Chapter 12

Non-degenerate Perturbation Theory

Verify the formula

oo t I l "'"' r< 47r "'"' } . ( )}' (" )

Ir - -r.,1 = L...J t+T 21+1 L...J 1.m r, l,m r2 ' 1 l=O r> m=-l

where

r1 = (0, ip) , r2 = (O', iy') Hint: Solve the problem

by a) expanding in spherical harmonics. b) realizing that i,.(i) is the potential for a unit charge located at ?, and com-paring the two solutions.

Solution We begin with Poissor.'s equation for the potential~ due lo a unit point charge

"\12 1/.J = &(r - r') . The solution of this equal ion is just

l l 1,'(i)= ----41i Ir- Pl

(12.1.l)

(12.1.2)

12.1 . EXPANSJO.V OF l/lf"1 - f21 233

Now, we also solve this differential equation in spherical coordinates. for this purpose we write

w(r) = L:a1,m(r)}im(O.;p). (12.l.3) I ,m

Jn spherical coordinates we can use tlw con1pleteness of the spherical harmonics to write

c5(r- ~) = _;.c5(r - r 1 ) '' } 1~m(r1)}i,,.(r) r- ~ Im

(12.IA)

where

r = (0, 'P) I ;I = (91 I ;p') (12. l.5) Substitut ing all this in the Pobson equation we get

l d2 l (l + 1 ) I I , --d .,[1'Cz1,,,.(r))- .,

rvnAYll;!;ff. Jj!. NUN-JJEGENERATE PERTURBATION THEORY

as well as

A1 m [-(l + l)r'21+1r'-(l+2) - lr,(l-l)] = __!_2 Y/"m (r') . r r' t Thus,

___ l ___ l_y; (r') A1,m - 21 + 1 r'(l+l) l,m .

So finally we get:

} { E1 m 21~1 r,j~I Yl~me)Yi,m(r) lr--r- 1=4rr ~ 1 r" v ( ')v () - L.....1,m 21+1 ;:r:rr r 1,m r I 1,m r

But, this is just the desired result.

r < r'

r > r'

12.2 Second Order Correction to State In the equation for the rth order correction to the state Im}

(12.1.13)

{12.1.14)

(12.1.15)

(Ol(mln}(r) = E~O) ~ E~) [(mlH' - E~ln}(r-1) - E~2) (Ol(mln}(r-2) - E~3) (Ol(mjn}(r-3) - ... - E~r-I) (Ol(mln) (ll] m ::j:. n (12.2.16)

set r = 2 and derive the equation

jn}(2) = ~ jm}(O) (Ol(mjH' lr}(O) (Ol(rjH'jn)CO) ~ (E(O) - E(O)) (E(O) - E(O)) m,r;tn n m n r

- L lm)CO) (Ol(mlH'ln)CO) (Ol(nlH'ln)CO) m;tn ( E~o) - E~?l) 2

- ~ L lm)CO) I (Ol(mlH' ln}(O)l2 2

m;tn ( E~o) - E~)f (12.2.l i)

for the second order correction to the wave function.

Solution We start with the equation (12.2.16) for the rth order correction to a state, namely

(Ol(mln}(r) = 1 [

J2.3. 1/2 ..\x2 PERTURBATION OF SHO 235

and we set r = 2. T hus, we get

(O)( I )(2) - (O}(mlH' - E~l}ln)(!} m n - (O) (O} En -Em

(12.2.19)

Inserting, from equation (12.2.16) the expression for C0>(mln)(ll, namely (O)(mln){l) 1 (O)(mlH' - El ln)(O}

E~o) - E~) n

E~o) ~ E~) (O)(mlH'ln) m-/: n (12.2.20) we find , a fter writing everything out, the desired result, namely equation (12.2.17).

12.3 1/2 .Ax2 Perturbation of SHO Consider the Hamiltonian

p2 l 1 ., H = - + -kx2 + -,\x k > 0

2m 2 2 a) Find the exact energy of the nth state of this Hamiltonian and expand it to order ,\2 assuming 1,\1 < k. b) Use perturbation theory, treating (1/2),\x2 as a perturbation, and find the energy of t he nth state to order ,\2 . c) Find a bound for the rth order correction and hence show that the pertur-bation series converges for l..\I < k. Hint: For part c) find a simple diagonal bound for the perturbation Hamiltonian.

Solution a) Here

p2 l ., l ? H = - + -kx+ -,\x- k > 1,\1.

2m 2 2 We now define

n2 = k + ,\ , w2 = !:_ 0'2 = ~ . m rn m

The exact energy eigenvalues are given by

En = (n + 1/2)M1 = (n + l/2)hJw2 + a-2 .

(12.3.2 1)

(12.3.22)

(12.3.23) If we now expand this in a binomial series in ,\ to get the corrections due to the perturbation we find

En= (n + 1/2)/iwjl + a-2/w2 = (n + 1/2)/iw [ 1 + 2: 2 - ;.,: + ... ]

1 1 ,\ 1 ,\2 (n + ;-)/iw + (n + -)Ii- - (n +-)Ii--+ .... 2 2 :.!mw :.! 8rn2w3

( 12.3.24)

236 CHAPTER 12. NON-DEGENERATE PERTURBATJON TIIEORY

b) Perturbat ion Theory We have that

E~o) = (n + 1/2)/iw Then the first order correction is given by

E~t) = ,\ (Ol(n!H'ln)(O) = ~ (Ol(nl.z:21n)(O)

2 >.Ii ? -- (Ol(nja2 +at-+ 2a1a + ljn)(o) 4mw >.Ii -(n+ 1/2) . 2mw

So this result agrees with the exact result to this order. Next we have

(2) _ 2 1(0l(nlH'ls)(O)l2 E,l - ,\ I: E(O) - E(O)

a;tn n

For s ::/; n we have

So,

Then,

,\ (Ol(nlH'ls)10l = ~ (O>(nja2 + a12 + 2a1a + lls)(o) 2mw

2~~ (Js(s- l)

12.4. 1/4 .\.r4 PERTURBATION OF SHO 237

But, as stated, we also have

JJ ' < fl" . (12.3.34) Therefore, the pnturbat1on herics is boundrd by the perturbation hcries for fl" as Jong ru; I.XI< k. On thr olhrr hand, thr rth order term for the perturbation series for H" is

= (>./k)r (Ol(nl 2 + ~k.r2ln)(r-1) 2m 2

(>./knn + 1/2)/i,.,.; 1>(nln)lr-l = (>./kr(n + 1/2)/i,.,.;or,t . (12.3.35)

From this we see that we can bound the perturbation serirs for H' by a conver-gent resu lt if l>.I < k. In fart,

r:I

(,\/k)(11 + l/'2)h..J . (12.3.36)

12.4 1/4 A.r4 Perturbation of SHO a) Find the approximate ground htate energy to second order for the Hamilto-nian

p2 1 ., I H = - + -kr + ->..r4 k > 0 2m 2 4

using the Rayleigh-SchrOdingcr perturbation theory. b) Find the ground state correct to order ,\.

Solution a) The Hamiltonian is

,,2 I J fl = Ho+ >.H' = - + -k:c2 + -,\.r-t .

2111 2 4 The unperturbed ground state energy is

E'0> = ~riw. 0 2 The first o rder perturbation gives

( 12.4 .37)

(l2A.:J8)

~ ~ ....... .e. . nv1v-u.cul!lN l!:RATE PERTURBAT I ON THEORY

The second order perturbation correction is given by

,\2 L I (Ol(Olx41n)(O)l2 ..\2 E~2l =

16 liw( l /2 - n - 1/2) n;tO

= - ,\2 (-h-)4 ~ I (O)(Ol(at + a)"ln)(O)l2 16 2mw ~ n/iw

n;tO

= __ 1_ (~)""" (v'4!c5n,4 + 6J2on,2)2 16/iw 2mw ~ n

rt

b) The wavefunction correct to order ..\ is given by

I )(0) (0) ( I H' IO)(O) IO) = IO)(o) + ..\ ~ m m ~ E(O) - E(O) m;tn O m

But, for m f: 0 we have

Therefore,

IO) = IO)(O) - ~ (__!!___) 2 [6J2 12)(0) + 2J6 14)(0)] 4 2mw 2riw 41lw

= IO)(O) - ~-h- [12J212)

12.6. TWO-LEVEL SYSTEM 239

Solution To first order the calculation is t.he same as for Rayleigh-Schrooinger perturba-tion theory. ln second order the appropriate formula is

).2 E(2) = ).2 '' (Ol(Olx"IO)COl o 4 L..., E E(O)

n;o!O 0 - n

Substituting the results from problem 12.4 this becomes

).2 el2l = A2 (-li-) 2 L

240 CHAPTER 12. NON-DEG ENERATE PERTURBATION THEORY

Solut ion The Hamiltonian is as always

II = Ho+ >.Fl'

where

0 ) H' - ( 2 ' -Exact solutions The eigenvalues are given by

( E1 - E i>.a ) det -i>.a E2 - E = O

0 -i>.a

(12.6.48)

(12.6.19)

( 12.6.50)

We ~ume that E2 > E1 so that I E2 - Ed = 2 - E1. Then we have to solw the quadratic equation

(12.6.51) The solutions are

( l 2.6.52)

Expanding in powers of). we get

E+ = l l >.2a2 2(1 + E2) + 2(2 - l~'i) + E2 - 1

E_

( 12.6.5:q

This result shows that

( 12.6.51)

so that the two r.nergy levels are 'r

12.6. TWO-LEVEL SYSTEM

so that -i>.a

a= E E lh. 1- :I:

We can now write

( -i>.a ) lJ' = A::1: E1 - E::1: where, with an arbitrary choicr of phru;e

1 IA:1:I = ? ? ?

..j(E1 - :1:)- +>.-a-

241

(12.6.57)

(12.6.58)

(12.6.59)

Now, to later compare with the perturbation theory we let>. -t 0 and find that >.2a2

E1 - E+ -t -(E2 - Ei) - ---

A+ -t

F2 -E1

>.2 0 2 (2 - Et)+ 2(2 - Ei) >.a .

So, with an arbitrary choice of pha. ... c we have

1/1+ -t ( E;~4E1 ) _ -t ( ,la.. ) E,-I, 1

Perturbation Theory The 0th order c-igenvalucs nncl eigenvectors arc

E1

242 CHAPTER 12. NON-DEGENERATE PERTURBATION T HEORY

So, to first order we have

1/Ji ( 1 ) -i.Xa ( 0 ) 0 + E1 - E2 1

( ;I., ) E2-E1 1/12 ( O ) i.Xa ( 1 ) 1 + E2 - E1 0

( E:is, ) . (12.6.67) To order .X we now have agreement with the exact solutions.

The energy to second order is

I ( V~o) ' H' t/J~o) J 2 a2 E(2J - ~ ""-----'-1 - ~ E(O) - E(o) E1 - E2

n;i!I I 2

J ( 1/J~o)' H'l/J~o) J2 a2 E(2J - ~ =--

2 - ~ E2(0) - E(10) E2 - E1 n;t2

( 12.6.68)

T herefore, the two energies to order .X2 are

E_

E+ (12.6.69) These results again coincide with those obtained by expanding the exact solu -tions in powers of .X.

12. 7 Approximate SHO A particle of mass m moves in a potential

1 V = -klxl2+c 11 < 1 . 2

Estimate the energy of the ground state. Hint:

1 2+ 1 2 1 2 2 1 ( 2klxl < = 2kx - 2k(x - lxl +c) :::::: '2kx2 + 2kx2 In I.xi .

Also,

where

c = 0.5772 l6 ... =Euler's constant.

12.8. T WO-DlMENSl ONAL SJJO 243

Solution Using the fir:;t hint and writing, as suggested.

~kl.rl2+( = ~kx2 + ~ka2 In l.rl c

_ --- ~~- , .,._,..-.1..n::.vr,1vi:;l{A"l'l:: PERTURBATION THEORY

where

w 12 = kifm , w22 = k2/m , a-2 = >./m.

a) Exact Solution We now rotate the axes to remove the cross term.

;c X cos8+Ysin8 y = -XsinO+YcosO

Then, 2V m

= wr [X2 cos2 () + }.2 sin2 0 + .'(} sin 20) + w~ [X2 sin2 0 + Y 2 cos2 8 - :o sin 28)

(12.8.78)

(12.8.79)

+ o 2 [-X2 sin20+Yzsin20+2XYcos28) (12.8.80) So, to remove the cross term , we choose the angle 8 to be given by

(wr -w~) sin 20 + 2o2 cos 20 = 0. (12.8.81) Also, since this is a rotation we have

., ., ., 2 p; + Py = Px + ~ . 2m 2m 2m 2m ( 12.8.82)

So,

p2. p2. l ., I ., ., H = ~ + _>_ + -mn-x2 + -mn-y-

2m 2m 2 1 2 2 (12.8.8:3) where

2 ') () ') . ') () 2 . 28 w 1 cos + w2 sin - a- s111 I ( 2 ") l /( ., ., ., 4

= ~ w, +w2 - 2V w2-wi)-+4o 04 ~ wi + ., ., as o ~ 0 .

Wj -w2 n~ = wi cos2 0 + w~ sin 2 8 + o 2 sin 20

~ (wr + w~) + ~J(w~ -wi)2 + fa4 ., o4 ~ w2 - 2 ., as o ~ 0 . w, -;.u2 (12.8.84)

The (exact) energy eigenrnlues are e,.,,n) = (111 + l/2)hf2, + (n2 + l/2)1if22. (12.8 85)

So, to lowest order in >., for later comparison with perturbation theory, we have

+

(111 + l/'/.)h...;1 + (112 + l/'/.)hw2 h>.2 (n2 + l/2)w1 - (n1 + l/2)w2 2m2 ;.u1w2(w~ - wf) (12.8.86)

J2.8. TWO-DIMENSIONAL SHO 245

b) Perturbation Theory The Hamiltonian Ho represents two uncoupled SHO's. The energy

ITI yW1W2

+ +

+

+

+

(Ol(111.1121\/(1111 +I )1112lm1+I,1112 - l)(OJ (Ol(111,112IJmi(m2 + l)lm1 - l,1112 + l)(O) (0)(111. 112l~J1111 - I, 1112 - 1)] Ii I [

:-;---- ~ v(m1 + 1)(1112 + 1) c5n1,m1+1 c5n,,m2+1 .::111 yWJW:?

J(1111 + l)m2 c5n 1,111 1+1 c5,.,,r11,-I Jm1 (m2 + l) c5n 1,m 1 -1 611, ,m,+1

So, after collrcling terms we get

(12.8.91)

(l:UUJL)

This is I he sanw r

246 CHAPTER 12. NON-DEGENERATE PERTURBATION THEO.Ry

12.9 Kuhn-Thomas-Reiche Sum Rule Classically the polarizability o of an atom is defined as the induced electric dipole moment elfl divided by the strength of the inducing electric field E. So

.

elfl a--- IEI

and for harmonically bound electrons takes the form

- __..:.:__ fj o - 47r2m L v~ - v2 .

j J

Here, fj are dimensionless constants called the "oscillator strengths" . In quan-tum mechanics these are defined by

47rm _ ? fj 3he2 v3olm;ol-

47rm E3 - Eo I _ 12 = 3he2 h erjo = :~ (Ej - Eo}lf}ol2 .

For N uncoupled electrons one then has the Kuhn-Thomas-Reiche sum rule

L li = N. j

This polarizability can be used to describe the absorption of light which carries an electron from its ground state IO) to an excited state In) in an atom.

If, rno = {nlflO) and the Hamiltonian for the bound electron is pi

H = 2m + V(r) . a) Show that

(En - Eo)rno = - iii (nlPlO) . m

Hint: Use the commutator [H. r] and work component by component. b) Use the commutators

together with the results of part a) to prove that for a single C'lectron ~2m(En-Eo) 2 2 2 ~ 3h2 [lxnol + IYnol + lznol ) = l

n

and hence deduce the Kuhn-Thomas-Reiche sum rule. For this problem and all subsequent problems dealing with sum rules it may

be usefu l to consult fl2.ll.

J2. 9. KUJJN-TJJOMAS-REIC'//E SUM RULE

Solution We start with the llarniltonian

pi H = 2m + V(Jl

Then.

l ih [.r, HJ= -2 [.r, P1) =-pr m m and similarly for (y, fl) and (:, /l] . Therefore.

ih -(nlPrlk) = (111[.r, ll]lk) m

Hence, we have

iii -(OIPrlk) = ( f:1t - /~o)(Olrlk) m

and by compll'X conjugation iii

--(klPrlO) = (Ek - Eo)(kl.rlO) . rn

It then follow~ that

L (Ek - l:.'o)l.r4012 = L(Ek - Eo)(Ol.rlk)(kl.rlO) k >-

I: ,11 =- -(OIPrlk)(kl.rlO}

m k ih

= -(OlprxlO} 111

Similarly.

L (Ek - Ho)lxkol2 = L(E1c - Eo)(Ol.rlk}(kl.rlO) k

" L: -ih = ~(01.rlk}(klPrlO)

m le

1h = - -(OlrPrlO)

m

Therefore,

L(E1c - Eo)lrkol2 ih = 2(0lpz.r - EPrlO) "

m

/j2 = '>~

247

( 12.9.93)

( 12.9.94)

(12.995)

( 12.9.96)

(12.9.97)

( 12.9.98)

(I 2.9 .99)

(12 .9.100)

248 CHAPTER 12. NON-DEGENERATE PERTURBATION THEORY

This means that for a single electron

L 2m ? -(Ek - Eo)l.rxol- = 1 . 112 k

(12.9.101)

[Jenee, since all three directions r , y: arc equivalent we have for a single

12.11. POSITRONIUM 249

12.11 Positroniun1 Positronium is a hydrogen-like :o;ystem consisting of a bound .,late of an electron and n positron (positive elPCtron). The ground state con:-i:-ts of a :-inglcl and thrN' trip let substates. The singlet 'it ate is the most st'lblc lying about 8 2 < I 0-4 cV blow thl triplet levels which nre degenerate Field theoretic calculations show that this 1s due to a spin-spin interaction of the form

II A - -0 =--,?bi S2 1-

a) 01'terruin

250 CHAPTER 12. NON-DEGENERATE PERTURBAT I ON T HEORY

So, we have

A=4.lx10-4 eV. (12.11.114)

b) If we introduce a magnetic field B pointing in the z-direction then the Hamil-tonian becomes

- B- A - - eh B- (- - ) H =Ho - = --si s2 + -- u1 - u2 h2 2mc (12.11.115)

where iii are the Pauli matrices and the index 1 refers to the electron and the index 2 refers to the positron. Calling

eB --=w 2mc

we find that in the representation already used we get that

(0 0 0 0 )

- - 0 0 2 0 -. B = liw 0 0 0 0 .

0 2 0 0 (12.11.116)

The eigenvalues of the total Hamiltonian are therefore (listed in the order in which the states are listed in (12.11.110)) A/2 , A/2 + 2/iw , A/'l. , A/2 - 2/"iw.

12.12 Rigid Rotator in Electric Field Consider a three-dimensional rigid rotator with moment of inertia I and electric dipole moment P parallel to the axis of the rotator. This rotator is placed in an uniform electric field E Compute, to lowest non-vanishing order in E, the strength of the electric field , the ground state energy of the rota.tor.

Solution The unperturbed Hamiltonian is

2 Ho= 2J .

The energy eigenvalues are

E _ l(l +l)h2 I - 2/ .

(12 12.117)

(12.12.118)

T he interaction Hamiltonian, if we take the a.xis of the rotator parallel to the z-axis, is

H' = -P E = -PEcosO . (12.12.119)

12.13. ELECTRIC DIPOLE MOMENT SUM RUL E 251

The effect of this perturbation is to produce (in lowest order) a change in energy

E1(l) = -PE(lmlcosOllm) . (12.12.120)

This yields no change. So we have to go to second order. In this case, the ground state is shifted by

E(2) = (PE)2~ {OOlcosOllm)(lmlcosOIOO) o w O-l(/+1)'12

lrn

The only non-zero matrix elements in this sum are

(OOI cos Ol 10} = (IOI cosOIOO}

This ill ustrates the selection rules

ill = 1 , Llm = 0

that apply to electric dipole transitions. Using the fact that

cos 0 = J. }'10 and Yoo= ~ we get

l (001 cos Ol 10} = ( 101cos0100) = J3 .

So, finally

(12.12.121)

(12.12.122)

(12.12.123)

(12.12.124)

12.13 Electric Dipole Moment Sum Rule Show I hat for a system of N particles with charges qJ and niru;ses Mi , j = l .. . N confined to a finite region of space we have the following sum rule (12.l] for the electric dipole moment.

T he sum here extends over a complete set of energy eigenstates.

252 CHAPTER 12. NON-DEGENERAT E PERT( RHATION THEORY

Solution For any op q-= ,,2~ --~\I

J J

(12.13.129)

12.14. ANOTHER SUM RULE 253

12.14 Another Sum Rule Use the double commutator

to derive the sum rule (see [12.1] 2

L (Em - En) (nj L riqr, Im) = N (~~~'.? m J

for a system of N interacting particle:-.. Here In) represents an energy eigenket of the Hamiltonian H.

Solution lf we first write out the double commutator explicitly we find

[[//. ~

254 CHAPTER 12. NON-DEGENERATE PERTURBATION THEORY

2

m j

On the other hand, if we evaluate the double commutator we fiud

= :i . L [Pj eiq ;:, + eiii ;:, Pi , e-iq ;:. ] jk

= -2 (fiq)2 L eiiir, e-;q;:,. t5 'k 2M ik ;

-2 (liq)2 N 2M .

This proves the desired result.

(12.14.131)

(12.14.132)

It is worth noting that if we make the dipole approximation by putting

(12.14.133) we get the result derived in the previous problem.

12.15 Gaussian P erturbation of SHO Bosons Two identical bosons move in the one-dimensional simple harmonic oscillator potential

1 v = -mw2 (xi +xD 2

and also interact with each other via the potential

Find the ground state energy correct to first order in Vo.

Solut ion The unperturbed ground state of the two bosons is

with energy

1 Eo = 2 x ;:liw = liw

(12.15.134)

12.16. GAUSSI AN PERTURBATION OF SHO FERMIONS 255

Here

( mw)l/4 ( mw 2) 0 (x) = 27rh exp - 2h x ( 12. 15. 135)

Also, the wavefunction 1,c>(x1, x2) is already properly symmetrized. The energy shift D. E d ue to the perturbation is given to lowest order by

D..E = (l/J, Vintt/J) = Vo - e-11 r,+:r, e-0 r,-:r, dx1dx2 . (12.15.136) ( mw) 1/2 Joo .. ~ ( 2 2) ( ), 211/i _00

We now change variables to

R = r

X1+2'2

2

The J acobian of this transformation is l. Therefore,

(12.15.137)

6.E = Vo c:~f'2 1: e-0 r 2 exp (- "'riw (2R2 + r 2 /2)) dRdr =

Vii (~) 1/2 ( 211/i ) 1/2 (..!!!!__) 1/2 2Trn mw + 2o-h 2mw

\'o ( 211/i ) 1/2 2 mw + 2crli (12.15.138)

12.16 Gaussian Perturbation of SHO Fermions Two identical spin l/2 fermions move 111 the one-dimensional simple harmonic oscillator potential

V = ~ mw2 (.r2 + .r2) 2 1 2 and also interact with each other via the potential

a) Find the ground state energy correct to first order in Vo for the case of the singlet spin state. b) Find the ground state energy correct to first order in \'o for the case of the t riplet spin state.

Solution a) In lhe singlet case, the spin wavefunction is antisymmetric and the space wavefunction is therefore symmetric in the interchange of the two coordinates. Thus, this cai;e is identical to the case of two bosons discussed in problem 12.15.

b) In the tripl

12.17 POLARJZABILJTY: PARTICL.t; JIV A l:1VA

Solution If we take the electric field pointing in the .r-dir('('tion then the perturbation is

I'= -er . (12.17.144) The p roblem is no". for all practical purpos

258 CHAPTER 12. NON-DEGENERATE PERTURBATION THEORY

After equating this to the expression for the energy shift in terms of the polar-izability we find that the polarizability is given by

4096 e2 ma4 00 n2 a= 7~ L (4n2- l)S

n:J

12.18 Atomic Isotope Effect Every nucleus has a finite radius R = roA l/3 where

ro = 1.2 x 10- 13 cm

( 12.17 .154)

and A is the atomic number of the nucleus. Thus, the potential energy experi-enced by an electron near a nucleus is not simply

Ze2 V(r) = -- . r

If we assume that the charge density in the nucleus is constant then we have instead the potential energy

r~ R r?. R

(12.18.155)

a) Use perturbation theory to calculate the i:;otope shift, that is the depen-dence on A of the K-electron {ls state) for an atom with Z protons and atomic number A.

b) Use this result lo compute the energy splitting for the K-electron between the heaviest lead (Z = 82) isotope A = 214 and the lightest A= 195. Neglect the presence of the other electrons.

Solut ion a) The unperturbed Ham iltonian is

p2 ze2 Ho=---. 2m r

The perturbation is

H' = ( ze2) V(r) - --r-{

Ze2 [ r 2 _ ;! + fl] R2JiY 2 r 0

T ~ R r?. R

The unperturbed ground state energy of the K-electron is

E~o) = - ~ Ze2 2 a/Z

(12.18.156)

(12.18.157)

(12.18.158)

12. 18. ATOMIC ISOTOPE EFFECT 259

where a= 5.292 x 10-9 cm is the Bohr radius. The corresponding wavefunction is

W(O)( ) - _l_ (2z)3/2 -Zr/o o r-vrs;r a e .

The fi rst order correction to E~o) is given by E~1> = (v~o>. H' v~o>) .

Thus,

(1) = ~ (2z)3 ze2 { R e-2Zr/a [~ - ~ R] r2 dr . 0 2 a R } 0 2R2 2 + r

We now let 2ZR 2Zr

n=-- z=-a a

Then,

=

( 12.18.159)

( 12.18.160)

(12.18.161)

(12. 18.162)

(12.18.163)

If we now make the dependence on the atomic number A explicit by writing 2Zro \1 /3 ,i/3 ( ) 0 = --j = "')'.'t 12.18.164 " we have the de.irNI dependence on ;\.

/ Cll l Ze2 2 1_113 [ 12A-2/3 3 A1/3 ~o = -- - . x - - + "f 2 a/Z / 12

(12 12 )]

-r:-i'''" 12

A-2/3 + -:yA-1/3 + :l . (12.18. 165)

b) If we tak1: Z = 82 and A = 195 we get that a = 0 238 Substituting these values we find that

.( 1) l Ze2 -3 l!0 (A= 195) = 2 a/Z x 9.91 x 10 . (12.18.166) Similarly, for Z = 82 and A= 214 we get that a= 0.245 . Thus, repeating the calculation we find that in this case

E~1 >(A = 214) = ~ :;; x 1.08 x 10-2 (12.18.167) Thus, r(.'('alling that

1 e2 2-; = l:l.6 e\' the energy difference in energy of the 1, electrons between the two isotopes is

I Zc2 .:lE = --1- x 9.4 x 10-

4 = 86 eV ( 12.18.168) 2a Z

260 CHAPTER 12. NON-DEGENERATE PERTURBATION T H EORY

12.19 Relativistic Correction to H atom The kinetic energy for a relativistic particle is

-2 l ( -2) 2 T = Jc2 ji 2 + m 2c4 - mc2 :::: !!._ - -- !!._ 2m 2mc2 2m (12.19.169)

Use the last term as a perturbation to calculate the first order correction to the energy levels of a hydrogenic atom. Hi nt: The fo llowing expectation values for a hydrogenic atom may be useful

(l/r)

(I / r2) n 2 a/Z

1 1 n3(l + 1/2) (a/Z)2

Here, a is the Bohr radius.

Solution To save labour we use the fact that

p2 l Ze2 1 Ze2 - = E-\f=-----+-2m 2 (a/Z) n2 r

Thus, the first order correction is givn by

1 ((I Ze2 1 ze2) 2) - 2mc2 2 (a/Z) n2 - -1-

(12.19.170)

(12.19.171)

--- ---- - --- 1 r + Z 2e2 1 r 2 I [ l Z2e

4 1 Z 2e4 l ] 2mc2 4 (a/Z) 2 n4 (a/Z) n2 ( / ) ( I ) Z

2e" [1 (z) 2 i (z) 2 1 (z) 2 1 ]

- 2mc2 4 ; n" - ; n" + ; n3(/ + 1/2) = z2e" (z)2 [ 3 1 ]

mc2 ; 8n4 - n3(2/ + 1) (12.19.172)

12.20 van der Waals' Interaction Two widely separated hydrogen atoms interact via a dipol~dipole interaction whose potent ial, known as the van cier Waals potential, is given by

I/ e2 [ - - ] ~ = RJ r1 r2 - z1 .:2

12.20. VAN DER WAALS' INTERACTION 261

where R is the separation of the centres of the two hydrogen atoms and rj = (x 1 ,y1,.:t) and r2 = (r2 ,y2 ,.::!) are rspc>ctively the coordinates of the electrons associated with atom l and 2. Using perturbation tlH'ory calculate the interac-tion energy of two widely Sl'parated hydrogen atoms. As an approximation for the unperturbed cnergie.-; use

Solution The total Hamiltonian for the two hydrogen atoms is

II = JI 0 I + H 02 + \ ! where

_., z., fl

_ p; - Je-o. - - - -2m r, i=l.2.

The unperturbed energies for H01 are

e2 l En=----;; ao n-

(12.20.li3)

(12.20.174)

(12.20.175)

'with the corresponding eigenstates lni,li,m;). We are interested in the energy shift of the ground state

(12.20.176) To lowest order this energy shift. is

E< 1> = ('-1( 1,2)1\'lt>'(l,2}) f2 R3 [(Il.rdl)(2l.r:zl2) + (Ilvd 1)(2lv212) + (ll.:ill)(21=212)] 0 . (12.20.177)

So we have lo go to second order. In this casl' we find

1 ~ 2 1 L (~1(1,2)1Vln1,l1,m1)ln2,/2,m2) n, n 2 ;1!1

x (n1,/1,mtl(n2,/2,m21\'li.(J,2)) l .,

2, (1!(1,2)1V-1~(1,2}) (12.20.178}

Here we have usl'd the indicated approximation to go to the second line and the completeness relation to obtain the last line. Thest> matrix C'lc1111'11ts are now

262 CIIA PTER 12. NON-DEGENERATE PERTURB.\TION T HEORY

easy to evaluate using the symmetry of the ground state wavefunction. In fact the only non-zero matrix elements are

But,

(ilr?li) = 3 e-2r/ao r4 dr 4100 ao o

- 3 2 - ao .

Thus, we finally have the de:;ire

Chapter 13

D egenerate Perturbation Theory

13.1 Stark Effect for n = 2 Level in H Find the shift in the energy of the n = 2 levels of a hydrogen atom, to fi rst order due to a comstant electric field (linear Stark effect). The potential is

V' = -eE r = -ef.:: .

Solution We first write out the n = 2 levels for the hydrogen atom in units such that length is scaled by the Bohr radius.

v12,o.o --1-(1 - r/2) e-r/2 vs;

1/12,1,0 1 -r/? 0

---re - cos v'321r

v2,1,1 = -1-re-r/2 sinOc1

v16

264 CHAPTER 13. DEGENERATE PERTURBATION THEORY

Here we have restored the usual units so that the Bohr radius a is no longer of unit length. The eigenvalues of this matrix are +ct:a, -ea, 0, 0. Therefore the degeneracy of two of the states is lifted in first order and the new energies a re E~o} + ea, E~O) - ea, E~0>, and E~o} where

2 E(O} - _:..___

2 - 2a

13.2 Perturbation of Particle in a Box A particle is in a two-dimensional box of sides a. If a perturbation

V' = >.xy

(13.1.2)

is applied, find the change in the energy of the first excited state to first non-trivial order.

Solution The eigenfunctions for the particle in the box are given by

2 . mrx . k1ry I/Jn k = - Siil -- Siil - 0 $ x, y 5 a . a a a

The corresponding energies are:

fi21r2 ? ? En,k = -2 ., (n- + k-). ma-

(13.2.3)

(13.'2..t) This shows that the ground state with n = k = 1 is the only nondegenerate stale. All other states are dearly degenerate. The first excited state is two-fold d

13.2. PERTURBATION OF PARTICLE IN A BOX 265

and

(13.2.9)

in our perturbation calculation. The first order correction is given by .>. times the eigenvalues above. The second order terms are given by

Now,

But,

(1.ryin, k}(O) 2J2 la [ . 1TX 21Ty -.,- d.x dy Sill - Sill -a- 0 a a

. '21T .r . rry] n1T .x . k1Ty sill -sill - ry sin -sin - .

a a a a

21a . 717rJ' . 1111TJ: ? .rd.r Sill -- Sill --a- 0 a a

1 la d [ (n - m)u (n + m)7r.r] = -;; ;c .r cos - cos

a- 0 a a

1 la [ a d. (n - m)u = -;; J: Sill

a- o (n - m)rr a a d. (n+m)7r.r]

Sill -'-----( n + m) 7r a l la . ( n - m) 7r .r d

1:>111 .r (11 - m)rra 0 a l la . ( 11 + m )7r ..r d Sill .r (11+111)7ra 0 a 1 ., .,[(-1)"-m -1)- 1 ., .,[(-l)"+m -1) (n-m)-7r- (n+m)-7r-

( 13.2.10)

(13.2.11)

[(-1)"+m _ q_!_ .,411111 ., ., (13 2 12) rr2 (n- - m-)-

With this rPsult wc> find

(1.rvln, k}(OJ = 16\1"2 { [(-1)"+1

- l) [(-l)k+2 - 1) 7r4 (n2 - 1)2 (k:! - 4)2 [(-l)n+2 - I) (( -I )k+' -1)}

2 . (n - 4)2 (k2 - 1)2 (13 2.13)

Sub::.tituling this in the sum for the perturbation sf'fies we have the desired result.

266 CHAPTER 13. DEGENE.RATE PERTURBATION THEOit,y

13.3 Perturbation of Isotropic Two-dimensional SHO

For the two-dimensional simple harmonic oscillator with

calculate the effect, to second order, of the perturbation

H' = >.(a!ata1a1 + a~a~a2a2) on the second excited states and to first order on the third excited states. What are the effects on the ground state and first excited states?

Solution In this case we have

Ho= !iw (a!a1 +a~a2) (13.3 .14) (13.3 .15)

and

( 13.3.16)

This shows that all states except the ground state are degenerate. For the second excited state we have three degenerate states:

All have the same unperturbed energy 2h..u. Jn this degenerate subspace, the perturbation Hamiltonian is

(13.3.17)

The eigenvalues are

2>., 0 , -2>. .

The corresponding eigenvectors are respectively

~ ( ~ ) ' (!) '~ ( j) ( 13.3.18)

TWO-DIMENSIONAL SHO WITH OFF-DIAGONAL TERM 267

Thus, we have as new basis states in this degenerate subspace

lt/12.+) = ~ [12, o) + 10, 2)] ll/J2,o) = 11. 1)(0)

lt/12.-) = ~ [12. O)(O) -10. 2)] Therefore, writing r == + 1, 0, -1 we find

I (Ol(n1. n2IH'll/J2,r)l2 = 2hw + r2,\ + L 2hw ( )Ii

- n1 + n2 w n1+n~;t2 = 2/i..iJ + r2,\ + 0 to second order in ,\ .

The third excited stales are 131 O)(O) I 12, l)(O) I 11. 2) t 10. 3) The perturbation Hamiltonian in this degenerate subspace is

The eigenvalues are:

0, , 0, Ji2,\, and, -/i2,\.

(13.3.19)

(13.3.20)

( 13.3.21)

Not all the degeneracies are lifted. f'or this reason we only compute to first order in ,\, To this order the eigenvalues are

E3,r = 3hw + rv'l2>. where r = 0, 0, 1.

13.4 Two-dimensional SHO with Off-diagonal Term

a) Repeat the previous problem with

H' = ,\((a!a2 + a;at)]. b) This problem can also be solved exactly by introducing operators

A2 = -a1 sin 0 + a2 cos8

( 13.3.22)

and choosing fJ appropriately. Do this and compare with the perturbation result.

268 CHAPTER 13. DEGENERATE PERTURBATION T H EORY

Solution a) Again we have

Holn1, n2)(o) = (n1 + n2)h ...... ln1, n2)(0) . (13.4.23) We fi rst solve for lhe eigenvalues to second order in A, then we solve the problem exactly and compare. For n1 + n2 = 2 (second excited state) we have the three degenerate states: 12, O)(o) , I l, 1 )(O) , and IO. 2). T he perturbation Hamiltonian in this degenerate subspace is

(13.4.2-1)

The eigenvalues arc: 2A , 0 , -2A. The corresponding eigenvectors arc respec-t ively

( 1/2 )

I I,/?, 1/2 (

I/,/?, ) -l~v'2 ( 1/2 ) -1/,/?, . 1/2 (13.4.25)

Thus, our new basis slates are

ltc12 +) = !12 O)(O) + - 1 I I t)

13.4. TWO-DIMENSIONAL SHO WITH OFF-DIAGONAL TERM 269

The eigenvalues for this matrix arc: :u . -:l>. . A , -A. The corresponding eigenvectors arc

( 1/../1 ) 1 ./3/2

2 ../3/2 I I .,/2

~ 1/../2 (

./3/2 ) 2 -1/./2

-/3/2

( 1/./2 ) 1 -../3/2

2 ../3/2 -1/./2

( /3/2 ) 1 -1/./2

2 -1/./2 . /3/2

(13.4.29)

We could now again compute the new ba..;i:-. \'ectors and go on to :second order, but becau:.c> [110 , ll') = 0 there is no second order contribution. Thu:;, correct to second order the energies art>

E2 = 3fiw 3A 3hw A . (134.30) b) \Ve next examine the exact solution. for this purpose we make the principal axis transformation

B -

270 CHAPTER 13. DEGENERATE PERTURBATION THEORY

13.5 Non-diagonal Two-dimensional SHO For a particle of mass m moving in the potential

Find to order ,\2 the energy of the ground state and first excited state. Compare your answer with the exact solution obtained in problem 12.8c.

Solution In this case k1 ~ k2 and all levels except the ground state are almost degenerate. Thus, for the fi rst two excited states we need to use degenerate perturbation theory. Ground state: n1 = n2 = 0

and

while

E(l) - 0 0,0 -

where we have

h 1 (Olxlm1)(0lylm2) = 2m Jwiw2 (80,m 1 -160,m2 -d Thus, we get

,(2) - >.2n2 1 1 Eo o - --------, 4m2 w1w2 h(w1 + w2)

Notice, that this expression does not involve

2 2 k1 - k2 Wl -W2 = ---

m

First two excited states: n 1 = 0, n2 = 1 ; n 1 = 1 , n2 = 0 The Hamiltonian is

2 2 J:r Pr + Py 1 ? 1 2 1 = + - k1r + -k2y + >.xy . 2m 2 2

Our perturbation is

H' = >.xy = ?h>. (a1 +al)(a2 +a~). m1.J

( 13.5.36)

( 13.5.37)

( 13.5.38)

(13.5.39)

( 13.5.40)

(13.5.41)

( 13.5.42)

(13.5.43)

13.5. NON-DIAGONAL T WO-DIMENSIONAL SHO 271

To do degenerate perturbation theory we diagonalize this perturbation in the degenerate subspace. That is, we diagonalize the matrix

( 13.5.44)

The eigenvalues are

E~> = h).. 2mJw1w2

(13.5.45)

and the eigenvectors are

( 13.5.46)

The energies are therefore

( 13.5.47)

To compare with the exact solution obtained in problem 12.8a we must use the fact that lk1 - k21 < 2>. and put w1 = w2. It is then easy to see that the results agree lo this order.

We can now do higher order perturbation theory ui:;ing that linear combi-nation of IO, 1) and p, 0) which diagonalizei:; H' In other words. to find the perturbation to second order of the levels IO, l) and 11. 0) we use ru> a basis the set : IO ,O), Ii+) .1~1-), ln1.n2) where n1+n2~2 . Thus, we get

Bu t,

1(1. n', n;IH'ln1. n:i)l 2 En~.n; - En, n1 ( 13.5.48)

2 ~ ~ [

v nAr i.eft l;J. Ut:vENERATE PERTURBATION THEORY

Combining lhese results we get

s- - +-(.,, >..2 ti [ i 1 ]

O,I - - 4m2w1w2 Wt +w2 2w1 (13.5.52)

and

(13.5.53)

Again, if we set w1 = w2 1 we find that the results agree to order >..2 with the exact results obtained in problem 12.8a.

13.6 Particle in a Box Perturbed by Electric Field

A particle of mass m and a charge q is placed in a box of sides (a, a, b) where b < a. ,\ weak electric field

E = E(y/a, .r/a, 0) is applied to this particle. Find the energy of the ground state and first excited states correct to order 11.

Solution for a box with sides (a, a, b) the nergy eigcn::.tates for a particle in this box arc

r(o) 2.J'i . (11rr.z:) . (mrry) (''rrz) Vn m r = /i' sin -- Siil -- Siil -' avb a u b (I :1.6 .51)

where n. 111, r = 1. 2, 3 . .... The corrcspo11di11g energies are:

I ., [ ( .. 2 ., ') ., ] (D) _ 1 11 + m )rr- r-11'-E,. ,m,r - -2 ., + -b.,

m a- -(I :l.6 .5fi)

Since b

13. 7. UNUSUAL PARTI CLE ON INTERVAL 273

f or the fi r:-.t excited states we nN'd to diagonalize the perturbation in the de-generate sub:-.pacP. T hus, we have to evaluate

e . _ r . . -(I. 2, !lr.yp, 2, 1) - -(:.! , l , l!r.1112.1, I} a a

rf. 414 ., (rr.r) 1/J . ., (2rry) = -; a 2 O Sill- 7 .r d.r O Sill- Q ydy

1.a 4 (l:l .6.58)

as w.H' = >..r . Hi nt: T iu ugensolutions for Jl0 arl' giv

vnArnn! lJ. DEGENERATE PERTURBATION T HEORY

Solution The eigenfunctions of the unperturbed Hamiltonian are, as in [13.l], given by

fn(x) = ~ ei1r(n+l/2):r/o n = 0, 1, 2, ... v2a

The energy eigenvalues are

1T2 h2 2 En = E-(n+l) = -2 2 (n + 1/2) . ma

(13.7.64)

(13.7.65)

This shows that every eigenvalue is two-fold degenerate. We thus need to diag-onalize the perturbation Hamiltonian in the degenerate subspace::.. This means we have to evaluate the matrix elements

(/n,Xfn) = U-(n+l).Xf-{n+l)) = 0 (13.7 66) as well as

Un. xf-{n+l)) = U-(n+l) x/n)r = 7r(2na + l) ( 13. 7 .67)

The energy corrections are now given by the eigenvalues of the matrix

(2 >.a ) ( 01 01 ) . n + l 1T

The eigenvalues are

>.a (2n+l)11"

( 13. 7 .68)

(13.7.69)

Thus, all the degeneracies are lifted and the energies corresponding to En and E-{n+l) are

1T2h2 ? >.a --(n + 1/2)- ---2ma2 (2n + 1 )11"

13.8 Rigid Rotator in Magnetic Field A system with moment of inertia I has the Hamiltonian

2 Ho=

21 .

a) What are the energies of the lowest and first excited states? b) A perturbation

H' eB = g-L:r Mc is applied. Find the splitting of the first excited states.

(13.7.70)

13.8. RJGID ROTATOR JN MAGNETJC FJELD 275

Solution We have

2 ' eB Ho = 21 , H = g Mc Lr . (13.8.71)

The eigenfunctions of L2 are Yi ,m with eigenvalue l(l + l)h2 . Therefore, we have l(/+l )h2

HoYi,m = 2I Yi,m ( 13.8. 72)

Every eigenvalue is (21+1)-fold degenerate Now, we can also write 1

Lr= 2"(L+ + L) . (13.8.73)

Thus, the first order corrections to the energy are given by

(13.8.74)

since

L Yi.m = )l(l + I) - m(m l)h}/ ml . (13.8.75) For the first excited state's we have to diagonalize the matrix with matrix

elements

(Y1 m, LrF1,m) = ~ [ )'2 - m(m + 1)6m' m+1 + J2 - m(m - 1)6m',m-l ] . {13.8.76)

Written out as a matrix this looks like

- I 0 l I ( 0 1 0) V2 0 l 0

(13.8.77)

The eigenvalues are: 0 , 1. Thus, the energies correct to second order are given by

fi:? I

li2 eB --g-r \le

( 13.8. 78)

Incidentally, since Lr commutes with L2 and is known to have the eigenvalues mh wi th m = -l, -(1- 1 ), ... , (l-1 ), I we can write down the exact eigmvalul'S of the total Hamiltonian. The~ are

'12 eB Et m = -1(1 + l) + g-rnh

' I Ale (13.8.79)

in complete agreement with the perturbation result.

276 CHAPTER 13. DEGENERATE PERTURBATION THEORY

13.9 >..xpy Perturbation of SHO Find the energy correct to order ..\2 for the second excited state of the Hamil-tonian

H =Ho+ ..\H'

where l 2 2 l 2( 2 2) Ho= 2m (Pr+ Py)+ 2mw x + y

and

H' = xpy.

Can this problem be solved exactly?

Solution The energy levels of the second excited stale are 12. 0) , ll. l) , IO. 2). They all have the same energy 3hw. We now diagonalize the perturbation part of the Hamiltonian in the degenerate subspace. To do this we use

th t . r;;r;:;; t x = v ~(a1 + ai) , Py= iv 2-2-(a2 - a2) {13.9.80)

Then,

\If' \ h[ t t t t ) " ='"2a1a2 +a1a 2 -a1a2-a1a2. (13.9.81)

In the degenerate subspace the perturbation matrix to be diagonalized is now easily computed to be

i- -1 0 l ,\h ( 0 1 0 ) 2 0 -1 0 (13 .9.82)

The eigenvalues and corresponding eigenvectors are ..\h/2 , 0 , -..\h/2 and

1 ( 1. ) 1 ( 1 ) 1 ( 1 ) v'6 ~ll t ../2 ~ ' v'6 -=-~l . (13.9.83)

Thus, the new ba:;is states wc use for starting the perturbation computation arc

It/I+> lt,'.>o)

= ~(l2,o)+2ill,l)-I0,2)) ~ ((12. O} +IO, 2))) 1

v'6c12,o}-2ip,1)-I0,2}) . (13.9.84)

13.10. PASCllEN-BACK EFFECT 277

\Ve now have to compute the terms H'l\.1+), H'ltbo) , and H'lv-) which do not contain terms 111 , m) with n + m = '2. T hen we have (arter omitting the terms with n + rn = 2)

>.(ri. mlH'l} = -i ).! [.J'i(on 30m 1 - O,, 10m 3) 2i(20n 20m., - On o6rn o)](t:J.9.85) v6 , , , , . - , ,

and

so that

(>./1)2 [ '}. '}. 16 4 ] -- ---+--+--+--16/i...; (3-5) (:l-!i :J-5 3-1

t>. 2 Ji =

Simila rly we find

F(2) - >.2/i 2,0 - --::;-

(I 3 .9.86)

( 1:U).8/)

(1:\.9.88)

Yes. t his prohlcrn c:u1 be .... ohed exactly since the total Hamiltonian is quadratic in thr. annihilation and creation oper:1tors.

13.10 Paschen-Back Effect Cak ulat 111ap,11rtic field 10 point i11 t lic> :-direct ion. In that cas ;,-(11, /l(l,. + 1S. l11. I} ,,,111c 1 r "''"r

( t:U0.90)

~ . v vnAYJ,t:;H. 13. DEGENERATE PERTURBATION THEORY

we have the weak field or Zeeman effect. In that case it is convenient to choose states of good j, mj and this situation is solved in general in problem 17 .10.

On the other hand if l 1 dV - eB (n,Ll--.,--d LSJn,l)< -2 -(n,llL: +2S:ln,l) (13.10.91) 2mc~ r r me

we have the Paschen-Back effect. In this case we want states of good m1 , m,. For this we use the results of problem 17.7 where we find

13/2, 3/2) = 11, 1)11/2, 1/2)

13/2, 1/2) = /i11, 0)11/2, 1/2) + ~11 , 1)11/2, -1/2) 13/2, - 1/2) = 1 A y13ll, -1)11/2, 1/2) + 311, 0)11/2, -1/2) 13/2, -3/2) = 11. -1)11/2, -1/2)

ll /2, 1/2) = ~11. 0)11/2, 1/2)- /i11. 1)11/2, -1/2) ll/2, -1/2) = A I 11. -1)11/2, 1/2) - .fill, O}ll/2, -1/2) .

T he inverse of these equations is

11. 1)11/2, l/2} = 13/2, 3/2) 11, O)ll/2, 1/2) = A 1 13/2, 1/2) + ./311/2, 1/2}

ll, 1)11/2, -1/2) = ~13/2, 1/2)- /ill/2, l/2) 11. -1)11/2, 1/2) = A 1 13/2, -1/2) + J31I/2, -1/2) II. O)II/2,-1/2) = 1 A v'313/2, -1/2)- 311/2, -1/2)

ll, -1)11/2, -1/2) = 13/2, -3/2) . For l = 0 the only possibilities are

IO, 0)11/2, m,) = 11/2, mj) IO, O}ll/2, !/2) = 11/2, 1/2)

IO,O)ll/2,-1/2} = ll/2,-1/2}. Thus, we can evaluate the magnetic part of the energy. If we call

h2(n,Ll-l-~ dVln,l) 2mc2 r dr = Cn,I eB 2mc = w

(13.10 .92)

(13.l0.9:l)

( J 3.10.94)

(13.10.95)

PASCHEN-BACK EFFECT

the result is

ll, m1)ll/2, m,) 11, 1)11/2, 1/2) 11. O)ll/2, 1/2)

11 . l )l l /2, -1/2) 11. -1)11/2, 1/2} ll , O)ll/2, -1/2)

11 , - 1)11/2, -1/2) IO, O)ll/2, 1/2)

IO, O)l 1/2, -1/2)

E Cn,l + 2fiw Cn,l + fiw 0 0 Cn,l - fiw Cn,l - 2/iw Cn,l + fiw

~ov

J3.11. If ATOM: WEAK FIELD STARK EFFECT 281

This shows that the term with the+ sign corresponds to the case m1 = 0, m, = 1/2 and the term with the - sign corresponds to m1 = 1, m, = -1/2.

Similarly, for mi= - 1/2 we find the eigenvalues

~ ( hw + lnt/2 J h2w 2 + 3

282 CHAPTER 13. DEGENERATE PERTURB. \TION THEORY

The energy levels in the unperturbed hydrogen atom are labc>lled by n, j. Under the perturbation the orbital angular momentum no longer commutes with the Hamiltonian and thus, I is no longer a good quantum number However , the z-component of the total angular momentum J 2 , namely m, continues to be a good quantum number. Thus, the matrix P)ements of the perturbation Hamilt onian V between stales with different m vanish. The diagonal matrix elements of \' also vanish

( '11n,l,j=I+1/2.m VIit n,1,j=I+ 1/2,m) ( IJT n,/,j=l-1/2.m V\Jfn l,j=l-1/2,m)

(~~1.VlJit)=O (t,12. V1P2) = 0 (13.11.l IL)

This means we must diagonalize V in I he degenerate subspace corre;ponding to the states 11'n,l=i+t/2,m and Wn l=j-1/2 m We call these matrix element::.

Vi:i = \'21 = (11'n.J-l/2,m \111' n.j+l/2,m) ef lc:-:i r 3 Rn (r)R .,(r) dr

2JjU + l) O ,J-1/2 n,J+I/. x [JU+ m)(; - m + 1) j Y/_ 112,m-t/2YJ+1/2,m-1/2 cosOdfl -JU- m)(j + m + 1) j Y/_ 1, 2,m+i/2Yj-1/2,m+1/2 cosOdn]

(13.11.113)

To perform th

BIBLIOGRAPHY 283

Combi ning these results we have

3 nJn2 - (j + 1/2)2m f' Vi2 =Vii= -4 j(j + 1 e ( 13.11.117)

The perurba.tion matrix to be diagonalized is now

v = ( ~1 Vi2 ) 0 . (13.11.118) The resulting energy shifts a.re

_ / _ 3nJn2

-(j+l/2)2m ~E-\12-4 j(j+l e. (13.11.119)

Finally we see that for a fixed value of n all tnms of the fine structure, except the term with j = n - 1/2. are split into 2j + 1 equidistant levels corresponding to m = - j,. . , j The term with j = n - 1 /2 is not split at all since I ha:; a fixed value I = j - 1/2 = n - l and is not degenerate as regards the quantum number I.

Bibliography [13.1] A.Z. Capri, .Vonrelatmstic Quantum Jfechamcs 3rd edition, World Sci-

entific Publishing Co. Pte. Ltd., section 6.8, (2002) .