Exercises in Quantum Mechanics: A Collection of Illustrative Problems and Their Solutions [2 ed.] 9401051720, 9789401051729

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Exercises in Quantum Mechanics

Kluwer Texts in the Mathematical Sciences VOLUME 6

A Graduate-Level Book Series

The titles published in this series are listed at the end of this volume.

Exercises in

Quantum Mechanics A Collection of Illustrative Problems and Their Solutions Second Revised Edition by

Harry Mavromatis Department of Physics, King Fahd University of Petroleum and Minerals, Dhahran, Saudi Arabia

Springer-Science+Business Media, B.V.

Library ofCongress Cataloging-in-Publication Data

"CIP-data available from the Publisher on request."

ISBN 978-94-010-5172-9

ISBN 978-94-011-2652-6 (eBook)

DOI 10.1007/978-94-011-2652-6

Printed on acid-free paper

AlI Rights Reserved © 1992 Springer Science+Business Media Dordrecht

Originally published by Kluwer Academic Publishers in 1992 Softcover reprint ofthe hardcover lst edition 1992

No part ofthe material protected by this copyright notice may be reproduced or uti1ized in any form or by any means, electronic or mechanical, including photocopying, recording or by any information storage and retrieval system, without written permission from the copyright owner.

To:

Vasso

and To:

Anthony and Blanche Tn f.p.a un f.uTiv

Table of Contents Preface Preface to Second Edition Schematic Illustration Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Index

10 11 12 13 14 15 16 17

IX Xl Xlll

Wilson-Sommerfeld Quantization Condition The Delta Function, Completeness, and Closure Momentum Space Wavepackets and the Uncertainty Principle Uncertainty Principle and Ground-State Energies of Quantum-Mechanical Systems Free Particles Incident on Potentials, Time Delay, Phase Shifts, and the Born Approximation Heisenberg Representation Two-, Three-, and N-, Versus One-Dimensional Problems 'Kramers' Type Expressions, The Virial Theorem, and Generalizations Upper Bounds and Parity Considerations Perturbation Theory Degeneracy The Inverse Problem The Dalgarno-Lewis Technique Angular Momentum and Coupled States Tensor Operators, and Evaluation of Matrix Elements Applications of Quantum Mechanics

1 16 34 47 68 74 102 111 124 133 157 186 214 227 262 291 312 326

Preface This monograph is written within the framework of the quantum mechanical paradigm. It is modest in scope in that it is restricted to some observations and solved illustrative problems not readily available in any of the many standard (and several excellent) texts or books with solved problems that have been written on this subject. Additionally a few more or less standard problems are included for continuity and purposes of comparison. The hope is that the points made and problems solved will give the student some additional insights and a better grasp of this fascinating but mathematically somewhat involved branch of physics. The hundred and fourteen problems discussed have intentionally been chosen to involve a minimum of technical complexity while still illustrating the consequences of the quantum-mechanical formalism. Concerning notation, useful expressions are displayed in rectangular boxes while calculational details which one may wish to skip are included in square brackets. Beirut June, 1985

HARRY A. MAVROMATIS

IX

Preface to Second Edition More than five years have passed since I prepared the first edition of this monograph. The present revised edition is more attractive in layout than its predecessor, and most, if not all of the errors in the original edition (many of which were kindly pointed out by reviewers, colleagues, and students) have now been corrected. Additionally the material in the original fourteen chapters has been extended with significant additions to Chapters 8, 13, and 14. Three new chapters, Chapters 15-17 were added in order to make this volume more comprehensive and more complete in its coverage of Elementary Quantum Mechanics, principally by including material on angular momentum coupling and tensor algebra. As a result of these additions and the revisions of the original fourteen chapters, the present monograph includes 228, as opposed to the 114 solved exercises in the first edition. The author would like to acknowledge the support of the King Fahd University of Petroleum and Minerals in this project.

Harry A. Mavromatis

Dhahran October, 1991

Xl

Schematic illustration of various approaches to calculating Energy Levels of Quantum Mechanical Systems

Energy Levels: Approach for High-lying States: Generally useful approaches:

1) Wilson-Sommerfeld Quantization Condition (Chapter 1)

1) Schrodinger equation in Momentum Space (Chapter 3)

2) Poles of Scattering Amplitude (Chapter 6)

3) Schrodinger equation in Coordinate Space (Chapters 8, 9, 12, 16)

4) Perturbation Theory (Chapter 11)

Approaches for Ground State: 1) Uncertainty Principle (Chapter 5)

5) Dalgarno-Lewis Technique (Chapter 14)

2) Variational Technique (Chapter 10)

Xlll

CHAPTER 1

Wilson-Sommerfeld Quantization Condition The hydrogen atom, when treated using Bohr's admixture of classical and quantum concepts involves an electron circulating about a proton (subject to the attractive Coulomb force - (e 2 147rfor2) r) in orbits which satisfy the condition: 27rr = nADe Broglie,

n = 1, 2, 3 ....

Since ADe Broglie = hlp this reduces to p27rr the Wilson-Sommerfeld quantization condition:

f

pdq

= nh

n

= nh, which may be generalized to

= 1, 2, 3...

(1.1)

,

where § implies a complete cycle, and p and q are conjugate variables. 'By construction' Eq. (1.1) gives the correct quantized energies for the hydrogen atom. But it also gives the correct energy spectrum for a particle in a box with infinite walls:

V(x) = {

0, 00,

0< x < a, x < 0, x> a.

EXAMPLE 1.1 Find the energy levels for a particle in a box with infinite walls:

V(x) = {

0, 00,

0< x < a, x < 0, x> a.

In the region 0 < x < a, E = p2/2m. Hence Eq. (1.1) in this case becomes

f .J2mEdx = nh where a cycle involves x varying from 0 --7 a and from a Integrating one obtains: 2a.J2mE = nh

or

--7

O.

n 2 h2 E = - 82 ' n = 1, 2, 3 .... ma

(1.2)

Chapter 1

2

By contrast the quantum-mechanical treatment of this problem involves solving the Schrodinger equation: "liZ dZ 2m dxz

[-

+ V(x) ] 'I/;(x)

(1.3)

= E'I/;(x),

V( xl

~----

______

o

~

____________ x

a

Figure 1.1: Potential in Example 1.1. for 0 < x < a with boundary conditions '1/;(0), 'I/;(a) = 0, 'I/;(x) being zero for x < 0, x> a. The properly normalized eigenfunctions of Eq. (1.3) which satisfy these boundary conditions are:

(:2

'I/;(x)=V~

sinkx

with

Thus '1/;(0) = and

'I/;(a) = while

If

sinka = 0

ka=mr, where

If if

k

"li 2 2

2m =E.

sinO = 0,

ka = mr,

n = 1, 2, 3... ,

Wilson-Sommerfeld Quantization Condition

3

by construction. Since

this implies

exactly the result Eq. (1.2). One can gain a little more insight as to the range of applicability of the WilsonSommerfeld quantization condition by studying slightly more complicated systems. EXAMPLE 1.2 Find the energy levels of a particle in the potential: 0,

V(x)=

1

Vo, a a + b.

Vo which corresponds to the interesting case.)

V( x)

Vo

~~----------~--------------~--------x

a

o+b

Figure 1.2: Potential in Example 1.2.

Chapter 1

4

The Wilson-Sommerfeld quantization condition can be immediately applied to this case: a+b fa 2 10 PI dx + 2 a P2 dx = nh

l

where

PI

= hkl = J2mE,

P2

= hk2 = V2m(E - Vo)

I.e.

( 1.4)

On the other hand, solving the Schrodinger equation in the regions 0 < x < a and a < x < a + b ('IjJ(x) being zero for x:::; 0, x ~ a + b) yields

0 < x < a,

'IjJ(x) = AsinkIx, 'IjJ(x)=Bsink2(x-a-b),

a Vo that is if the total energy is large compared to the potential energy does Eq. (1.5) reduce to Eq. (1.4) since then tan kla rv - tan k2b which is satisfied if kla + k2b = mr, n = 1, 2, 3 ....

A second example which illustrates the range of applicability of the WilsonSommerfeld quantization condition is the finite square-well problem. EXAMPLE 1.3 Find the energy levels for a particle in a square well

vwhere E


a,

-IVoI, Ixl < a,

Wilson-Sommerfeld Quantization Condition

5

The kinetic energy p2

T = -

2m

=

IVoI - lEI

for

Ixl < a.

The Wilson-Sommerfeld quantization condition therefore yields in this case

l.e.

(1.6)

v (x)

o

-0

----------------.---------r--------.---------------- - - - - - - - - - - - - - - - 1 - - - - - -lEI

x

T

L---_ _

+--_~ t -IVol

Figure 1.3: Potential in Example 1.3. On the other hand, solving the Schrodinger equation in the regions a yields respectively

Ix I >

1/Je(x) = Acoskx where

Ixl
a,

_J2m EI l

k=

K-

~'

or

1/Jo(x)

=

Asinkx

(where the subscripts e,

x

---+

-x.)

0

Ixl < a, 1/Jo(x) = ±Bexp {-Klxl}

x~

± a,

stand for even and odd solutions under the interchange

Chapter 1

6

The continuity of 'l/J'f,(x) and d'l/J'f,(x)/dx at x = ± a (independently for the odd and even solutions) implies: tan ka = and cot ka =

K

(even solution),

k

( 1.7) k

(odd solution).

As in the previous example, Eq. (1.6) and Eq. (1.7) are not identical. Only if the total energy is large compared to the potential energy, i.e. IVoI » lEI (since these quantities are negative) i.e. K/k -+ 0 do they agree, since then tanka =

0

cot ka =

0

and or ka =

n7r

2'

i.e. ka = {

7r,27r ... 7r /2, 37r /2 ... ,

n = 1, 2, 3... ,

yielding

I.e.

which is identical to Eq. (1.6). Though the Wilson-Sommerfeld quantization condition was superseded by Quantum Mechanics (with the Schrodinger and Heisenberg formulations in the early twenties), as a calculational aid it has the advantage over the Schrodinger equation that it is easier to work with since it involves an integral rather than a differential equation. However, as indicated in Examples 1.2-3, it generally gives results which are reasonably accurate (i.e. in agreement with Quantum Mechanics) only when the energy is large compared to the potential under consideration. If V(x) = Alxl P one can obtain the form of the energy sequence according to the Wilson-Sommerfeld quantization condition Eq. (1.1) as follows:

can be written (for En

> 0)

as

Wilson-Sommerfeld Quantization Condition

7

Hence

or generally:

lEn I = n 2p /(p+2)

where

AII/Ph { I

n = 1, 2, ... (p

V2m I(p)

f -Iul du I(p) = f Vlul du I(p) =

and

}2P/(P+2)

)1

(1.8)

if En > 0,

p

P -

> -2),

1

if

En


= n 2Aoh 2/32m = n 2h 2 /32m. This corresponds to a = 2 in Eq. (1.2), i.e. V = 0, Ixl < 1, V = 00, Ixl > 1.] In several cases I(p) can be easily evaluated directly. EXAMPLE 1.4 Find the energy levels for a particle in the well V(x) = Ax2,

-00


O. E - n 2/ 5 { n -

A2h

V2rri 1(1/2)

}2/5 ,

where 1.e.

En = n 2/ 5 (

A2h 15

V2rri 2m 32

)2/5 '

n = 1, 2, 3....

EXAMPLE 1.11 Find the energy levels for a particle in the well:

V(x) = {

Ax1/2, x> 0, 00,

One proceeds as in Example 1.10 except

x < O.

(1.17)

Wilson-Sommerfeld Quantization Condition

11

I.e.

n = 1, 2, 3 ....

(1.18)

EXAMPLE 1.12 Find the energy levels for a particle in the well:

V(x) = Here p = -1/2 and Thus

E

IAI -lxI 1/ 2

all x.

'

E < O.

- n- 2 / 3 {

I nl-

IAI-2h

}-2/3

.,fiiTi 1(-1/2)

n

,

= 1, 2, 3, ... ,

where 1.e.

En = _n- 2 / 3

IAI- 2h) -2/3 ( .,fiiTi 7r ,n =

1, 2, 3 ....

(1.19)

EXAMPLE 1.13 Find the energy levels for a particle in the well:

V(x) =

{

-IAlx-l/2, x> 0, x

00,

(E < 0)

< O.

One proceeds as in Example 1.12 except 1(

-~) = 2101 VU- 1/ 2 -

1.e.

_ -2/3 En - -n

Generally I(p)

=

=

f VI -Iul ±B (~ p

p

du =

(IAI-22h) -2/3 , .,fiiTi 2m

7r

4101 viI -

1 du =

i'

n = 1, 2, 3 ....

uP du =

~ 101(1 -

~) _ 2v1iT ( {I / p} + 1 ) 2' p - r( {1/p} + {3/2} ) .

W)1/2 w {1!P)-1

(1.20)

dw

( 1.21)

Chapter 1

12 EXAMPLE 1.14 Find the energy levels for a particle in each of the following wells:

Vi(x) = Ax 1 / 3 ,

V2(x) = Ax 1 / \

ltJ(x) = AX 1/ 5 ,

all x,

using Eq. (1.21). From Eq. (1.21): I(1/3) = 64/35, I(1/4) = 512/315, I(1/5) = 1024/693. Hence A3h 35 }2/7 { (1.22) En(P = 1/3) = n2/7...;2m ,

2m 64

E ( = 1/4) = n2/9 { A4h 315 }2 / 9 n P ...;2m 512 '

(1.23)

E (p = 1/5) = n2/11 { A5h 693 }2/11 n ...;2m 1024

(1.24)

The results of Example 1.4-14 are summarized in Table 1.1. For two-dimensional systems (or three-dimensional systems where a particle moves in a plane chosen for convenience to be the x-y plane), where the potential only depends on p,

P~

E = -2 m

f

Hence

P~ + -mp 22 + V(p).

P¢ de/> = n¢h,

which implies P¢

(1.25)

n¢ = 1, 2, 3 ... ,

(1.26)

= n¢n,

and fpp dp= nph, I.e.

f

np = 1, 2, 3 ... ,

2mE - (n¢h)2 - 2mV(p) dp = nph, p2

np = 1, 2, 3... .

(1.27)

(1.28)

The integral in Eq. (1.28) can be evaluated analytically for certain problems. EXAMPLE 1.15 Find the energy levels for a particle of mass m in the potential

Wilson-Sommerfeld Quantization Condition

13

TABLE 1.1

Energy levels for various potentials Vex)

p

Range

1

x ;:: 0

1

-00

-00

-00

1/2

k(X) involves continuous functions, where

an =

4>~(x)'IjJ(x)

dx, a(k) =

4>;;(x)'IjJ(x) dx.

(2.5)

Expanding the delta function 6(x - x') in terms of a complete set of discrete functions implies:

6(x - x') = I>n4>n(x) n

where i.e.

00

6(x - x') =

I: 4>~(x')4>n(x),

n=O

(2.6)

17

The Delta Function, Completeness, and Closure

and similarly expanding the delta function in terms of a complete set of continuous functions implies

(2.7) EXAMPLE 2.1 Suppose one uses as a complete discrete set the eigenfunctions of a particle in an infinite square-well potential (a box with infinite walls):

V(x)

={

0,

-a/2 .:::; x.:::; a/2,

00,

x

< -a/2, x > a/2,

for the expansion in Eq. (2.6), with the choice x' = 0. The normalized even subset of the above eigenfunctions (the rest, i.e. the odd subset is zero at x =:= 0, and does not contribute to the integral in Eq. (2.5) for an, and hence to the sum in Eq. (2.6) ) is " ~a2 cos (2n a+ 1) ",.x,

Hence a possible representation of the delta function is

h'(x) =

~

f

a n=O

=

0,

a

cos (2n + 1)n:x

--2 < -

a

x

x

a

'),

n,l,m

J8(r' - iJ)

where I.e.

roo 8(r -

Jo

r') dr

11

-1

dr= 1,

8( cos 0 - cos 0') d cos 0

If one multiplies both sides of Eq. (2.20) by over dO. dO.' one obtains

8(r - r')

~--:-2~ r

r

Jo

21f

6(4)-

4>') d4> =

~q(O, 4»~*q(O',

= ,,* L.J Rnq () r Rnq (') r , n

4>'),

l.

and integrates

Chapter 2

24

and for three-dimensional systems quite generally one can thus write for the delta function (for any choice of orbital quantum number 1):

6(r - r') =

00

E u:l(r')Unl(r),

(2.21)

n=O

where unl(r) is the solution, subject to the condition Unl(O) Schrodinger equation:

0, of the radial

The choice of central potential V (r) determines the detailed form of Unl (r). The corresponding one-dimensional expression is Eq. (2.6) where x extends over all space, and ¢n (x) is the solution of the one-dimensional Schrodinger equation:

1i2 cP { -2mdx 2

+ V(x) -

E

}

¢n(x) = 0, (-00 < x < 00),

with the choice of V (x) similarly determining the detailed form of ¢n (x). For continuum states Eq. (2.21) is replaced by (2.22) which is analogous to the one dimensional Eq. (2.7). Eqs. (2.6), (2.7), (2.21), (2.22) are illustrations of the 'closure' property of quantummechanical wave functions. EXAMPLE 2.7 Obtain an expression for 6(r - r') if V(r) is the three-dimensional harmonicoscillator potential: 1i 2r 2/2mb\ r 2:: 0,

V(r) = {

undefined,

r

< 0.

Inserting the detailed solutions Unl( r) for this potential (d. Eq. (8.13) ) in Eq. (2.21) one obtains:

( 2.23)

The Delta Function, Completeness, and Closure

25

for each permissible value of 1 (1 = 0, 1, 2... ), and where r, r' If I = 0 for example

> O.

( 2.24)

Substituting Eq. (2.12), the solutions of the one-dimensional harmonic oscillator, in Eq. (2.6), one has analogously: (2.25)

H2n+1(e) = (-nIt (2n

+ 1)!2e [1F1 ( -nj ~j e)] .

Substituting this expression into Eq. (2.24) and using Legendre's duplication formula 2

'r( n + 3/2) = (2n 2+2nI)! -..Ii +1'

n.

The Eq. (2.24) can be written in a form similar to Eq. (2.25), namely 00

Hn (rib) Hn (r'lb)

n=1, 3,5

(2.26)

e

The Eq. (2.26) involves a sum only over odd n terms, for which Hn(O is zero if = o. Thus 8(r - r') in Eq. (2.26) vanishes if either r or r' is zero. Also, an additional factor 2 arises in Eq. (2.26) as compared to Eq. (2.24) because x, x' extend from -00 to +00 whereas r, r' extend from 0 -+ 00. Using the Hille-Hardy formula3

+ 1)y} /2) ~ r(n ( xy)a/2 exP (-2({x + ~

r

a:

n=O

_ r

+ a:, + 1) 1 F1 (-n., a: + 1., x) 1 F1 (-n·, a: + 1., y) t n n.

a/2

- - exp -l-t one can sum Eq. (2.23). Here

Ia(x) = i-a Ja(ix) =

({I l+t}) I (2(x yt)1/2) - -(x + y)-2 1-t 1-t ' a

J ~:-1 7r

ja-1/2(ix).

Chapter 2

26

Thus Eq. (2.23) can be cast into the form:

8(r

_

1

_.

r ) - ~~

{2(rr/)1/2 r(21+1)/4 (_ [(r2 + r/2) ~]) ( 2rr't 1/ 2 )} b2 1 _ t exp 2b2 1 _ t 11+1/2 b2(1 - t) ,

where (r, r' > 0), which can be written in terms of spherical Bessel functions:

8(r -

1

r ) =

(2.27)

• {4rrl t- I/ 2 ([(r2 + r/2) 1 + t]) . (i2rrltl/2)} ~~ i l b3Vi (1 _ t)3/2 exp 2b2 1 _ t Jl b2(1 - t) .

If one makes the substitution

f

= (1 - t)b2 , this becomes:

(2.28)

or

For I = 0, since4

But as

f

-+

0,

2rr/) 1 (2rr/) sinh ( - f - -+ 2exp - f - , and Eq. (2.31) reduces to

1

1 {I

/ uC( r-r') = - 1.1m {2 -exp (r2+r/2)}eXP(2rrljf) = - 1.1m -exp {(r-f r )2}}

Vi b.

00,

The potential is suddenly changed to

Vex) = {

0,

Ixl < b,

00,

Ixl > b.

What is the probability the particle is in the ground state of this potential? The normalized ground state of this new potential is just

fl fl Vb cos kx = Vb cos 2b ' 7rX

where Hence

I.e.

ao = -8v15 3 - '" 0.9993, 7r

thus

laol 2 =

0.9986 '" 1,

as it should be since the wave function forms are very similar (d. Example 13.1, Figure 13.1), as too their energies:

1

Chapter 2

30 EXAMPLE 2.11 Consider a particle bound by the potential:

-Ivai

c5(x).

The potential suddenly changes to:

={

V(x)

0,

Ixl < b,

00,

Ixl > b.

Find the probability the particle will be in one of the even-parity states of this new potential. Here (cf. Eq. (13.7)

if;o( x ) --

JlVolm h2

exp

(_IValmlxl) h2 ,

and one wishes to evaluate the the overlap: an

=

l

=

2

b -b

J'Valm

-,;,z-

exp

(IValmlxl) {l h2 Vb

(IValm)1/2 [be (_IValmx) Jo xP h2 h2b

cos

cos

(2n + 1)7rX d 2b x

(2n+1)1rx d 2b x.

( 2.35)

Evaluating this integral one obtains: an = 2

(IVaJm) 1/2 [ IValm/h2 + (_l}n [ (2~ + 1}1r/2b] exp (-IValmb/h 2) ] hb [ IValm/1i2] + [ (2n + 1)1r/2b ]2

As b --+ 00, an --+ o. There is no probability the particle will be in one of the odd-parity states of this potential, namely:

if;m(x) =

(l sm . (m1rx) Vb -b- .

EXAMPLE 2.12 A particle is originally in the ground state of the well

V(x) =

00,

x < 0, x > b,

0,

0

{

< x < b.

The Delta Function, Completeness, and Closure

31

Suddenly the wall at x = b is shifted to x = 2b. Find the probability the particle will be in the ground state of the new potential which results from this shift. The wave function is originally:

4>o(x) =

{

{iib

sin (7fX/b),

0,

°< x

x < b,

< 0, x> b,

and one is interested in the overlap:

rb f2 . 7rX {1. 7rX ao = Jo V"b sm T V"b sm 2b

d

x

Thus The original energy is

E =

n, 2 7r 2

/2mb 2 •

The energy of the new ground state is n,2 7r 2 /8mb 2 while that of the new first-excited state is n, 2 7r 2 /2mb 2 , which is just the original energy. There is therefore a 32/97r 2 probability the energy will be less than before. The probability the energy will be unchanged is related to the overlap:

rb f2 . 7rX {1 . 7rX d 1 Jo V"bsmTV"bslllT x = V2. Thus the probability the energy is unchanged is 1/2. The probability the new energy is more than the original energy is 1-

32 97r2

(

1)

+ 2" .

One can also use complete set expansions in related problems. For instance: EXAMPLE 2.13 Given a particle with wave function:

1jJ(x) =

{

J3/2c 0,

(I-lxi/c),

Ixl < c, Ixl > c.

Chapter 2

32 It is placed in a well

Vex) = {

0,

Ixl < c,

00,

Ixl > c.

Find the probability the particle will be in the various states of this well. As can be easily verified, '¢( x) is properly normalized. However it has a discontinuous derivative at x = 0, implying a potential which is not even piecewise continuous at that point. The wave function '¢( x) can be expanded in terms of the complete set of eigenfunctions of the well it is now placed in i.e.

All the bn 's are zero since '¢( x) is an even function, whereas the 1

. (

;::;Slll

yC

n

+ 1) -7I"X c

,

s

are odd functions. Meanwhile an =

Thus

laol 2 =

l 1 ( 21) -,¢(x)dx= c

-c

;::;cos n+-

7I"X

yC

C

(n

v'6

+ 1/2)

2

71"2

0.986 etc.

EXAMPLE 2.14 Verify the sum of the probabilities

la n l2

in Example 2.13 is unity.

From Example 2.13: 6

2

96

lanl = (n + 1/2)471"4 = (2n + 1)471"4 But (d. Eq. (17.18) ), 00

~

1

71" (2n + 1)4 = 96 .

Hence, as expected

The Eq. (4.22) of Chapter 4 lists some other representations of the delta function. These can be compared to Eqs. (2.9), (2.13), (2.15), (2.17), (2.19), (2.23), (2.25), and (2.30), which were discussed in this chapter.

The Delta Function, Completeness, and Closure

33

References

1. Erdelyi et al., Higher Transcendental Functions, McGraw-Hill (1953), v. 2, p. 194. 2. Op. cit., v. 1, p. 5. 3. Op. cit., v. 2, p. 189, v. 3, p. 272. 4. Op. cit., v. 2, p. 79. 5. Op. cit., v. 2, p. 193. 6. Op. cit., v. 2, p. 195. 7. 1. B. W. Jolley, Summation of Series, Dover (1961), p. 16. 8. Op. cit., p. 32. 9. E. Merzhacher, Quantum Mechanics, Wiley (1970), p. 60.

CHAPTER 3

Momentum Space

Working in momentum space involves taking the Fourier transform of the eigenfunction 'Ij.;(x, t) of the Schrodinger equation. Thus if:

1 *(p', t) exp (_ i~x)dP' n(x)'1f;(x,t) dx =

1~~~: ¢>*(p', t) exp ( - iP~X)

It

dp' 1:~: exp

1t,21"'=00

2~1t

"'=-00

C~X)n ( -~ ;p) ¢>(p, t)dpdx (3.8)

1:~:1~~~: ¢>*(p',t)n(-~ ~) ¢>(p,t) [2~ 1:~: exp C(p~p')x) d;] X

dp'dp:;;1 = It

IP=oo ¢>*(p, t) n (--;--8 h 8) ¢>(p, t) Tdp p=-oo

Z

P

It

.

36

Chapter 3

One also notes that if 'IjJ(x, t) is normalized ¢(p, t) is also normalized since, using Eq. (3.8) with n = 1 :

dp 100 'IjJ*{x, t)'IjJ{x, t)dx = 100 -00 ¢*{p, t)¢{p, t)r; . -00

The above formalism generalizes to three dimensions by replacing

n abyn-;-'\7 --a i x z

etc.

Six simple problems follow which illustrate the usefulness of the momentum representation and the fact that for certain potentials it is easier to work in momentum, rather than coordinate space. EXAMPLE 3.1 Consider the case of a free particle (V{ x) = 0). In momentum space the Schrodinger equation, (Eq. (3.7) ) for this system is:

(3.9) I.e.

¢(p) = 8 (~_

Since generally

'IjJ{x) =

vk J:

v'2;:E) .

¢(p) exp

C~X)d (~) ,

if one suppresses the time variable in Eq. (3.2), in this case

'IjJ(x)

= v'211i" exp (.v'2mE z n x) '

as it should be. EXAMPLE 3.2 Consider a particle in the momentum-dependent potential

V{x) =

a~~ = z ax

ap.

In momentum space the Schrodinger equation (Eq. (3.7) ) for this potential is:

{:: + ap -

E} ¢(p) = 0,

(3.10)

Momentum Space

37

i.e.

0, (A> 0), 00,

x < 0,

(illustrated in Figure 3.1), with the help of the momentum representation.

v (x)

~----------------------~~x

Figure 3.1: Potential in Example 3.3. In momentum space the Schrodinger equation (Eq. (3.7) ) for this potential is: 11, d } { -p2 - A -;-- E 0),

(3.11)

Chapter 3

38

Integrating this expression yields: i {p3 } { 0 in Examples 3.3 and 3.4 (Figures 3.1 and 3.2). One can proceed here using methods similar to those used in Examples 3.3 and 3.4. However, instead one can compare (cf. Eqs. (1.12) and (1.10) ) the solutions of the problem of the standard harmonic oscillator potential restricted to x > 0 i.e. V (x) = AX2, 0:::; x < 00, V(x) = 00, x < 0, to the oscillator extending over all space i.e. V(x) = Ax2, -00 < x < 00. The former potential's allowed energies are

En = (2n - ~) htf., n=

1, 2, 3 ... ,

and correspond to odd-parity solutions (see Eq. (10.12) ), while the latter's allowed energies are

En = (n - ~) htf. ' n=

1, 2, 3 ... ,

and both odd and even parity solutions are allowed. One can thus obtain the solutions in the latter case from the solutions in the former by letting n -+ n/2. Similarly if one considers the infinite well V(x) = 00, x < 0, x > a, V(x) = 0, 0 < x < a, its solutions are the odd-parity wave functions tjJ( x) = )2/ a sin kx (where ka = mr) i.e.

44

Chapter 3

whereas if one considers the infinite well V(x) = 00, x < -a, x > a, V(x) = 0, -a < x < a, its solutions are tf;{x) = V1/asinkx (odd parity), and tf;{x) =

V1/a cos kx (even parity), where ka

= mr or (n -1/2)11",

2 E = h k 2 = ~ (~)2 11"2, 2 2m

2m

= 1, 2, 3... , i.e.

n = 1, 2, 3... ,

a

2

n

Again the solutions for the latter case can be obtained from the former by letting

n

-+

n/2.

A third example which illustrates the fact that this procedure may be applied generally is the potential weIll

{

Vex) =

-110 cosh- 2 ax, x> 0, x < O.

00,

The (odd-parity) solutions for this case are:

tf;(X)= ( 1-e )

(s-2n+1)/2

2Fl(1-2n,2s-2n+2;s-2n+2;{1-0/2),

with corresponding energies 1 + 8mVo a h 2 2

]2 , n = 1, 2, 3... ,

while the symmetric potential

V(x)

= -Vo cosh- 2 ax,

(all x),

has solutions

tf;(x)= ( 1-e )

(s-n+1)/2

2

Fl(1-n,2s-n+2js-n+2j{1-0/2),

where 1 (with the symbols s, and

~

s -2 =

+

8mVo a2 h2

n = 1, 2, 3 ... ,

edefined as

(-1 +

8mVo )

1 + a2 h2

,

e== tanh ax

).

Thus in this case as well one obtains the solutions for the symmetric well by letting n -+ n/2 in the solutions for the well which extends only from 0 < x < 00.

Momentum Space

45

The solutions for Example 3 are

(n - D2/3 (A~2r/3 (2~r/3

En =

(2n - ~r/3 (A~2) 1/3 (4~) 2/3, n= 1, 2, 3....

=

These correspond only to the odd-parity solutions of Example 3.5. By analogy with the above three problems one thus expects that the odd- and even-parity solutions for Example 3.5 have energies: (3.31 )

an expression which for large n, n rv (n - 1/2) agrees with the Wilson-Sommerfeld result (Eq. (1.13) ). Similarly the solution for Example 3.4 is

and one expects that the energy in Example 3.6 is just n = 1, 2, 3... ,

(3.32)

and that this includes both even- and odd-parity solutions. This result agrees precisely with the Wilson-Sommerfeld result (Eq. (1.15) ). EXAMPLE 3.7 Obtain the ground-state wave function in coordinate space for the Hamiltonian in Example 3.4 starting with the momentum-space wave function Eq. (3.24). From Eq. (3.24) one notes that the ground-state function (n = 1) is: '" ( ) =

'l'IP

(2h

(.l )

1/2 exp (-2i arctan {flIP})

7r1-'1

1 + (fllp)2

,

Using Eq. (3.2) and suppressing the time variable:

~1 (x)

1 Ii = v'2i

1

00

-00

¢1(P) exp

(ipx) T d (P) 1i,

(3.33)

Chapter 3

46 Substituting Eq. (3.33) in this equation

1/J ( ) - (2/i 1 X

-

(3 )1/2

-:; 1

With the change of variable

1 V2i n 1

00

-00

u

exp (ipx//i - 2 arctan {(31P}) d 1 (filP)2 p.

+

== arctan /3IP this becomes

1/Jl (x) = ~ ((3~/i f/2

1:/:

exp

{i (/i~1

tan u - 2U) } du .

Writing the exponential in terms of sines and cosines, only the cosine term survives because of parity considerations when one integrates from -1r /2 to 1r /2, i. e.

2 ( 1 ) 1/Jl(X) = -; (31/i

1/2 1:/2/2 cos (/i~1 tan u - 2u) du.

Evaluating this expression3 one obtains finally:

1/Jl(X)

= 2 ((3~/ir/2 xexp (-/i~J,

(3.34)

a wave function which one can easily show to be normalized. With this expression one can for example calculate:

(x)

=

(X2) =

(1)3 10roo exp (2X) 3 3(31/i 3/i2 - /i(31 x dx = -2- = 2mlAI '

4 (31/i 4

(f3~/ir 1 exp (00

:;J

X4 dx = 3 {f31/i}2 =

m~~~12 '

results consistent with Eqs. (3.27), (3.28) (in the case n = 1).

References 1. 1. Landau et I. Lifchitz, Mechanique Quantique, Mir (1966), p. 94.

2. I. S. Gradshteyn and I. M. Ryzhik, Tables of Integrals, Series and Products, Fourth (English) Edition, prepared by A. Jeffrey, Academic Press, New York (1980), p. 292. 3. Op. cit. p. 405.

CHAPTER 4

Wavepackets and the Uncertainty Principle The properly normalized free-particle wave function is: 1 exp {.Z (PX Et)} J2i h - t;

'ljJp(x, t) =

(4.1)

(cf. Eq. (2.18) ). One problem with this function is that it has no spatial localization, i.e. though the momentum is precisely known (in other words tlp = 0), tlx = 00, where tlA is rigorously the square root of the variance of A (see Eq. (3.26) ), or more loosely speaking represents the uncertainty in A. This shortcoming may be easily removed by constructing a wavepacket

where A(p') is a function concentrated about p' (cf. Eqs. (2.1), (2.2) ):

= p. Thus if A(p') is a delta function,

A(p') = 8 (pI ; p) Eq. (4.2) reduces to Eq. (4.1). The function

A( p') exp {iE(pl)t ---'h- } is in fact the wave function in momentum space (d. Eq. (3.2) ). Any detailed functional form A(p') gives the same general properties to w(x, t), namely localizes the particle, (i.e. ensures that tlx =J 00). EXAMPLE 4.1 Suppose

A(p') =

.fi

exp ( _Ip'

~ pI)

,

(4.3)

48

Chapter 4

a normalized function peaked about rI = p, and illustrated in Figure 4.1. At first sight it appears this gives an uncertainty in momentum

The gain however is that W(x, t) is now localized. Thus

W(x,O)

= ~ 28 7f

W(x,O)

=

or

A ( pi) =

ft.

pl =oo lp'_oo (Ip' - pi +T iP'X)

837f exp

J~

exp-

8

Ct:PX ) x + h /8

(d{) ,

1

2

2

(4.4)

2 •

e -I p' - p I/6'

____~--~~--~-+--~~~~------~~pl p-26' p-6' p p+6' p+26'

Figure 4.1: The momentum distribution in Example 4.1.

[ The normalization of W(x,O) can be confirmed by noting roo

Loo (x 2

dx

+ 1)

2

=

~ 2

.]

The amplitude of the function given by Eq. (4.4) (ignoring the phase) is drawn in Figure 4.2. There is a spread about x = 0 given approximately by tJ..x = 2h/8. Thus tJ..x tJ..p '" 2h/8 28 = 4h. A more accurate estimate of tJ..x tJ..p is possible since

49

Wavepackets and the Uncertainty Principle

A02

~

=

(02) _ (0}2

h {O} = , were -

J \lI*O\ll dr J \lI*\lI dr

=

J A*(p)OA(p)dp J A*(p )A(p )dp

.

Using the function of Eq. (4.4), {x} = 0, while in this case:

.6.x2={x2)=

21i31x=oo 63 7r "'=-00

x 2 dx

(x2+n2j6 2)2

1i 2 = 62

•

Figure 4.2: Amplitude of \lI(x,O) in Example 4.1. Similarly

!!.l

P '=oo, (_ 21p' ( )= P fJ p'=-oo P exp fJ

pI)

(dP') = fi P,

while 62 .

A (pi) = 0

=

Ipl_pl

>s

J ~~3 (6-l pl_pl >

Ipl_pl(x)=

1 tn=

v21r

1 A(k')exp{i(k'-k)x}dk' 00

(4.16)

-00

where according to Eq. (4.9)

A(k') =

.~

v21r

1

00

\J!(x', 0) exp (-ik'x')dx' =

-00

Hence

A(k + k')

=

1 tn=

v21r

1 ¢>(x') 00

exp (-ix'k')dx'

-00

~

v21r

=

1

00

¢>(x') exp {-ix'(k' - k)}dx'.

-00

1 tn=

v21r

1 ¢>(x') 00

exp (ix'k')dx',

-00

if ¢>(x) = ¢>(-x), or

A(k'+k)=

.~1°O ¢>(x'-x)exp{i(x'-x)k'}dx'.

v21r

(4.17)

-00

Comparing Eqs. (4.16) and (4.17) one notes that if one identifies A(k') with a particular (even) function ¢>( x' - x) one can then identify ¢>( x) with the corresponding

A(k + k').

Wavepackets and the Uncertainty Principle

55

In Example 4.2 for instance,

4>(x)

=~

[38 [sin (8X/21i)]

2Y;r;,

2

and

8x/21i

A(k+k')

! [8-Jk'J1i].

= J23

v

Hence if one identifies A( k') with

~ [38{sin{(k'- k)8/21i}}2

2y;r;,

(k' - k)8/21i

'

one obtains the corresponding

directly from A(k' + k). Letting 8 __ 2Ti2 / 8, if

A(p')

=J

3Ti {sin {(p' - P)/8}}2 , (p' - p)/8

27r8

then

~4YT {3i [2 _ ~] Ti·

\If(x 0) = exp (ipx) , Ti EXAMPLE 4.4 Consider the case A(')

p

e3

= J2---;-Ti

1

(4.19)

(p' --p""-')2"'-+-e-=-2 .

For this distribution of momenta:

On the other hand the wave function at time t = 0, i.e.

\If(x, 0) =

~JeTi 7r

Je Ti

1

00

-00

exp

exp (ip'x/Ti) dp' (p'-p)2+e 2 Ti

(iPX) .!. Ti 7r

1 exp(p'{i(p'- p)2- p)x/Ti} d( , _ )/Ti +e p p 00

-00

(4.18)

2

2.Je 3 Ti ex (ipx) roo cos (p' - p)x/Ti d( , _ )/Ti 7rTi 2 P Ti Jo ({p'_p}/Ti)2+(UTi)2 P P .

Chapter 4

56

. Hence

W(x,O)

2 W (ipx) roo cosux du ([ = ;Ytl exp T Jo u2 + (e/n)2 = Y"h

exp

(iPX)

T

exp

(elxl)

-r;- . (4.20)

For this wave function

=

(x)

0,

I.e.

~x~p=

n

V2'

The form of these distributions is given in Figures 4.1 and 4.2 with the appropriate identifications. The symmetry between the A(p') and W(x, 0) in this example and the W(x,O) and A(p') in Example 4.1 can be understood in the light of the remarks in Example 4.3. What is of some interest is that one can extract a representation of the delta function from this example, namely:

6(u) =

.!.lim~ 'K ......

1

Ot (u/t)2+ 1

.

(4.21 )

Thus if one defines

one notes that as

e

--+

0,

A'(p')

--+

6 (p, ; p) .

This follows from the observation that when W(x,O) is evaluated using A'(p) rather than A(p), then

and as

e

--+

W(x 0) = _1 exp (iPx) exp , y"h n

(_~) n

0 this expression for W(x, 0) becomes just a plane wave

1

y"hexp

(ipx)

T .

This simplification occurs whenever a delta function 6( {p' - p} In) is substituted for A(p') in Eq. (4.2), since then W(x, t) in this equation reverts to a plane-wave expression. Hence the A(p') above is a particular delta-function representation in the limit

Wavepackets and the Uncertainty Principle

57

e

- t O. The procedure thus involves essentially determining the right overall factor by which to multiply A(p') such that in some limit the resulting w(x, 0) reduces to a plane wave. The A(p') multiplied by this factor is then a delta function in that limit. By a similar analysis one can extract several other delta-function representations from the A's used in this Chapter's examples. Some of these are listed in Eq. (4.22):

(i)

8(u)

= (l/2)limf..... o (l/f) exp (-lui/f)

(Ex. 4.1)

(ii)

8(u)

= (1/4)limf..... o (l/f)(l + \ul/f)exp(-Iul/f)

(Ex. 4.6)

(iii)

8(u) = (1/'rr) limf-+ OO f [sin (W)/W]2

(Ex. 4.3)

(iv)

8( u) = (4/37r) lim. ..... oo f [sin (w)/ W]3

(Ex. 4.5)

(4.22) lui < f

(vi)

8(u) = lim.-+o (1 -Iul/f)/f,

(vii)

8(u) = (3/2) limf-+o (l/f) (1 -lul/f)2,

(Ex. 4.2) lui < f

(viii) 8(u) = (2/7r)limf.....o (l/f) (U 2/f2 + 1)-2

(ix)

8(u)

(Ex. 4.6)

< lul/f < 1

= (1/8) limf.....o (l/f) (3 - u 2 / (2) ,

0

= (1/16) lim......o (l/f) (3 -lul/f)2,

1 < lul/f < 3

=0,

lul/f> 3

(Ex. 4.5)

(Ex. 4.11)

(xi)

8(u) = (2/7r)lim f ..... oo f exp (-w)[sin (w)/w]

(Ex. 4.10)

Chapter 4

58

EXAMPLE 4.5 Consider

A(p') =

V201i

Ibr5

[sin (p - p') /

(P' - p)/5

8]

3

(4.23)

Obtain the corresponding W(x,O) by a straightforward calculation.

W(x,O) =

_1 J 201i .j2; 1l1r8

1

(sin(pl- P)/8)3 exp (ipIX)d (pI) .

00

-00

(pl-p)/8

Ii

Ii

pex ,0)

--3~~~----~~---O+----h~--~--~~~------~X

-T

-0

~

T

Figure 4.5: Plot of W(x,O) in Example 4.5.

°< Ixl < n/8, W(x,O) =

J58/881i exp(ipx/Ii) (1/2) [3 - 8Ixl/Ii]2, n/8 < Ixl < 3n/8,

0,

Ixl > 3n/8.

Hence1 :

W(x,O) =

J~~~exPC~X)[~(3_8~~2)],

Wavepackets and the Uncertainty Principle

59 (4.24)

Expression (4.24) is plotted in Figure 4.5. Besides localizing \fI(x, t) the A of Eq. (4.23) is proportional to another representation of the delta function. In particular u!:() x

= -311"4

l'1m

- 2(3n+2)/(n+2) EXAMPLE 5.1 Consider the case n = 2,

E =

{ 'li 2 A2/n }n/(n+2) nm

-00

a.

The wave function is continuous at x = ± a and its derivative satisfies equations analogous to Eq. (6.10), namely: lim{d'lj;j _ d'lj;j }=_2ml;01'lj;(x=±a). 5.....0 dx x=±a+5 dx x=±a-5 h Hence, defining

_ m/vol

1I=~,

one obtains that:

A+B

C+D

A(1+2ill) +B(-l +2ill)

C-D

C exp (2ika)

+ D exp (-2ika)

Cexp(2ika)-Dexp(-2ika)

F =

F(1-2ill),

which imply

A

(1 - ill)C - illD

C =

exp (-2ika)(1 - ill)F

D

ill exp (2ika) F.

99

Free Particles Incident on Potentials, ... Solving for F / A one obtains F A

{(1 - i//)2 exp (-2ika) =

+ //2 exp (2ika) }-1

{ (1 - i2//) cos 2ka + i(2//2 - 1 + 2i//) sin 2ka} -1

.

The scattering amplitude is thus Fexp (-ika) Aexp (ika)

= ~ exp (-2ika) =

exp (-2ika) [cos 2ka - 2// sin 2ka + i {(2//2 - 1) sin 2ka - 2// cos 2ka}] (6.43)

=

with

exp {-i(2ka - 8)}

J{cos 2ka - 2// sin 2ka}2 + {(2//2 - 1) sin 2ka (' = -arctan [(2// 1) tan 2ka - 2//] . 2 -

u

If a

--4

2// cos 2ka}2 '

1 - 2// tan 2ka

0 Eq. (6.43) reduces to Eq.(6.13) as it must (but with Vo replaced by

2Vo), namely T = 1/(1- i2//), 8 = arctan (2//). The poles of Eq. (6.43) occur when:

" 1 - i2// - dan2ka = 2 2 2"

// -1+ w

(6.44)

This equation is satisfied if E < 0 since then

mlVoI " // = --:=-U:. 1i 2 i", Eq. (6.44) then reduces to the equation for the allowed bound states 2f -1

tanh2",a= 1 - 2 f + 2 f 2 ' which is satisfied when either tanh ",a = 2f - 1

(6.45)

or (6.46)

100

Chapter 6

V( X)

-0 0 ------.-----~----_T----------.

X

Figure 6.10: Potential in Example 6.15. Eq. (6.45) corresponds to even-parity bound states while Eq. (6.46) to odd-parity bound states. If a -+ 0 Eq. (6.46) has no solution, while Eq. (6.45) has only one solution, namely 2c: -1 = 0 i.e. c: =

~,

that is

lEI = m(~~1)2

which is just Eq. (6.15), with Va replaced by 2Va, while if a -+ 00, both Eq. (6.45) and Eq. (6.46) reduce to c: = 1 which is just Eq. (6.15). These results are as expected since in the former case the delta-function potentials coalesce, while in the latter they essentially uncouple. If k becomes large, v tends to zero and 8 to 2ka. This means if the incident wavepacket is i.e. x

= -a at t = 0, then the transmitted wavepacket

Wtrans is

i.e. x = a at t = 2a/vg which as expected is the classical transit time for crossing these two potentials.

Free Particles Incident on Potentials, ... References 1. E. Merzbacher, Quantum Mechanics, Wiley (1970), p. 234 f.

101

CHAPTER 7

Heisenberg Representation Starting with the expression for the expectation value of an operator Os in the Schrodinger representation (at some time t) :

(O}t

=

L: L:

W*(x, t)OsW(x, t) dx,

(7.1)

and the fact that generally (Eq. (4.12) )

W(x, t) =

G(x, x', t)W(x', 0) dx',

(7.2)

where for a free particle,

1

1 (iP'X') 1 (iP'X) --exp - - - --exp exp (iE(P')t) - - - - d (p,) V2i fi V2i fi fi fi whereas for a particle in a potential (which is assumed time-independent) (7.3)

G(x,x',t) =

00

- 00

G(x,x',t) =

fo(x) = exp -2

etc. Hence

(d;; ) t

= -

i~ + i~ = 0,

and the expectation value of the kinetic energy of the ground (and by a similar calculation of any) state of a simple harmonic oscillator is independent of t. EXAMPLE 7.4 Generalize the result of Example 7.3 for any state of the simple harmonic oscillator. Defining ±= - ~ 1 [p H ±.zmwxH, 1 aH v2m one can quickly show, using Eq. (7.13) that {a1f, all}

2~i

XHPH =

= -nw,

and that

[(a1f)2 - (all)2 + {at, all}]

Thus

jdX'k) \ dt t

-.zmw1_1

00

-00

4>~(x) [(at)2 -

h

h

mz

mz

(alif - nw] 4>n(X) dx

+~ mz

= --. +-. =0, since (allY operating on 4>n(x) produce states orthogonal to 4>n(x). For non-diagonal matrix elements (i.e. min), it may happen that (d/dt) (nl mw 2x'k/2Im}t i 0 (d. Example 7.7). EXAMPLE 7.5 Show

dTH _ PHFH _1_ {F dt - m + 2m H,

PH

}

.

where TH is the kinetic-energy operator in the Heisenberg representation.

(7.17)

Chapter 7

108 From Eq. (7.15)

d(PH )2 = d(p2)H = 2 dPH dt dt PH dt

+

} {dPH dt' PH .

But from Eq. (7.11)

Therefore:

dp2 d: = 2PHFH + {FH, PH},

or

dTH _ PHFH _l_{F } dt - m + 2m H, PH . Thus for example if VH = mw2x~/2,

aVH -dPH = --= FH = -mw 2 XH dt ax

,

and

EXAMPLE 7.6 Evaluate

d

dt (x}t for a particle in the ground state of a simple harmonic oscillator potential:

Using Ehrenfest's expression one obtains:

d{x}t dt

= (dXH) = i{{H }) dt Ii H, X H t· t

From Eq. (7.10):

i ({HH' XH }}t

= =

(~)t = ~ (PH}t ~ ~~(x)exp (i~t)~ :x exp (- i~t)~o(x)dx

i: i:

= i~ since

~~(x) [- m;x] ~o(x)dx = iw

J::'oo l~o(xW xdx = 0 from parity considerations.

i:

~~(x)x~o(x)dx = 0,

109

Heisenberg Representation EXAMPLE 7.7 Evaluate

11

2 21) t =

d \2 / 2'mw xH 0 dt

d dt

1 4>2 00

-00

2

*( x ) 2'mw 1 2 () xH4>o x dx,

where ~(X)-2-4>0(x)dx

'hw + -2. Z 2

'hw 2 _ 0

+2i- .

Combining the results of Examples 7.3 and 7.8,

in this case, where

P'i:, H H = 2m

1 2 2 + '2mw xH = TH + TJ' vH,

0) an d 'T,( 'I" x,

A.. ( ) = '1'0 X.

Thus the expectation value of this Hamiltonian does not change with time. This is not surprising since H does not depend explicitly on time and generally using Ehrenfest's expression:

What is more surprising is that (TH}t and (VH}t do not change with time. This is because one is evaluating the time evolution of the expectation or "average" value of these operators, not of the operators themselves.

CHAPTER 8

Two-, Three-, and N-, Versus One-Dimensional Problems

The one-dimensional Schrodinger equation for a particle in a potential

Vi (x)

is

(8.1)

L:

where

l1l!n(xWdx

= l.

The radial equation for a particle in a three-dimensional radial potential Vj(r) is (8.2) where

Unl(r) I 1 = 0, 1, 2, ... , 1l!nlm(r, (), 4» = - - Ym«(),4», r

1 unl(r?dr

and

00

-I

~ m ~

I,

= l.

If 1 = 0 and Vi(x) = Vj(x) Eqs. (8.1) and (8.2) are identical. However,1l!n(x) in Eq. (8.1) extends from -00 ~ x ~ 00 while Unl(r) must be zero at r = 0 in order that Unl(r)Jr be finite at the origin. In addition Unl(r) extends only from r = 0 to r = 00. A consequence of the above is that some solutions acceptable for Eq. (8.1) are not acceptable for Eq. (8.2). Radial solutions of Eq. (8.2) with potential Vj(r) are identical to solutions of the one-dimensional problem Eq. (8.1) if:

Vi(x) V1(x)

=

00,

=

¥() 3x

x < 0,

+

h 1(1+1) 2mx 2 ' 2

x> O.

(8.3)

If

V,()-¥() h2 1(l+1) x> 0, lX-3 X + 2mx 2 ' and moreover Vi(x) = Vi(-x) then the odd-parity solutions (i.e. those solutions which vanish at x=O) of 1l!n(x) (n = 1, 3, 5 ... ) are identical to Unll(X)J...ti (n' = 0, 1, 2, ... ,1 = 0, 1,2, ... ).

Chapter 8

112

EXAMPLE 8.1 Consider the potential Vi. (x) = (h 2 /2mb4 )x 2 • For this potential the solutions ofEq. (8.1) for the energy En are En = (n+ 1/2)h 2 /mb2 (n = 0, 1,2 ... ). The lowest energy is ;,,2 /2mb2 and the ground-state wave function wo(x) = {l/(bJ"i )} 1/2 exp (_x 2 /2b 2). Consider now the potential ¥J(r) = (;,,2 /2mb 4 )r2 , and the case 1 = O. The minimum energy Eoo for a particle in this potential is 31i? /2mb 2 • Examine why this is so. The three-dimensional radial function uoo(r) cannot be proportional to (the even function) Wo(r) since wo(x) does not vanish at x = o. The second energy eigenvalue for the one-dimensional problem is E1 = 3h 2 /2mb2 , with (odd) eigenfunction 2 WI (x) = ( b.ji

)1/2 (i) exp (2 ) -;b2 .

This is acceptable as the lowest eigenfunction of Eq. (8.2) (to within an overall factor) since WI(O) = o. The normalization must be modified however since:

while Thus uoo(r) =

(bV: (7;) )1/2

exp

(-;~).

Similarly all odd-parity solutions W3(X), ws(x) are acceptable (see Example 10.3) solutions since for these solutions W(O) = o. Thus

etc. and E(n-1)/2,o =

(n + 2"1) mbh 2 ' 2

n = 1, 3, 5 ... ,

i.e.

EXAMPLE 8.2 Consider a particle in the three-dimensional well: 0,

V3 (r) =

-IVoI, 0,

0< r < a, a < r < b, b < r < 00.

Two-, Three-, and N-, Versus One-Dimensional Problems

113

Compare this and the analogous one-dimensional system:

< a, < x < b, b < x < 00.

x

0,

Vt(x) = -IVoI, 0, V(

a

r)

V( X)

J J

--+------a~-----r.b----~-x

~-----aT-----~~b----~r

Figure 8.1: Diagram in Example 8.2 The allowed 1 = 0 (negative) energy levels are easy to obtain. In detail the solutions of Eq. (8.2) which satisfy the boundary conditions at the origin (uno(O) = 0) are

uno(r)

= Asinhkr,

0< r < a,

uno(r) = B sin (Kr + 6),

< r < b, b < r < 00.

a

uno(r) = C exp (-kr),

(k = V2m1EI/h2 ) (K = V2m(IVoI-IEI)/h

2 )

Matching the wave functions and their derivatives at a and b yields tanh ka

k

=

tan (Ka

t.e.

tan(Ka+6)

and

+ 6)

K K = Ttanhka,

tan (Kb + 6) K

=-"k'

tan(Kb+6)

= -T.

1.e.

(8.4)

1

K

(8.5)

Chapter 8

114 One can rewrite Eq. (8.5) as tan {K a

+ h + K (b -

K

a)}

tan(Ka+h)+tanK(b-a) tan (K a + 15) tan K (b - a)

= - k = 1-

=

(K/k) tanh ka + tan K(b - a) 1- (K/k) tanh ka tan K(b - a) .

Thus the energy levels for this system are obtained by solving the transcendental equation K (K/k)tanhka+tanK(b-a) (8.6) -k= l-(K/k)tanhkatanK(b-a)'

If a If a

= 0 Eq. ---+ 00,

(8.6) reduces to -K/k = tanKb. b ---+ 00, c == b - a, Eq. (8.6) reduces to:

K

K/k+tanKc

-k =

1- (K/k) tan Kc .

These results can be compared with the energy levels for a particle in the onedimensional well given in this example. The solutions of Eq. (8.1) for this potential are Aexp (klxl), x < a, \li'(x) = Bsin(Kx + 15), a < x < b, Cexp (-kx), x> b. Matching boundary conditions one obtains

1

"k=

tan(Ka + 6) K

tan(Kb+h) K

1

--"k'

(8.7) (8.8)

One can rewrite Eq. (8.8) as tan {(Ka

+ 15) + K(b -

an

K

tan (Ka + 15) + tanK(b - a) tan (K a + 15) tan K (b - a)

= - k =1=

Kjk + tanK(b - a) 1 - (Kjk) tanK(b - a) .

Thus the energy levels for this system are obtained by solving the transcendental equation K K/k + tanK(b - a) (8.9) - k = 1 - (K j k) tan K (b - a) .

Two-, Three-, and N-, Versus One-Dimensional Problems

115

This is not equal to Eq. (8.6) except in the limit a

-+ 00

b -+

c

== b-a

00

where c is a constant, in which case the boundary condition at the origin is unimportant. The two-dimensional Schrodinger equation for a particle of mass M in a potential

V2(p) can be written:

(8.10) where ,T, ( "-) = Wnm(p) exp (±imcP) ~nm P,o/

vp

"fi1r'

m

= 0,

1, 2, 3... ,

and

Here

wnm(p)

must be zero at p = 0 so that wnm(p) will be finite at the origin.

vp

Comparing Eqs. (8.2) and (8.10) and noting that

indicates that if one has a solution Unl(r) of Eq. (8.2), then wnm(p) is Unm -l/2(P). One notes both u(O) and w(O) must be zero, as too u(oo), and w(oo), (if the particle is bound) i.e. u(r) and w(p) have identical boundary conditions. EXAMPLE 8.3 Suppose one has a particle of mass M in a two-dimensional Coulomb potential well: Ze 2 V2(p) = --4. '!rEoP

Find the corresponding negative energies and wave functions.

Chapter 8

116 Since, as standard texts report, for the three-dimensional well

Va(r) Unl(r) =

Ze 2

= --47rEor -

,

(2Kr)I+lexp(-Kr) [ K(n+l)! ]1/2 (21+1)! n{n-I-l)! tFt{-n+l+li21+2i2Kr), (8.11)

and

Z2a 2 Me 2 ZMea e2 ---::---, K = , a = - , n=l, 2... , 1=0, 1, ...n-l, 2n 2 nn 47r€one one can immediately write:

_ (2Kp)m+1/2 exp (-Kp) [K (n + m -1/2)!] 1/2 Wnm{P) (2m)! n (n _ m _ 1/2)!

X

(8.12) 1Ft

where

Z 2 a 2 Me 2 2n 2

Enm =

K

~i2m+ 1;2Kp) ,

(-n +m +

= ZMea, nn

1

n = 2"

3

1

2' ... , m = 0, 1, ...n - 2' . -n + m + 1/2 is a negative integer

Here n must be a positive half-integer so that or zero, making 1Ft a terminating series. Otherwise wnm{p) blows up as p Thus 9

00.

, ....

EXAMPLE 8.4 Suppose one has a particle of mass M in a two-dimensional oscillator well: 1 V;(p) = 2'MW 2p2. Find the corresponding energies and wave functions. Since, as standard texts report, for the three-dimensional well: 1

Va () r = 2'Mw r , =

{2r (n

+ 1 + 3/2) b3n!

x

1Ft

2 2

}t/2 r l+ exp (_r2 /2b2) l

bl r(l + 3/2)

(-nil+~; ~), (8.13)

Enl =

(2n+l+~)nw,

b=J);w' n=O, 1,2... ,1=0,1,2... ,

Two-, Three-, and N-, Versus One-Dimensional Problems

117

one can immediately write _ {2 (n + m)! b2 n.I

wnm(p}-

}1/2 pm+1/2 exp (_p2 /2b 2) bm I

m.

IFI

(_. . p2) n,m+1'b2'

(8.14) Enm = (2n+m+1)'liw, b=J:;w' n=O, 1, 2 ... , m=O, 1, 2 .... EXAMPLE 8.5 Suppose one has a particle of mass M in a two-dimensional well:

V=

0,

0 < p < a,

00,

a

< p < 00.

Solve this problem. To find the eigenvalues and eigenfunctions of this system one notes that for the analogous three-dimensional case the standard result involves spherical Bessel and spherical harmonic functions:

0,

a

0) for all x.

If P, the so called 'parity' operator is such that when operating on any function

f(x)

P f(x) = fe-x), Then

PH(x)tfJ(x) = H(-x)tfJ(-x) = H(-x)PtfJ(x) = H(x)PtfJ(x) for this case since for the V of this example Vex) = V( -x),

V (X)

-----------------&--------------------x Figure 10.3: Potential in Example 10.3. while quite generally

T(x) = T(-x). Hence in this case H(x) = H( -x). Thus for this problem (indeed whenever Vex)

V(-x) )

[PH(x) - H(x)P]tfJ(x)

= 0,

I.e.

[P, H]

= O.

=

Upper Bounds and Parity Considerations

139

But this has consequences on the eigenvalues and eigenfunctions of H. Assuming

[H, Pj = 0 consider the Schrodinger equation Eq. (10.1) :

Premultiplying by P

Assuming the system is non-degenerate (see Chapter 12), if Wn is an eigenfunction of H and at the same time PW n is an eigenfunction of H with the same eigenvalue En, this implies (10.11) i.e. PWn(x) is proportional to Wn(x). But Eq. (10.11) is just the eigenvalue equation for P. Thus whenever [H, Pj = 0, the eigenfunctions of the Hamiltonian are also eigenfunctions of P! Consider now the eigenvalue problem for P. If Po are the eigenvalues of P,

Pf(x) = Pof(x) = f(-x) P 2 f(x) = p~f(x) = f(x).

(10.12)

Hence Po = ± 1. Thus in Eq. (10.11) A = ± 1, and

Thus whenever, as is the case in this problem, V(x) = V( -x) and the system is nondegenerate, the eigenfunctions of H have either even parity or odd parity. Going through the derivation for upper bounds Eqs. (10.1) - (10.5), one sees that in this case Eq. (10.5) applies independently to the odd- and even-parity solutions since Eq. (10.3) will be an expansion either in terms of the even or the odd eigenfunctions 'lin. Hence working with odd-parity trial wave functions one gets an upper bound to the lowest odd-parity state energy and similarly for even-parity trial wave functions an upper bound to the lowest even-parity state energy. A simple (though inadequate, since its derivative is not continuous at x = 0) even-parity trial wave function for Example 10.3 is:

'1Pt(x) = {

J3/2b3 (b 0,

Ixl) ,

Ixl < b, Ixl > b,

Chapter 10

140 with free parameter b, yielding,

The condition 8E+

7ib =

O· Ii b3 = 121i2 Imp es Am

(where one uses the representation of the delta function

d?

dx21xl = 28(x) ). For optimal b,

~ (3A 21i 2) 1/3 + ~ (3A21i2)1/3 4

2m

2

~ (A 21i2) 1/3 '"

=

m

122/3

2m 0.86

(A 21i2) 1/3 m

A better even-parity trial wave function one can use is

~t(x) = Jb~ exp (-;;2)' with free parameter b, yielding

If 8E+

7ib =

b3 = 1i2../i .

2mA

0,

For optimal b, E;round exact :::; E+(boptimal)

=

~ (A21i2) 1/3 + (A 21i2 ) 1/3

=

(~:) 1/3 ( A~2) 1/3 '" 0.81 ( A~2) 1/3

2

21Tm

21Tm

An odd-parity trial wave function one can use is 2 )

~t(x) = ( bv'i

1/2 ~ exp (2 ) - ;b 2

'

141

Upper Bounds and Parity Considerations

with free parameter b, yielding

If

and for this optimal b :

=

(3h 2A2)1/3 +2 (3h 2A2) 41rm

=

1/3

41rm

(!~r/3 (h~2r/3 ~ 1.86 (h~2r/3

One thus has roughly determined two energy levels with the help of parity considerations in this case. One notes

E- (boptimal ) = E+ (boptimal)

1.86 ~ 2.30, 0.81

as opposed to

E- ground exact = 1.84 E+ground exact 0.89

rv

2.07

(c£. Eqs. (3.18), (3.31) ). EXAMPLE 10.4 Consider the system H = p2/2m - Vo6{x), Vo > O. V( x)

Figure 10.4: Potential in Example 10.4.

Chapter 10

142

Estimate the energy of this system using as trial wave function

'ljJt(X)

( mw)1/4 h7r exp

=

(mwx2) -21i

(the ground-state wave function of the one-dimensional harmonic oscillator), with free parameter w. One obtains directly hw (mw)1/2 E(w)=--Va -

4

h7r

Requiring

oE(w) = 0 implies ow

~ _ Va (~)1/2 _1__ 4 2 h7r W1/2 - 0,

I.e. Woptimal

=

4mVo2

7rh3 .

Hence for this trial wave function,

mVo2 7rh2

=

_

(~)1/2 (mVi) 1/2

Vo h7r

m Vo2 2mVo2 2 7rh - 7rh 2

2

=-

7rh3

m Vo2 h 27r '

vs. the exact ground (and only) state and only energy of this system (cf. Eq. (13.8»:

Eground exact --

mV,2

°

- 2h2 .

One again notes mv,2

E (Woptimal) = -0.318 h20 > Eground exact = -0.5

t/ '

mv,2

as required by Eq. (10.5). EXAMPLE 10.5 Consider the system H = p2/2m where

Vex) = {

+ V (x)

mw2x2/2 + a/x 2 , x 2: 0, (where a is a positive or negative constant), 00,

x

< o.

Estimate the ground-state energy of this system using as trial wave function:

Upper Bounds and Parity Considerations (1)

143

(mw')3/4 2x

T

_

'l/;t (x) -

'/r1/4

(This is the lowest eigenfunction of H'

exp

(mw'x 2)

--v;- .

= p2/2m + V'(x) where

V'(x) = { 00,

x

2:: 0,

x

< 0,

with eigenvalue 3nw' /2.) Using this trial wave function,

= ~nw' + 2mw'a + (w2

E(1)(w')

n

2

_

W,2

1) ~nw' 4

3 /{ 8ma} 3 nw2 4"nw 1 + 3n2 + 4"7 . [This result can easily be obtained by rewriting p2 H = -2m

and noting that

1

1 ,2 2 + -mw x + -xa2 + ( W,2 -w 2 2

00 01.(1)*( )

o

a

x ""2

'f/t

01.(1)( ) 'f/t

X

~mw'2 x2 Jo[00 ol't(l)*(X) 'f/ 2

ol't(l)(X) 'f/

X

dx = ~ 2

1

)

/2 2

1 -mw x 2

d _ 2mw'a . -~- ,

X -

1£

(~nw') 2

(f c. Eq. (9 .5) ) . 1

From the constraint one obtains

, w

optimal =

w V1 + 8ma /

~2

31£

•

Hence

E(l)

3 nwV1 -4

(w' optimal) =

If a

= 0, E(1)

(W~Ptimal)

+ 8ma/3n2 + ~1iwV1 + 8ma/3n? 1

~nwV1 + 8ma/31i 2 .

= 3nw/2, as expected for this trial wave function.

Chapter 10 144 2 If a = n /m, the Hamiltonian above admits of an exact solution since n2/mr2 = 2n2/2mr2 is then just the centripetal term in the three-dimensional simple harmonic oscillator (V(r) = rnw 2r2/2) Hamiltonian, for angular momentum quantum number 1 = 1. The exact energies in this case are (2n + 1 + 3/2) nw, n = 0, 1, 2... i.e. the lowest exact solution is Eground = 5hw/2. In this case

).J33

I E (1) ( Woptimal = -2-

i:

nW

'V

2.873 nw > 2.5 nw,

as expected. One may use instead:

(2,8)2n+l (2n)!

x

n

exp (-,8x),

as trial wave function. For this value of a,

For optimal ,8 R

•

,voptImal

=

(~)1/2

n

(2n -1)n(2n + l)(n 2( n + 4)

+ 1))1/4

,

one obtains E(2) (~. n) = nw optImal>

(2n

+ l)(n + l)(n + 4) 2n(2n - 1)

which has values 2.556 nw and 2.535 fiw for n = 3 and 4 respectively, while for optimal noptimal 3.74, 'V

E(2) (,8optimal, noptimal)

'V

2.533 nw > 2.5 fiw.

As expected one again gets an energy greater than the exact energy, but ¢F) (x) is seen to give a better upper bound than ¢(l)(x). If a = 3n2/m, 6fi 2 /m, 10fi 2/m etc. again one knows the exact solution which one can compare with the upper-bound results that arise in these cases. Thus for a = 3fi 2 /m, and 1 = 2, Eexact ground = nw/2, while E(l) (W'optimal) = 4.5 nw

> 3.5 fiw.

Upper Bounds and Parity Considerations

145

For general a ,82 (n1i?

+ 4am)

2m(2n - l)n 8E(2) (,8, n)

,8 (nfi2

=

8,8

+

+ 4am)

4,82

a . _ Jmw [(2n t"Optlmal 1i

'

+ l)(n + 1)

mw2(2n

2,83

m(2n - l)n

and

+ l)(n + 1)

mw2(2n

+ 1)(2n -l)(n + 1)nfi 2] 1/4 2(n1i2

+ 4am)

Thus

~ {(2n + l)(n + 1)(n1i2 + 4am) }1/2 2

+ =

2n(2n - 1)

~ {(2n + l)(n + 1)(n1i2 + 4am) 2

2n(2n - 1)

w {(2n

}1/2

+ l)(n + 1)(nfi2 + 4am) }1/2 2n(2n - 1)

For a = 0 this reduces to Eq. (10.8) E(2)

(a.

) _ fi

t"optlmal, n -

w

{(n

+ 1)(2n + 1) }1/2 2(2n -1)

For a = 3fi 2/m E(2)

(,8.

) _ fi

optlmal,n -

For this value of a, noptimal

rv

w

{(n

+ 1)(2n + l)(n + 12) }1/2 2n(2n -1)

5.74 and

E(2) (,8optimal, noptimal)

If

rv

3.52 fiw.

~~3\X) = C~f/2 (m;'r/4x2exp (- m;~x2)

is used as trial wave function, E(3)( ') w

2 = (!..2am) fiw' 12 + 3fi2 + ~4 fiw Wi ,

,

Chapter 10

146 with

,

w

Woptimal = J7/15

+ 8exm/15h 2

Substituting this value of w',

If E(3) (W'optimal) = 5/2

hw = Eexact ground,

which implies that ~P)(x) is the exact ground-state wave function for this value of ex, while if

_ 3h 2

a -

m' E

(3) ( '

.

) _

~J155 3

WoptImal - 2

hw

rv

~_

3.59 hw > 2 hw - E ground exact·

Thus ~~3\ x) gives a better upper estimate for Eground than ~?) (x) while the best upper limit for this case is given by ~}2). EXAMPLE 10.6 Consider the system H = p2/2m

V(x) = {

+ V( x), where -ahe/x,

x

2': 0,

=,

x

< 0,

with ground-state eigenvalue of E ground exact

=

-mc 2 a 2 /2.

Estimate the ground-state energy of this system using as trial wave function: ~)

_

~t (x) -

2

1

b ( byf1r )

1/2

(2 ) X

x exp - 2b2

•

Performing the integral in Eq. (10.2) one obtains:

Imposing the condition 8E(1) (b)

8b

..

= 0, ImplIes

b

.

_ 3hyf1r

optImal - 4mea .

147

Upper Bounds and Parity Considerations Hence

=

>

4 -_mc2a 2 '" -0.424 mc2a 2 37r Eground exact'

as it must be. This result is quite close to the exact ground-state energy. Other trial wave functions do not do as well. Thus, if instead one uses:

J2

(2) _ 2 2 (X2 ) '!f;t (x) - b2V~ x exp -2b2 '

one obtains 71i2 41ica E(b) = 12mb2 - 3b..,fi .

Imposing the condition 8E(2)(b) b . _ Th..,fi 8b = 0, implies optimal - 8mca . Hence =

16 32 _mc2a 2 - _mc2 a 2 2b 2b

=

-217rmc2a2 '" -0.243 mc2a 2.

16

If one uses:

one obtains

Imposing the condition 8E(3) (b)

8b

..

3amc

= 0, ImplIes boptimal = ~ .

Chapter 10

148 Hence

3

3

-mc2 a 2

=

_mc2 a 2

=

3 _gmc2a 2 = -0.375 mc2a 2•

8

_

4

Though both E(2) (boptimal) and E(3)(bopt imal) are above Eground exact, neither is as close to it as is E(1)(boptimal)' A fourth trial wave function: .1,(4)( ) _ 'f't

X

-

V2 3/2 b2 x

(x2 )

exp - 2b2

'

yields

and imposing the condition 8E(4)

--m;- =

0, results in

b . _ 51i optImal - 2macV; .

Hence for this trial wave function:

EXAMPLE 10.7 Consider the Hamiltonian: (all x). Given the trial wave function:

tPt(X,w') =

1

7r 1 / 4

(mw') T

1/4

exp

(mw'x ----y;-

2)

,

evaluate E (w') and find the minimum upper bound to the ground-state energy with this trial wave function for the case a = 1.

149

Upper Bounds and Parity Considerations

Evaluating

I.,p; H.,pt dt, one obtains:

hence

8E(w')

---::a:-'-w""'"-"-

liw 2

liw'

4" + 4w' +

E(w') =

Ii

= '4 -

3aliw3 4W,2 ,

liw 2 6anw 3 4W,2 - 4w,3

The condition 8E(w')/8w' = 0 yields the result:

w' ----6a=0. 3 w w

W,3

Thus if a = 1, w' = 2w, and E(2w) = 13/16 nw = 0.8125 nw. This can be compared to the exact answer (obtained by numerical methods) in this easel =0.803771... liw . If one substitutes w' = w then E(w) = liw/2 + 3aliw/4, which is the zero (nw/2), plus first order (3anw/4) perturbation contribution if one writes H = Ho + V with 2x 2/2, V = aliw Ho = p2/2m + /Ii x) 4, a decomposition which for a = 1

mw

(Jmw

results in a much worse upper bound E(w) = 1.25 liw, and a diverging perturbation series 2 (for all values of a). EXAMPLE 10.8 Consider the Hamiltonian (cf. Example 10.7): H= : :

+~mw2x2+anw(J~w X)4

(all x).

Use the trial wave function:

.,pt(x,

b) = {

J3/21? (b -Ix!),

Ixl < b,

0,

Ixl > b,

to evaluate E(b). Find the minimum energy for this trial wave function in the cases a = 0, 1. By a straightforward calculation (cf. Example 10.3)

E(b)

3li2

=

8E(b) 8b

2m~ +

3li2 - ml?

am 2w3 b4 + 351i mw 2b 4am 2w3 fi3

mw2~

----w-

+ 10 +

35ft

The optimization requirement is :

8au6

+ 7u 4 -

210 = 0, where

u=

J~w

b.

Chapter 10

150

If

a=O,

t h e con d ·, ItIon

8E(b) = ---a;;-

b = (30)1/4

O'ImpIies:

J

Ii

mw

rv

2.340

J

Ii

mw

.

Hence

f[; liw

E (boptirnal) =

If a = 1, boptirnal

rv

1.645

rv

0.55 liw

Jm~'

VS.

0.5 Tiw, the exact result.

hence E (boptirnal)

rv

0.899liw.

Similarly

E(b = 2.0 Jm~) E(b = 1.7 J!)

rv

1.032 liw,

rv

0.902 liw,

E(b = 1.4 J~w) 0.973 hw, as compared to 0.803 hw, the exact (numerical) ground state energy. rv

EXAMPLE 10.9 Consider the Hamiltonian :

(all x). Given the trial wave functions: 1/4 (mw'x 2 ) 'IjI~1) (x,w') = ( mw') 1i1r exp -----v;-

Ixl
b, find E(l)(w',a), E(2)(b,a) and the minimum values for E(1), E(2) for arbitrary a.

151

Upper Bounds and Parity Considerations By a straightforward calculation (d. Example 10.7) one obtains: =

aE(1) (w', a)

ow'

-

nw'

4

nw 2 + 4w'

a15nw 4 + 8W,3

4" -

4W,2 -

8W,4

The minimization condition is thus: 2

(~)

4 _

2

(~r

45a = 0,

,

I.e.

Woptimal = w

J1 + VI + 90a 2

.

J5 w, and

For example if a = 8/9, then W'optimal =

11 v~ E (1) ( w, optimab 8 /9) = 30 5 nw ~ 0.820 nw.

Similarly

and OE(2)(b a)

ab'

I.e.

= 0,

5au8

implies

+ 7u 4 -

210

=0

where u

Jmw

= b T'

uoptimal . = Jmwn (-7 + V49lOa+ 4200a )1/4 If a

= 8/9,

boptimal

rv

1.57 Jmw/n, and E(2) (boptimab 8/9)

indicates 'I/'~1) is a better trial function than

rv

0.910 nw. This

'l/'F).

EXAMPLE 10.10 Consider H = p2/2m + Alxl n , Evaluate Emin for

'l/'t(x,

-00

< x


o.

Find E(b, n) and E(boptimab n) if 'l/Jt(x, b) = 2/b (1/bVi)1/2xexp(-x2/2b2). One obtains

3)

31i 2 2Abn f (n + 4mb2 + Vi 2 ' 3h 2 2nAbn- 1 (n + -2mb3 + Vi f -2-

E(b, n)

3)

8E(b, n) 8b The requirement 8E(b, n)/8b

= 0 implies (

boptimal =

3h2 Vi ) 1/(n+2) 4nmAf([n + 3] /2)

Thus E(boptimab n)

[

3n/ 2Ahnnf([n + 3] /2)] 2/(n+2) m n / 2 2 n Vi

+ ~ n

[

E (boptimab 1)

[3 n/ 2Ahnnr([n + 3] /2)] 2/(n+2) m n / 2 2n Vi

3n/ 2Ahn(n

+ 2)(n+2)/2f([n (nm )n/22n Vi

G~r/3 (A~2r/3

,

+ 3] /2)] 2/(n+2)

154

Chapter 10

E(boptimal, 2)

=

Eground exact(n

= 2)

3

(

Ah2)1/2 2m

'

etc. EXAMPLE 10.13 find E(boptimal)

'¢;t(b, x)

=

{

.jlib

cos (7rx/2b) ,

where

-b < x < b,

Ixl > b.

0,

By a straightforward calculation one obtains:

oE(b)

fib Hence

and E (boptimal) =

(7r 2/6 _1)1/2 (7r2/6 _1)1/2 ( 7r2/68 _1)1/2 hw + 8 hw = 2 hw '" 0.568 hw.

EXAMPLE 10.14 Consider the system (cf. Example lOA)

Estimate the energy of this system using as trial wave function:

'¢;t(x, b) = {

J3/2b3 (b -Ix!),

Ixl < b,

0,

Ixl > b.

Upper Bounds and Parity Considerations

155

One obtains:

E(b)

=

8E(b)

fib = This implies boptimal Hence

E(b

.

311. 2 3Vo 2mb2 - 2b' 311. 2 3Vo - mfi3 + 2b2 .

= 211.2 1m Yo. ) - 3mVo2 811. 2

optImal -

_

3mVo2 __ 3mVo2 _ -0375 mVo2 411. 2 8h 2 - . h2'

This is to be compared with Eground exact = -0.5 mVo2 /ti?, and E (Woptimal) = -0.318 m Vo2 /h 2 in Example 10.4. EXAMPLE 10.15 Consider the system (d. Example 10.3)

p2

H = 2m

+ Vex),

Vex) = Alxl ,

for all x.

Estimate the energy of this system using as trial wave function:

¢t(x, b) =

{

f1ib

Ixl < b,

cos (1rX/2b) ,

Ixl > b.

0, By a straightforward calculation one obtains:

E(b) =

h2 7r 2 Ab 8mb2 + 27r 2 [7r 2

-

4] .

Requiring 8E(b)/8b = 0 implies io2 4 )1/3 b- ( n7r 2 - 2Am (7r - 4)

Hence

This result can be compared to the results 0.81 ( A2'/i2/m ) 1/3 and 0.86 ( A 2h2/m )1/3 of Example 10.3.

Chapter 10

156

One notes in passing3 that for all the Hamiltonians discussed in this Chapter of the form

H=T+Axn,

the Virial Theorem (cf. Eq. (9.6) ) (T) = (n/2) (V),

is also satisfied if etc. provided b is chosen to minimize

This situation arises in Examples 10.1-4, 6, 10-15.

References 1. F. T. Hioe and E. W. Montroll, J. Math. Phys. 16 1975, p. 1945. 2. C. M. Bender and T. T. Wu, Phys. Rev. 184 1969, p. 1231. 3. H. A. Mavromatis, to be published.

CHAPTER 11

Perturbation Theory

Consider a system whose Hamiltonian H contains a 'perturbation' V such that

H= Ho+ V.

(11.1 )

If one knows the eigenfunctions (cPn) and eigenvalues (En) of H o, i.e. (11.2) and wishes to find the eigenfunctions (1/Jn) and eigenvalus (En) of the Schrodinger equation: (11.3) where 1/Jn reduces to cPn as V terms of the complete set cP :

--+

0, one can formally expand each eigenfunction 1/J in

(11.4) m=O

where the cP's are assumed orthonormal, i.e.

One can then rewrite Eq. (11.3) as 00

L

(H - E)amcPm = 0,

(11.5)

m=O

where for notational simplicity the subscripts n have been omitted from the 1/J's and E's. Premultiplying Eq. (11.5) by cP; and integrating over the relevant variables one obtains: 00

L

m=O

where

(Hpm - E Opm) am

= 0,

p

= 0,

1, ... ,

(11.6)

Chapter 11

158 and

Vpm ==

J4>;V4>m dr

=

(4)plVl4>m).

(11. 7)

Writing Eq. (11.6) in matrix form one has:

Hoo-E H11-E

Vio

Vi2

=0.

"V:!o The eigenvalues E of Eq. (11.6) are the exact energies of the system Eq. (11.3) and the corresponding coefficients am when substituted into Eq. (11.4) give the corresponding eigenfunctions. Consider for simplicity the special case when Eq. (11.6) is a 2 x 2 matrix. The determinant of this matrix must be zero i.e.

E2 - E (Hoo + H11 ) - VOl Vio + HooH11 =

o.

(11.8)

Hence

E

(Hoo + H11 ) +_ (Hoo - H 11 ) ( 1 + 4VOl Vio/{Hoo - H11 )2 )1/2

=

2

{Hoo + H11 ):!: (Hoo - H 11 ) (1

-

+ 2V(n Vio/(Hoo -

2

H11)2

+ ...)

(11.9)

Keeping only the first two terms in this expansion results in the following expansions:

IT

Eo

flOO =

Eo

+ HooVOl-VioHu = Eo + VOO + - -V(n-Vio- - fO + Voo - fl - ViI TT

+

volVio 00 + (fO - fl)( 1 + (Voo - Vid/(fo - fl) ) ,

V,

(11.10) H 11 =

fl

+

Vio VOl Hu - Hoo

= f1

Vio VOl + V;11 + ---::-:--"-~':---::-::,f1 + ViI - fO - VOo

Viovol + V;11 + ..,..--~--,----,-----:---:-:---~ (fl - fO) { 1 + (ViI - VOO)/(fl - fo) )

Upon expanding the denominators in these expressions one obtains:

159

Perturbation Theory

Eo

=

fo

+

TT

voo

VOl Vlo fo - f1

Y(ll Vll Vlo )2 Eo - f1

+ -- + (

-

TT

Voo (

VOl Vlo )2 fO - f1

+ ... , (11.11)

El =

fl

VlOY(ll + Vll + - + VlOVoOY(ll2 f TT

f1- O

(f1-fO)

TT

vll

VlOVo1 2 (f1- fO)

+ ....

From the 2 X 2 matrix one can also obtain expressions for tP by solving for the eigenvectors ao and aI,

tPo

=

tPl =

{

ao mw 2 a2 /2 for reasons similar to those mentioned in Example 11.1.

168 EXAMPLE 11.4 Consider a particle subject to the Hamiltonian:

H

= T + Alxl P

Chapter 11

(all x) .

(11.24)

This may be written:

where

n2 x2 2mlfl

Alxl P -

__

can be treated as the perturbation V. With this decomposition the ground state energy of this sytem can be immediately calculated using first-order perturbation theory, with

Ho

= T + 2!2b4

x 2,

Eo

= 2:b2

If p is odd one has

(oIAlxl

n2 x

__ 2 2mb4

P-

I) 0

The choice 11'+2

..mw 2xy .

Suppose one wishes to find the (non-degenerate) ground-state energy of this system. The unperturbed ground-state energy of this system is nw with corresponding wave function