74, 81, 90 central potential, 65 charmonium, 92 hydrogen atom, 85 hydrogen ion, 162 hydrogen molecule, 171 simple harmonic oscillator, 38 Schwartz space, 15 Screened Coulomb potential, 127 Screening of charge, 150 Separable potential, 146 Simple harmonic oscillator, 32, 69 in one dimension, 41 in two dimensions, 42, 126 in three dimensions, 32, 42, 126 isotropic oscillator, 41, 69 Singlet state, 104, 143 Slater determinant, 162 S-matrix, 139 phase shifts, 139 resonances, 140 Spectroscopic notation, 161, 172
Spherical Bessel function, 75, 82 Spherical Hankel functions, 78 Spherical Neumann function, 77, 82 Spherical harmonics, 54, 57 Spin, 100 Spin-orbit coupling, 161 Spin-and-statistics theorem, 144, 155 Square integrability, 66, 68 Square well potential, 71 bound states, 72, scattering states, 74, 141 Stark effect, in harmonic oscillator, 39 in hydrogen atom, 120, 128 Step function, 18 Strangeness, 91 Superposition of states, 125 Symmetry operator, 91 Test functions, 15 Thomas-Reiche-Kuhn rule, 41 Triplet state, 104, 142 Tritium, 98 Two-electron system, 147 Uncertainty, 41, 62, 63 Unitary operators, 22 Unitary transformation, 184 Variational method, 117 Vector spaces, 9 Vibrational-rotational spectra, 80 Wave function, 20 Width of resonance, 141 WKB approximation, 95 X-ray photon, 108 Yukawa potential, 75, 146, 168 in Born approximation, 136 Zeeman effect, 60 Zero-point energy, 33