Constantinescu--Problems in Quantum Mechanic

PROBLEMS IN QUANTUM MECHANICS BY

F. CONSTANTINESCU A.v. Humboldt Fellow, University of Munich, and University of Cluj, Rumania

AND

E. MAGYARI University of Craiova, Rumania

TRANSLATED BY

V. V. GREeD

EDITED BY

J. A. SPIERS

PERGAMON PRESS Oxford New York .Toronto Sydney Braunschweig

Pergamon Press Ltd., Headington Hill Hall, Oxford Pergamon Press Inc., Maxwell House, Fairview Park, Elmsford, New York 10523 Pergamon of Canada Ltd., 207 Queen's Quay West, Toronto 1 Pergamon Press (Aust.) Pty. Ltd., 19a Boundary Street, Rushcutters Bay, N.S.W. 2011, Australia Vieweg & Sohn GmbH, Burgplatz 1, Braunschweig

Copyright 1971 Pergamon Press Ltd. All Rights Reserved. No part of this publication may be repro-duced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, record-ng or otherwise, wi thout the prior permission of Pergamon Press Ltd.

First edition 1971

Library of Congress Catalog Card No. 73-116573

PRINTED IN HUNGARY

08006826 X

Contents

FOREWORD vii A NOTE ON THE LAYOUT OF TIllS BOOK vii

CHAPTER I. THE MATHEMATICAL FORMAUSM OF QUANTUM MECHANICS 1. Hilbert Spaces 1 2. Operators in Hilbert Space 2 3. The Matrix Representation of Vectors and of Operators 3 Problems 4 Solutions 8

CHAPTER II. SIMPLE QUANTUM SYSTEMS 20 1. The Eigenfunctions and the Energy Spectrum 20 2. The Transmission of Particles through Potential Barriers 21 3. Motion in a Central Field 21 Problems 23 Solutions 32

CHAPTER III. MEAN VALUES AND UNCERTAINTY RELATIONS 75 1. The Mean Values of Dynamical Variables 75 2. The Uncertainty Relations 76 Problems 77 Solutions 79

CHAPTER IV. THE SEMI-CLASSICAL ApPROXIMATION 93 1. The Wavefunction in the WKB Approximation 93 2. Formulae for connecting WKB Wavefunctions on Opposite Sides of Turning Points 94 Problems 95 Solutions 98

CHAPTER V. PICTURES AND REPRESENTATIONS 112 1. The SchrOdinger, the Heisenberg and the Interaction Pictures 112 2. Representations 113 3. The Density Operator 115 Problems 115 Solutions 119

CHAPTER VI. ORBITAL ANGULAR MOMENTUM AND SPIN 143 1. Properties of Angular Momentum Operators 143 2S~n 1~ 3. Angular Momentum and Rotations of Coordinate Axes. The Addition of Angular Momenta 144 Problems 146 Solutions 152

v

Contents

CHAPTER VII. SYSTEMS OF IDENTICAL PARTICLES. SECOND QUANTIZATION 1. Symmetry and Anti-symmetry of State Vectors 2. Isotopic Spin 3. Second QUantization Problems Solutions

CHAPTER VIII. PERTURBATION THEORY. THE VARIATIONAL METHOD

1. Stationary State Perturbation Theory 2. The Variational Method Problems Solutions

CHAPTER IX. TIME-DEPENDENT PERTURBATIONS. RADIATION THEORY

1. Time-dependent Perturbations 2. Radiation 3. The Interaction of Radiation with Atomic Systems Problems Solutions

CHAPTER X. CoLLISION THEORY

1. Potential Scattering 2. The Lippmann-Schwinger Equations Problems Solutions

CHAPTER XI. ATOMS AND MOLECULES

Problems Solutions

CHAPTER XII. RELATIYlSTlC QUANTUM MECHANICS I. Definitions, Notation, Conventions 2. Elements of Relativistic Mechanics 3. The K1ein-Gordon Equation and the Dirac Equation Problems Solutions

APPENDIX. CERTAIN FUNCTIONS USED IN QUANTUM MECHANICS I. Hermite Polynomials 2. Legendre Polynomials and the Associated Legendre Functions 3. Spherical Harmonics 4. Laguerre Polynomials 5. The Gamma Function 6. Bessel Functions of the First Kind 7. Spherical Bessel Functions 8. The Hypergeometric Function 9. The Confluent Hypergeometric Function

GENERAL REFERENCES

[~DEX

OTHER TITLES IN THE SERIES

VI

180 180 181 182 183 18i

203 203 204 205 207

225 225 226 227 229 232

263 263 268 270 275

321 322 323

337 337 338 339 341 348

396 396 396 398 399 399 400 401 402 404

405

407

411

Foreword

WHEN Dr. Constantinescu and Dr. Magyari approached me about the possibility of the pub-lication of an English edition of their collection of problems on quantum mechanics I was very pleased to arrange this. Although there exist a few good collections of this kind, none of those has the same extensive coverage which the present volume gives, and I felt that it would be a valuable addition to the existing literature. Quantum mechanics courses are getting more and more advanced, and topics which only a few years ago were deemed to be fit only for graduate courses now appear in undergraduate courses. It is thus very much to be welcomed to have a problems book which covers such topics as the Dirac equation, second quantization, many-body problems, and the density matrix. I hope and expect that quantum mechanics teachers as well as students will find the present book a great help, and I wish it every success.

Magdalen College, Oxford D.TERHAAR

A Note on the Layout of this Book

EACH chapter is essentially self-contained, and is in three parts. In the first part, basic propositions relating to a particular topic are given and some consequential theorems are stated without proof. In the second part, a set of graded problems leads the reader to prove the theorems already stated, and to derive further theorems and applications. Detailed solutions of the problems are given in the third part. Equations in the first part are labelled with the number of the chapter (in roman numerals) followed by a serial number; those in the second and third parts with the number of the problem followed, respectively, by a serial letter or a serial number. The letter A followed by a number is a reference to an epuation in the Appendix.

vii

CHAPTER.

The Mathematical Formalism of Quantum Mechanics

1. Hilbert Spaces A set of entities E is said to constitute a linear (complex) space if, on the elements of the

set, operations of commutative addition, and of multiplication by a complex number, are defined and are such that the results of these operations are also element of E. The term "vector" is often used to denote the elements of such a space; these vectors should not be confused with the vectors of ordinary geometric space.

Let Hbe a linear (complex) space such that, to every pair of elements x, y of H, there can be made to correspond a complex number, called the scalar product (x, y) of the elements x and y, and having the following properties:

(a) (x, y) = (y, x)*, (b) (Xl+X2, y) = (Xl. y)+(X2' y), (c) (AX, y) = A*(X, y); (x, AY) = A(X, y), (d) (x, x) ;;;,. 0, the equality holding only for the "null element" x = O. The "norm"

of an element is defined to be II x II = + V (x, x). H is then said to be a Hilbert space provided it is complete. t

In any Hilbert space a set of elements can be found, such that any element of the space can be written as a linear combination of these. Such a set is called a basis.

In quantum mechanics, every state of a physical system is associated with a vector, some-times called a "ket", and denoted by the general symbol I ), into which may be inserted symbols denoting eigenvalues, quantum numbers, etc., which specify the state in question. From the principle of superposition of states, it follows that the ket vectors of a system together form a linear space.

t An ordered set {x,,} of elements of H is said to converge to the limit xo, if II x" - Xo II - 0 as n _ 00. If a y ordered set {x,,} of H, which is such that II x" - Xm II - 0 as n, m - 00, converges to a limit which is an -lement of H, then H is said to be "complete".

1

Problems in Quantum Mechanics

The scalar product of any two ket vectors I u) and I v) of a system is a complex number, written as (v I u). The symbol (v I is sometimes called the "bra" vector corresponding to the ket I v), since a "bra" and a "ket" together form a "bra(c)ket". This somewhat whimsical terminology is due to Dirac.

The set of all ket vectors of a system form a Hilbert space.

2. Operators in Hilbert Space Any procedure A whereby each vector I u) of a Hilbert space H is related to one and only

one vector I v) = A I u) of that space is called an operator of that space. An operator A is said to be "linear" if, for any vectors lu) and Iv), and any complex

numbers A and {t, A(Alu)+{t I v = AAlu)+{tAlv). (1.1)

Two operators A and A+ are said to be Hermitian adjoints if, for any vectors I v) and u) of H,

(vIA+ lu) = (uIAlv)*. An operator A is said to be "Hermitian" (or "self-adjoint") if A = A+. An operator U is said to be "unitary" if UU+ = U+ U = 1. If A is an operator and I u) is such that

Alu) = alu),

(1.2)

(1.3) where a is a real or complex number, then I u) is said to be an "eigenvector" of A with the "eigenvalue" a. The eigenvalues of Hermitian operators are always real numbers. There may exist more than one linearly independent eigenvector of A with the same eigenvalue a. Any linear combination of these is then also an eigenvector of A with the same eigenvalue a. The number of such linearly independent eigenvectors (if greater than one) is called the "degener-acy" of the eigenvalue a. The set of all eigenvalues of A is called the "spectrum" of A.

The spectrum of an operator may be discrete, or may be continuous, or may have a dis-crete part and a continuous part. Let us denote by I nr) an eigenvector of A with the eigenvalue an belonging to the discrete part, and by I ve) an eigenvector of A with the eigenvalue av belonging to the continuous part of the spectrum. (Here rand e are indices which distinguish between degenerate states having the same eigenvalue.)

For the eigenvectors of Hermitian operators, the following orthonormalization relations are satisfied for all values of the indices (provided that for degenerate eigenvalues a suitable choice of eigenvectors is made):

(nr I n'r') = 0nn,orr', (nr I v' e') = 0, (1.4) (ve I v' e') = o( v - v')O(e - e'),

where 0nn" is the Kronecker delta and o(v-v') is the Dirac delta function.

2

Ch.l. The Mathematical Formalism of Quantum Mechanics

If the set of all eigenvectors I nr) and I ve) of the Hermitian operator A spans (by linear combination) the entire space H, this set is said to be complete, and the operator A is said to be an "observable" (the operator A is then capable of having a physical interpretation).

The operator [A, B] = AB -BA is called the "commutator" of the operators A and B. If [A, B] = 0, A and B are said to commute. If two observables A and B commute, they have a complete set of eigenvectors in common, and conversely.

3. The Matrix Representation of Vectors and of Operators For any given vector lu), the number (n lu), where In) is an eigenvector of some observ-

able A, can be regarded as an element having "row number" n ofa one-column matrix (u); the whole matrix is then said to be a "representation" of the ket lu). The bra vector (ul is represented by the Hermitian adjoint matrix whose elements are (u In).

For a given operator Q, the eigenvectors 1m) and I n) of A enable one to define the num-ber Qmn = (m I Q I n), which can be regarded as the element with "row number" m and "col-umn number" n of a square matrix (Q). The whole matrix is then said to be a representation of the operator Q. Note that, in this representation, A itselfis represented by a diagonal matrix.

Properties of representations: (a) The Hermitian adjoints of operators are represented by the Hermitian ad joints ofthe

matrix representations of the original operators. (b) Algebraic relations between vectors and operators lead to equivalent algebraic rela-

tions between their representative matrices (thus, in particular, the equations which define the eigenvalues and eigenvectors of operators become matrix eigenvalue equations).

(c) The trace Tr(Q) of any HermitianoperatorQ, defined as L Qmm' is independent of the m

representation used to define it.

Let us denote by I n), and by I w), eigenvectors of the observables A and B, respectively. The two sets of eigenvectors define two possible representations, {A} and {B} say. The transformation from the first representation (of vectors and operators) to the second can be made with the help of the unitary matrix S = (w I n), thus

(U)B = S(U)A (Q)B = S(Q)A S+.

(1.5) (1.5')

If there is a one-to-one correspondence between the vectors In) and Iw), the matrix S is square and corresponds to a unitary operator U such that In) = U I w). U is then said to define a unitary transformation of vectors and operators. Under a unitary transformation of vectors and operators, all scalar products remain unchanged and all Hermitian adjoints transform into Hermitian adjoints.

If 14 is the state vector of some physical system at a given time t, then the "matrix" (r I 4 = 4>(r), say, with "row number" r = the position vector(s) ofthe particle(s) constitut-

3


ing the system, is called the "wavefunction of the system in coordinate representation". Note that the scalar product of two state vectors can be written, in terms of wavefunctions, as follows:

Problems

1. Derive the Schwartz inequality

l(ulv)1 ~v(ulu)v(vIV), in which lu) and Iv) are any two vectors of a Hilbert space H.

2. Derive the triangle inequalityt

3. Show that the validity of the relation

(3a)

for any arbitrary vectors 1 IX) and 1,8), is a necessary and sufficient condition for the system of orthonormal vectors I U1), I U2), ... , I u.), ... , to be complete.

4. Let S1 be a subspace of a Hilbert space Hand S2 its complementary orthogonal sub-space. Any vector 1 u) can be written as a sum of its projections in the two subspaces, 1 u) = lus)+ I us.) Show that the projection operator PSI' which is such that I us) = PSI I u), is Hermitian, and satisfies the equation P~, = P St.

S. Show that if the subspace S1 of the preceding problem is taken to be the subspace spanned by a single normalized vector I a), then the corresponding projection operator is given by

6. Show that if 1 nr) denotes the eigenvectors of an observable, the following "closure relation" is valid

L Inr)(nrl = 1. n,r

7. A projection operator PM is said to be greater than or equal to another projection operator PN, i.e. PM "'" PN, if the subspace N is contained in M. Show that

(a) The relation PM"'" PN satisfies the axioms required of any relation of inequality. (b) [PM, PN] = 0, t To simplify the writing, the notation I A1U+ A2v) is used instead of All u)+ 1.21 v), Al and ).2 being com-

plex numbers. Then (Alu+Azvl = Ar(ul+A~(vl.

4

Ch. I. The Mathematical Formalism of Quantum Mechanics

(c) The relation PM ~ PNis equivalentto the statement that (uIPMI u) ~ (uIPNlu)forany vector I u) of the Hilbert space.

S. Consider a set of vectors I v), in which v is a continuous index which can take all values in the interval (Vh V2). Show that if the vectors I v) are orthonormal in the sense that (v'I v) = b(v' - v), then the operator

VI

P = f I v)dv(vl V1

is the projection operator of the subspace spanned by the set of vectors I v ). 9. Show that if a unitary operator U can be written in the form U = 1 + ieF, where e is a

real infinitesimally small number, then the operator F is Hermitian. 10. A Hermitian operator A is said to be positive-definite if, for any vector I u), (u I A I u) "'"

O. Show that the operator A = I a) (a I is Hermitian and positive-definite. 11. If A is a Hermitian positive-definite operator (see problem 10), then

l(uIAlv)l".; V(uIAlu)v(vIAlv). Show that Tr (A) ~ 0, and that the equality holds if and only if A = O.

12. Show that the operator F defined by the relation

F = If Is) k(s, 1)(/1 dl ds, in which the kernel k(s, I) is real, is a linear Hermitian operator.

13. Show that the differential operator

Ii d P=Ydx'

(lla)

is linear and Hermitian in the space of all differentiable wavefunctions (x I ) = (x), say" which vanish at both ends of an interval (a, b).

14. The translation operator Q(a) is defined to be such that Q(a) (x) = (x+a).

Show that:

(a)Q(a) may be expressed in terms of the operator P = ~ dd , l x

(b) Q(a) is unitary. 15. Given three operators A, Band C, express the commutator [AB, C] in terms of the'

commutators [A, C] and [B, C].

5


16. Let H(r) be an operator acting on the wavefunctions tp(r), and 0 a coordinate trans-formation operator which acts on the wavefunctions in such a way that Otp(r) = tp(r'). Show that if H(r) is invariant under the coordinate transformation 0, i.e. H(r') == H(r), then [H, 0] = O.

17. Find the expansion of the operator (A - )'B)-l in a power series in )" assuming that the inverse A- l of A exists.

18. Show that if A and B are two operators satisfying the relation [[A, B], A] = 0, then the relation [Am, B] = mAm-l[A, B] holds for all positive integers m.

19. Show that [p, x] = -in, [p, xn] = -ninxn-l, n:> 1,

. dA [p, A] = -ill dx

where p = ~ ~ and A = A(x) is a differentiable function of x. I dx

(l9a)

(l9b)

20. If the characteristic equationf(),) = o for an observable A is given, showthatf(A) = O. 21. Let J u) and J v) be two vectors of finite norm. Show that

Tr (Ju)

Ch.l. The Mathematical Formalism of Quantum Mechanics

Show that d dA dB d)' (AB) = d).-B+A d)' , (25a)

d dA _(A-I) = _A-I_A-I d)' d)" (25b)

26. Show that the operator B(t) defined by the expression

B(t) = eiA/Boe-iAt,

where A and Bo are operators independent of the parameter t, is a solution of the integral equation

B(t) = Bo+i[ A, i B(-r:)d-r: l 27. Show that, for any two operators A and L,

eLAe- L = A+[L, A]+ i! [L, [L, AJ]+ ;! [L, [L, [L, AJ]]+... (27a) 28. Show that eAeB = eA + Be-(1/2) [A, B] if [[A, B], A] [[A, B], B] = O. (28a) 29. Verify Kubo's indentity

P [A, e-PH] = e-PH f e'H[A, H] e-i.H d)', (29a)

o

where A and H are any two operators.

30. Show that a necessary and sufficient condition for two linear operators A and B to be equal (to within a phase factor), i.e. for A = Beia, is that

l(uIAlv)1 = l(uIBlv)1 (30a) should hold for any pair oflinearly independent kets I u) and I v).

31. Show that a necessary and sufficient condition for a linear operator U to be unitary is that the matrix elements (i I U I k) of this operator in a given representation should obey the following equations:

II(iIUlk)12 = 1, I (31a)

I(il Ulh)(il Ulk)* = 0, i

The convergence of the sums may be assumed.

7


Solutions 1. Let us write for a startt (ul u)+(vlv)-2 Re(ulv) = (ul u)+(v Iv)-(ulv)-(v 1 u) =

(u-vlu-v) ~ O. Hence Re (ul v) """ tul u)+(vl v. (1.1)

Ifwe substitute e 1 u) and (lIe) 1 v) for 1 u) and 1 v) (e being a positive real number) then the left-hand side remains unchanged and the right-hand side becomes t(e2(ul u)+ I/e2(v 1 v). Since this expression is greater than Re (u 1 v), its minimum also satisfies the inequality. This mini-mum is attained when e2 = V(v 1 v)/(u 1 u). Hence

Re (ulv) """ v(ulu) v(vlv). (1.2) Let us now subsitute e-ia.1 u) for 1 u), where IX is a real number. The right-hand side of (l.2)is unchanged, while the left-hand side becomes

Re (eia.(ul v = cos IX Re (ulv)-sinlX Im(ulv). This expression has a maximum when

v(Re (ul v)2+(Im (ul V2 = 1 (ul v) I,

and thus the relation to be proved is obtained, viz.,

l(ulv)1 """ v(ulu)v(vlv). (1.3) From the proof it can be seen thatthe equality holds only if 1 u) or 1 v)is zero, or I u)and 1 v) are linearly related.

2. We have that

(u+vlu+v) = (ulu)+(ulv)+(vlu)+(vlv) = (ulu)+(vlv)+2Re(ulv). Using (1.2) we obtain

(u+vlu+v) """ (ul u)+(vlv)+2V(ulu) v(vlv) = (v(ulu)+ V(VIV)2 and thus

v(u+vlu+v) ~ V

Ch. I The Mathematical Formalism of Quantum Mechanics

we suppose that there exists such a vector, we can choose I rx) = 1,8) = I u) and we have (u I u) = 0, whence I u) = 0 which is contrary to the supposition we have made.

From the completeness of the set I U1), I U2), ... , I uv)' it follows that 00

Irx) = L a.lu.), a. = (u.lrx) ,=1 (3.1)

00

1,8) = L b.lu.), bv = (u.I,8) .=1

and thus that 00 00

(rxl,8) = L (rx I Ul') (u.I,8) (UI'I u,) = L (rx I u,)(u.I,8). ~,'=1 ,=1

(3.2)

Remarks: Taking 1,8) = Irx), we have 00

(rxlrx) = L l(rx lu.)1 2 (3.3) ,=1

The equality (3.3) is called Parseval's relation. It is valid for any complete set I U1), ... , I u.), . .. If the set is not complete it can be seen that

00

(rxlrx)".; L I (rx 1u,)l2. ,=1

4. Take the scalar product with I v) of both sides of the equation Psilu) = lusi )

so that (v I PSi I u) = (v Ius}

Similarly

for any I u) and I v), i.e. PSi is Hermitian. Applying the operator PSi to both sides of (4.1) we have that

Plil u) = PSi I USi) = I USi) = PSi I u), for any lu), hence ~i = PSi'

(3.4)

(4.1)

5. In this case we can write I u) = I ua) + I us,). By hypothesis (a I us.) = 0 and I ua) = c I a), c being a complex number. Taking the scalar product of lu) and I a) we have c = (a I u), i.e. I ua) = I a) (a I u), whence the relation to be proved follows.

6. Let P A be the projection operator on the subspace spanned by the vectors I nr). Since A is an observable, we have P A = 1. On the other hand the projection on a subspace is equal to the sum of all projections onto vectors forming a base in this subspace. In particular,

PQM 2 9


PA = L Pnr where Pnr is the projection operator on the vector I nr). Thus we have that n,r L Pnr = 1. But, from the preceding problem, P;.r = I nr) (nr I, and hence

n,r L I nr) (nr \ = 1. (6.1) n,r

Remarks: From the solutions of problems 3 and 6 it follows directly that, for any ortho-normal set of vectors, the closure relation and the completeness condition are equivalent.

7. (a) From the definition it follows immediately that, if PM;;;" PN and PN ;;;.. PM' then PM = PN' and that, if PM;;;" PN and PN ;;;.. PrJ' then PM;;;" PQ'

(b) the inequality PM;;;" PN implies that PMPN = PN and PNPM = PN' and hence that [PM,PN] = O.

(c) In general, for the projection operator Ps" we have that Ps,l u) = I us,), I u) = I us,) + Ius,).

Consider the operator 1 -Ps" then

(l-Ps,)lu) = lu)-Ps,lu) = lu)-Ius') = Ius,) = ps.lu), and consequently I-Ps, is also a projection operator. Now

(uIPs,lu)+(uIPs.lu) = (ul (Ps,+Ps,) I u) = (ulu). But (u I Ps,l u) = (u I p~.1 u) is non-negative, and

(uIPs,lu)",;;: (ulu). We can therefore write (uIPNlu) = (UIPJNPMlu)",;;: (uIPMlu).

(7.1)

Conversely, if I u) is a vector of the complementary space of M we have PM I u) = 0, and, in accordance with the relation (u IPMlu) ~ (uIPNI u), we have PNlu) = O. Then, if lu) belongs toN,(l-PM)lu) = OandPN(l-PM)lu) = O. If lu) does not belong to M, (l-PM)lu) = lu). ButPNlu) = O,and therefore for any lu)we havePN(1-PM)lu) = 0, and hencePN = PNPM' which is equivalent to the inequality PM;;;" PN'

S. For any arbitrary vector I u), we have that v,

Plu) = f I v) dv(v I u), "

and hence ~ VI

(v'I (l-P)lu) = (v'I u)- J (v'I v) dv(vl u) = (v'I u)- J !5(v' -v) dv(vlu) = o. ~ ~

Since PI u) is a linear combination of the I v) vectors, and, as we have just seen, (l-P) I u) is orthogonal to all of these vectors, it follows that P is a projection operator onto the sub-

10


space spanned by the 1 v) vectors (see problem 4). Note that it is sometimes more convenient to write the orthonormalization condition in the form (v 1 v') = f(v)o(v - v'), where f(v) is a real and positive function of v. The vectors so defined are equal to the normalized vectors previously defined, except that each vector 1 v) is multiplied by a constant of modulus vi f(v ). The projection operator on the subspace spanned by the 1 v) vectors is, in this case, given by

(8.1)

9. Since U is unitary, we have that

(l-ieF+) (1 +ieF) = (1 +ieF) (1-ieF+)

and, retaining only the terms to first order in E', it follows that F = F+.

10. We have that

(uIAlv) = (ula)(aiv) = (vla)*(alu/ = (vIAiu)* and also that

(ul AI u) = (ul a)(ai u) = I (uja)12 ~ o.

11. The inequality to be proved can be deduced from the fact that A is positive-definite, by using the same procedure as was used to prove the Schwartz inequality (see problem 1).

Note that Tr (A) = L (n I A I n), where the In) form the (complete) set of eigenvectors of an /I

observable. It follows directly that Tr (A) ~ o. The equality is true if, and only if, for alII n), (n I A I n) = o. If this is the case, then, taking I v) = I n) in (1la), it follows that (u I A I n) = 0 for any I u), and hence that A I n) = O. Since the I n) vectors form a basis we have that A X any linear combination of the 1 n)s = 0, i.e. A = O.

12. To prove the linearity of F is straightforward. To prove that it is Hermitian we write, using the notation (tl4 = 4>(t), (sl1J') = 1J'(s),

(1J'1 F 14 = f 1J'*(s) [f k(s, t) 4>(t) dt] ds = f k(s, t) 1J'*(s) 4>(t) dt ds. Also

(4) I FI1J')* = [f 4>*(s) (f k(s, t) 1J'(t) dt) ds]* = f k(s, t) 1J'*(s) 4>(t) dt ds, whence it follows that (1J' I FI 4 = (4) 1 FI1J')* for any 14 and 11J'), i.e. that Fis Hermitian.

13. We have that b

(1J'lpl4 = ~ f 1J'* ~: dx. a

2* 11


Integrating by parts we obtain

b b

Ii Ii f drp* Ii f d1p (1plp I cP) = ---;- (cP1pY:,--;- -d cP dx = --;- -d cP dx = (cP Ipl1p)* I I x I X

a a

14. (a) Using Taylor's theorem, we can write

00 a" d" 00 I (iap )" ~ap Q(a)cP(n) = cP(x+a) = I - --cP(x) = I - -- cP(x) = eli cP(x). "=0 n! dx" ,,=0 n! Ii

It follows thatQ(a) = exp (iap/li). (b)Q+(a) is defined to be such that, foralll1p) and I cP),

(1pIQI4J)* = (cP IQ+ 11p), whence f cP*(x) Q+(a) 1p(x) dx = f cP(x+a)1p(x) dx. By making the change of variable x+a-->-x in the second integral we obtain

f cP(x)Q+(a)1jJ{x)dx = f cP(x)1p(x-a)dx, for any 4J(x). It follows that Q+(a) 1p(x) = 1p(x-a), hence that Q(a) Q+(a) = Q+ (a) Q(a) = 1.

15. We have [AB, C] = (AB)C-C(AB) = ABC-CAB+ACB-ACB = A(BC-CB)+(AC-CA)B,

i.e. [AB, C] = A[B, C]+ [A, C]B. (15.1)

16. From the statement of the problem we can see that for any 1p(r), O[H(r)1jJ{r)] = H(r')1p(r') = H(r)1jJ{r') = H(r)01p{r)

whence [H, 0] = O. 17. Let the expression we are seeking be written as a series in powers of A, thus:

00

(A -AB)-l = I A"L", (17.1) ,,=0

in which the operators L" are to be determined. Multiplying on the left by A - AB one ob-tains

00 00

I = I A"(A-AB)L" = ALo+ I A"(AL,,-BL,,_l). "=0 ,,=1

12


By equating coefficients of powers of A, it is found that Lo = A-I, Ln = A-1BLn_1, n = 1, 2, ., ., and hence that

(A-AB)-l = A-l+AA-lBA-l+A2A-lBA-lBA-1+ ... (17.2) If [A-I, B] = 0, (17.2) becomes

(A -AB)-l = A-l+ ABA-2+ A2B2 A-3+ .. , (17.3)

18. To prove this theorem we shall use the method of induction. Note that for m = 1, the relation is obvious. Let uS suppose that it is true for an arbitrary m and then deduce its valid-ity for m + 1. Thus, let us suppose that

AmB-BAm = mAm-l(AB-BA).

Multiplying on the left by A, we find that

Am+1B-ABAm = mAm(AB-BA)

and, on adding Am(AB - BA) to both sides, we obtain Am+1B-ABAm + Am(AB-BA) = (m+ 1) Am(AB-BA).

But, according to the conditions of the problem, we have that

Am(AB-BA) = (AB-BA) Am.

(18.1)

By making use of this fact, a relation of the form (18.1) is finally obtained with m replaced by m+1.

19. The first relation (19a) follows directly by applying the commutator to an arbitrary function of x. To establish the second one we use the identity

n-l [A, Bn] = L Bk[A, B] Bn-k-l,

k=o (19.1)

which can be obtained by repeated use of (15.1). Relation (19b) is then verified if A(x) is a polynomial in x or, more generally, ifit is a convergent series in powers of x.

20. We shall show that any vector I u) is transformed by the operator J(A) into the null vector, i.e. that J(A) I u) = 0, and hence that J(A) = is true as an operator relation.

Let I n) be an eigenvector of the observable A with the eigenvalue An' i.e. A I n) = An In). By applying the operator A r times to both sides we find that A r I n) = (AnY I n) and hence,

for an arbitrary polynomial peA), peA) I n) = P(An) \ n). Since any functionJ(A) can be approxi-mated arbitrarily closely by polynomials peA), we conclude that J(A) I n) = J(An) I n) = 0, since the An are the solutions ofJ(A) = 0.

Let I u) be an arbitrary vector. This can always be written in the form I u) = L un In). n

13


Applying the operator f(A) to both sides we have f(A) I u) = L: unfO'n) I n) = O. (20.1)

n

Remarks: The reader can now solve the first part of problem 14 by using the relation (20.1).

21. Let us represent the operator Q = I u)(v I by a matrix, by using the eigenvectors of some observable. The elements of this matrix will then be

Qmn = (mIQln) = (mlu)(vln). Then

Tr(Q) = L:Qnn = L:(nlu)(vln). n n

Using the identity (3.2), the desired result follows. 22. Using the properties of A+ we have that

(uIA+Alv) = (vIA+Alu)*, which expresses the fact that A+ A is Hermitian.

Taking now lu) = Iv), and writing 1(0) = Alu), we have that (ul A+ Alu) = (010);;;" 0,

and hence A+ A is positive-definite. Let I n) be a complete set of eigenvectors of an observable. We have then that

Tr(A+ A) = L: (nl A+ A In). n

Using the closure property L: In) (n I = 1, we find that n

(22.1)

Tr(A+ A) = ~ (njA+ ~Im) (ml Aln) = kn (nIA+lm)(ml Aln) = knl(n l Alm)12;;;.. o. The equality is true only if all the elements (n I A I m) = 0, which is equivalent to the operator relation A = O.

23. Let I n) be an eigenvector of the observable B. Then Tr(AB) = L(nIABln) = LAn(nIAln). (23.1)

n n

But since the observable A is positive-definite, we have (n I A I n) ~ O. To study the sign of An we write

(nIBln) = An(nln);;;.. O. But (n I n) >- 0 and hence An .,.. O.

14


By (23.1) it then follows that Tr(AB);;",O. (23.2)

24. Since the matrix e is Hermitian we can write e12 = A+iB, e21 = A-iB where A and B are real numbers. On expanding the exponentials in power series, the following expression is obtained for U

U = { e(i/2)tp cos ~ e-(i/2)tp sin ~} - e(i/2)tp sin t r(i/2)tp cos t

2 2 The transformed matrix e = UeU+ has the form

I-(1-2en) cos cf> (I-2en) sin cf> +2 sin cf>(A cos 1jJ-B sin 1jJ) +2i(A sin 1jJ+B cos 1jJ)

+ 2 cos cf>(A cos 1jJ -B sin 1jJ) e= (1 - 2en) sin cf> 1 + (1-2en) cos cf>

-2i(A sin 1jJ+B cos 1jJ) -2 sin cf>(A cos 1jJ-B sin 1jJ) +2 cos cf>(A cos 1jJ-B sin 1jJ)

This is diagonal if

i.e. if

(1-2en) sin cf>+2 cos cf>(A cos 1jJ-B sin 1jJ) = 0, A sin 1jJ + B cos 1jJ = 0,

B tan1jJ=-A' 2yA2+B2 tan cf> = . 1-2en

With these results the matrix U is determined and e becomes

where 25. We can write

A(A+e)B(A+e)- A(A)B(A) e

A(A+e) B(A+e) -A(A+e) B(A) + A(A+e) B(A)-A(A) B(A) e

A(A+e) [B(A+e)-B(A)] [A(A+e)-A(A)] B(A) = + .

e e

(24.1)

(24.2)

(24.3)

(24.4)

15


Taking the limit as e tends to zero, we obtain

d dA dB d)' (AB) = d)' B + A d)' .

For the other relation we start from AA-1 = I and differentiate it with respect to

i.e. A dA - 1 __ dA A-I

d)' - d)' .

Multiplying both sides on the left by A-I we obtain

26. The integral equation can be written as t

B(t) = Bo+i f [A,B(r)] dr, o

which is equivalent to the differential equation

~ = irA, B(t)] with the initial condition B(O) = Bo

Using the expression for B(t) we obtain

~ = iAeiAtBoe-iAt -ieiAtBoe-iAtA = iAB(t) -iB(t)A = irA, B(t)]. The initial condition is also satisfied.

27. Consider the operator A(s) = esL Ae- sL ,

where s is a parameter. We have then that

16

(25.1)

(25.2)

(26.1)

(26.2)

(27.1)


Let us differentiate once more, then

d2 ~~s) = [L, d~;S)] = [L, [L, A(s)J], and so on. We shall now write the operator I-Ae-L = A(l) as a Taylor series expansion about the origin

I dA(O) I d2A(O) A(l) = A(O)+TI ~+ 21 dSJ + ....

Since A(O) = A, the required identity follows directly. It can be obtained also in another way, by using the result of the preceding problem.

28. Consider the operator (28.1)

and differentiate it with respect to s:

(28.2)

Since the operators [B, AJ and A commute, we find, by using the results of problem 18, that

[B, AnJ = nAn-1[B, AJ, sn sn [B, e-AsJ = ~ ( _l)n liT [B, AnJ = ~ ( _l)n (n-l)1 An-1[B, AJ = -e-As[B, AJs

and hence that eAsBe-As = B-[B, AJs. (28.3)

The relation (28.3) could have been obtained directly from (27a). From (28.2) and (28.3) we have that

~~) = (A+B+[A, BJs)T(s), (28.4) and T(s) is thus the solution of this differential equation with the initial condition T(O) = I. Since the operators A + B and [A, BJ commute, equation (28.4) can be integrated as if they were merely numbers, to give the solution

T(s) = exp [(A + B)sJ exp {f[A, BJs2}. (28.5) The identity (28a) follows by putting s = I.

29. Let us denote by C(fJ) the left-hand side of (29a) and by D(fJ) the right-hand side. We have evidently

C(O) = D(O) = O. (29.1)

17


Ifwe can also prove that C(f3) and D(f3) satisfy the same first order differential equation, then the identity (29a) is valid. This is in fact the case, since

dC - = -AHe-{JH+He-{JHA = H(e-{JHA-Ae-{JH)-(AH-HA)e-{JH d{J

= HC-[A, H]e-H{J, (29.2) and

dD = HD-e-{JHefJH[A H]e-{JH = HD-[A H]e-{JH d{J , ,. (29.3)

30. The necessity of the condition is evident. To prove its sufficiency, let us consider a representation in which Aij = < i I A Ij) and Bij = < i I B Ij) are the matrix elements of the opera-tors A and B.

By (30a), with I u) = I i) and I v) = Ij), we have that IAul = IBul (30.1)

for any Ii) and Ij). On the other hand, with lu) = Ii) and Iv) = Xjlj)+x1Il), where Xj and x are arbitrary complex numbers, from (30a) we find that

(30.2)

Taking into account (30.1), (30.2) can be written as

Re [xjxi(AjjAil-BjjBil)] = O. (30.3) Since the complex number XjX; is arbitrary, it follows from (30.3) that

(30.4)

From (30.1) and (30.4) we then have that A .. IJ T=-B'l' IJ I

Ail (30.5)

which means that the ratio Aij/Bjj does not depend on j. On repeating the same argument after interchanging rows and columns, we find that the ratio Aij/Bjj does not depend on i either. Taking into account (30.1), we conclude that

(30.6)

where oc is a real number independent of i and of j; i.e. the two operators A and B are equal to within a constant phase factor.

18


31. Let us consider first the necessity of the condition. Let U be a unitary operator. Then for the matrices representing U and U+ we have

UU+ =1 (3l.l) where I is the unit matrix. Since the matrix U+ is the complex conjugate transpose of U, the relations (3Ia) follow directly from (31.1). Conversely, from (3Ia), valid in a particular re-presentation, (31.1) follows. We have to prove only that (31.1) is valid in any representation. Now the transformation from any representation to another is made by a unitary matrix S in the following way (see 1.5'),

U = SUS+, U+ = SU+S+. (31.2) Hence it follows that

UU+ = (SUS+)(SU+S+) = I, (31.3) i.e. (31.1) is valid in any representation. This fact ensures the unitarity of the operator U.

19

CHAPTER II

Simple Quantum Systems

1. The Eigenfunctions and the Energy Spectrum

In quantum mechanics the non-relativistic motion of a particle of mass m, in a potential V(r, t), is described by the time-dependent Schrodinger equation for the wavefunction 'P(r, t), of the particle

.J:. alPer, t) _ [ 112 2 ( ] ITF( I" at - - 2m '7 + V r, t) rl r, t). (II.1 )

For conservative systems, V = V(r), equation (IU) has solutions of the form

(II.2)

which describe dynamical states with a well-defined energy E. Then, because of the linearity ofthe equation, the most general solution 'P(r, t) can be written as

(II.3)

where the values of E are the eigenvalues ofthe time-independent SchrOdinger equation

2m '72.rpE(r)+ (E- V(r))1fJE(r) = 0, (II.4)

and the functions 1fJE(r) are the corresponding eigenfunctions (continuous, differentiable, and bounded at infinity), often called the wavefunctions of the stationary states, or the time-independent wavefunctions, of the system. The coefficients C E are arbitrary constants. The set of all eigenvalues E is called the energy spectrum of the system. The eigenfunctions 1fJE(r) are subject to the orthonormality condition:

f 1fJE(r) 1fJE'(r) dr = beE, E'), (II. 5) 20

Ch. II Simple Quantum Systems

where b(E, E') is the Kronecker delta or the Dirac delta function, according to whether E andE' belong to a discrete ("bound state") or to a continuous ("free state") part of the spec-trum. In the former case, but not in the latter, 1jJE is square integrable. If lJf(r, t) is square inte-grable, the quantity IlJf(r, t)1 2 dr then gives the probability of finding the particle described by YJ(r, t) in the volume element dr at time t, provided lJf(r, t) is normalized to unity, i.e. that

J Il[I(r, t) 12 dr = 1 (11.6)

2. The Transmission of Particles through Potential Barriers According to classical mechanics, a particle approaching a "potential barrier" (i.e. a

region of space in which the potential energy of the forces acting on the particle has a maxi-mum, V max' say) will pass through this region if its total energy E is greater than V max' and will be reflected back if E is less than V max. According to quantum mechanics, on the other hand, such a particle, whatever its energy, has in geneml a finite probability of passing through the barrier, and a finite probability of being reflected from it.

The probability of transmission (or of reflection) can be conveniently expressed in terms of the transmission coefficient T(or the reflection coefficient R), defined as the ratio of the probability flux of the transmitted (or reflected) wave to the probability flux of the in-cident wave, thus

where

R = liRI liIl

j(r, t) = Re [ i: 1jJ*(r, t) 91jJ{r, t)] By conservation of particles, iiI I = liT I + IiR I, and thus T + R = 1.

3. Motion in a Central Field

(11.7)

If the potential energy has spherical symmetry, i.e. if V(r) = V(r), it is convenient to change from Cartesian coordinates x, y, z to spherical coordinates r, 0, cf>. The time-in de-, pendent SchrOdinger equation (11.4) then becomes

{ P~ F } H1jJ(r, 0, cf == 2m + 2mr2 + V(r) 1jJ(r 0, cf = E1jJ(r, 0, cf, (11.8) where Pr is the so-called "radial momentum" operator

P =-ih~ ~r r r ar '

(11.9)

21


and 2 _ 112 [. a (. a ' a2 ] ) - - sin2 () sm eo sm 0 eo) + a4>2 (ILl 0)

is the square of the orbital angular momentum operator, the latter being defined by the rela-tion

1 = -ilzrXv. (ILl 1 ) The Hermitian operators H, 12 and lz = -il1(a/a4 commute among themselves and form a complete set of observables for the motion of a (spinless) particle in a central field.

The simultaneous eigenfunctions of)2 and of lz are the spherical harmonics Yi(O,4, which satisfy the following eigenvalue equations

12Yi(O, 4 = 1(1+ 1) 1z2y,{,(O, 4, lzY'{'(O, 4 = mIzY7'(O, 4.

(II.l2) (ILl 3)

The orbital quantum number I may have any integral value 1= 0, 1,2, ... , and, for each value of l, the "magnetic" quantum number m has the possible values m = 0, 1, ... , l.

The simultaneous eigenfunctions of all three observables H, )2, lz are solutions of the SChrOdinger equation (11.8), and have the form

RE1(r)

"PElm(r, ,4 = --Y'{'( , 4, r

(II.l4)

where the functions RElr) are the solutions of the "radial equation"

d2REl 2m [E- (V(). 1(1+1)112)]R = 0 dr2 + 112 r -f- 2mr2 El , (II.15)

which are bounded at infinity and have the boundary condition at the origin that

REl(O) = O. (11.16) Thus, for motion in a central field, the solutions of the three-dimensional SchrOdinger equation can be found by solving a one-dimensional problem, with an effective potential

1(1+ 1)1z2 Velf(r) = V(r)+ 2 2 '

mr (II.l7)

in the range (0, + 00 ), with the boundary condition (II.16) at r = O.

22

Ch.1I Simple Quantum Systems

Problems 1. Show that if the potential energy VCr) can be written as a sum of functions of a single

coordinate, VCr) = Vl(Xl)+ V2(X2)+ V 3 (X3), then the time-independent Schrodinger equa-tion (II.4) can be decomposed into a set of one-dimensional equations of the form

i = 1,2, 3,

with 1p(r) = 1pl(Xl) 1p2(X2) 1p3(X3) and E = El + E2 + E3. 2. The time-independent wavefunctions, i.e. the solutions of the Schrodinger equation

correspond to bound or to unbound states according to whether they vanish or are merely bounded at infinity. Supposing that lim Vex) = V exists, and that V + V _; (2) if V _ >- E >- V +; (3) if V+ :>E.

3. Show that, in one-dimensional problems, the energy spectrum of the bound states is always non-degenerate.

4. The well-known "oscillation theorem" states that if the discrete eigenvalues of a one-dimensional Schrodinger equation are placed in order of increasing magnitude, El


V(x)

-00 00

-0 o o

FIG. ILL

8. A particle is enclosed in a rectangular box with impenetrable walls, inside which it can move freely (Fig. II.2). Find the eigenfunctions and the possible values of the energy. What can be said about the degeneracy, if any, ofthe eigenfunctions?

I I Ie I I .1----------- -----:J---

0 ... / 0 b y ...

......

...

FIG. 11.2.

9. Find the energies of the bound states of a particle in the symmetrical potential well given by V(x) = -Yo if Ixl - a (Fig. II.3), where Vo is a positive quantity.

10. Find the energies of the bound states of a particle in the potential well given by Vex) = + 00 if x


V(xl

@ -0 o ------0 ----r

r FIG. II.3.

V(x)


12. Solve the SchrOdinger equation for the potential shown in Fig. 11.6. Write down the condition which gives the possible energy eigenvalues of a particle in such a potential.


14. Show that the energy spectrum of a particle in the periodic potential of Fig. 11.8 has a band structure.

v(x)

1 -b 0 Q

FIG. 11.8.

15. Solve the SchrOdinger equation for the Poschl-Teller potentialt

Vo V(x) = , cos2


Compare this probability with the quantum-mechanical one

where _ ( h )11" xo- - .

mw

V(x)

---\-----+------t--E1 ct n w

-a

FIG. 11.9.

19. Calculate the possible energy values of a particle in the potential given by V(x) = 00 if x ... 0, and V(x) = m;2 x2 if x >- o.

20. A system described by the Hamiltonian

h2 m H = -2m V'2+2(co~x2+roiy2+co~Z2) (lOa)

is called an "anisotropic harmonic oscillator". Determine the possible energies of this system, and, for the isotropic case (COl = CO2 =

COs = co), calculate the degeneracy of the level En' 21. Find the wavefunction of a particle in the homogeneous field Vex) = -kx. What

can be said about its energy spectrum?

22. The conduction electrons in metals are held inside the metal by an average potential called the inner potential of the metal. Calculate, for the one-dimensional model given by V(x) = - Vo if x - 0 (Fig. 11.10), the probability of reflection and of transmission of a conduction electron approaching the surface of the metal with total energy E, (i) if E >- 0, and (ii) if - V 0 -< E -< o.

28


Vacuum

x

-c

FIG. II.IO.

23. A beam of mono-energetic electrons strikes the surface of a metal at normal incidence. Calculate the reflection probability of these electrons if E = 01 eV and Vo = 8 eV.

24. In problem 22 it was supposed that, at the metal-vacuum interface, the potential energy of the electrons jumps from - Vo to O. Actually this change is continuous over an interval a, whose dimensions are of the order of the interatomic distances in the metal. The potential energy near the surface of the metal can thus be written approximately as

Vex) = Vo l+eXla (24a)

(Fig. 11.11), which, as a -+ 0, approximates to the previously used discontinuous potential. Calculate the reflection probability of a conduction electron approaching the surface of the

~WWro-~


25. Calculate the transmission and the reflection coefficients of a particle having total energy E, at the potential barrier given by V(x) = 0 if x -< 0, V(x) = Vo if 0 -< x -< a, V(x) = 0 if x:> a, for the cases E:> Vo and 0 -< E -< Vo (Fig. 11.12).

vex)

Vot---"I

ECD ----------

x

o a

FIG. 11.12.

26. A particle of total energy E enters the barrier V = V(x) (Fig. 11.13) at the point x = Xl and leaves it at the point x = X2 (the "tunnel effect"). Assuming that the potential energy curve V(x) is SUfficiently smooth, let us divide the interval [Xl, X2] into intervals of length Llxi, large compared with the relative penetration depth di = h[8m(V(xi) _E)]-1/2 of a particle in the rectangular barriers so obtained. Calculate in this approximate way the transmission coefficient T for the whole barrier V = V(x), knowing that Ti ~ exp [ - ! V 8m(V(x) -E) LlXi] for the ith rectangular barrier (see problem 25).

vex)

E----:~

x

FIG. 11.13.

27. It has been shown experimentally that, under the influence of a strong electric field normal to the surface of a metal, there is a flow of conduction electrons out of the metal (the "cold emission" effect). According to classical electrodynamics, those electrons can leave the metal which have enough energy to surmount the potential barrier produced jointly by the electrical image force -e2 j4x2 (which acts upon an electron at a distance x

30


outside the metal) and by the force et due to the external applied field. Although this expla-nation is qualitatively correct, the quantitative results of classical calculations based upon it are in complete disagreement with the experimental data. This disagreement is resolved when a quantum phenomenon-the tunnel effect-is taken into account. Supposing, for simplicity, that the electric field is homogeneous, determine the dependence of the cold emission current (at a given temperature) on the magnitude of the applied field.

28. The fact that ex-particles having energies of a few MeV can leave potential wells with depths of tens of MeV (inside which they find themselves in radioactive nUclei), can be explained by the tunnel effect. Using a simplified model, let V{r) = - Vo if r -< Ro, VCr) = e1e2 if r :> Ro (Fig. 11.14), and calculate Gamow's factor for this barrier, i.e. the

r

transmission probability for ex-particles of energy Ethrough the barrier. Express the result in terms of the final velocity of the ex-particle, and estimate the mean life of an ex-emitting nucleus.

VCr)

FIG. 11.14.

29. Find the eigenfunctions of a free particle, in the limiting case of motion in a central field in which the potential VCr) -+ O. Compare these eigenfunctions, based on the complete set of observables H, )2, iz , with the "plane wave" eigenfunctions, in which the motion is

characterized by the observables Px, PY' Pz' and H = in, , which also form a complete set of observables for a spinless free particle.

30. Find the possible energies of a particle in the spherical potential well given by VCr) = -Vo if r -< a and VCr) = 0 if r :> a (Fig. 11.15).

31. Find the energy levels of a particle in the central field

VCr) = A +Br2 r2 ' (3Ia)

where A and B are positive constants.

31


VIr)

o ----3r-------__

-V. 1-----'

FIG. 11.15.

Show that, in the particular case A = 0, B = j-nu.o2, the levels are the same as those found in problem 20 for an isotropic three-dimensional oscillator.

32. Show that in quantum mechanics (as in classical mechanics), the problem of the motion of two interacting particles of masses ml and m2 can be reduced essentially to the problem of the motion of a single particle of effective ("reduced") mass m = mlm2/(m1 + m2) moving in the potential V(r) of the mutual interaction of the two particles.

33. Calculate the energy levels and eigenfunctions of a hydrogen atom. Discuss the degeneracy of these levels.

34. By analogy with classical mechanics, a system described in quantum mechanics by the Hamiltonian

(34a)

is called a "rigid rotator". Here I is the (constant) moment of inertia of the rotator. Deter-mine the corresponding energies and eigenfunctions. What is the degeneracy of the energy levels?

Solutions

1. Substituting the trial function tp(r) = tpl(Xl) tp2(X2) tps(xs) into equation (11.4), and dividing by tpltp2tpS, we obtain

s (1 cAp; 2m) 2m .r -: d;!-~VI =-u;E. ,=1 tp, X,,, "

32


Because the terms in each bracket of the sum contain independent variables, the equality can be valid for all (Xl, X2, xs) only if each bracket is a constant, i.e. if

1 tf2tp i 2m 2m - d 2 -""""i2 Vi = -""""i2 E;, "Pi Xi n n

i=1,2,3,

where the Ej are constants, and E = EI + E2 + Es. In particular, if VCr) = VI(XI), the function "PI(XI) is given by the one-dimensional

SchrOdinger equation cJ2"P1 2m dx~ + 7z2 (EI - V l)tpl = 0,

and "P2(X2) and "Ps(xs) are solutions of the free particle equations cJ2!pj 2m

dx~ +7z2Ei"Pj = 0, I

i = 2,3.

2. In the first case the difference E - V(x) is positive at both ends of the interval ( - 00 , + 00 ), and thus the eigenfunctions in these regions oscillate indefinitely between finite bounds and therefore correspond to unbound states. The energy spectrum is continuous and doubly degenerate. In the case V _ :> E :> V +, the difference E - V(x) is negative in the limit X -+ - 00, so that in this asymptotic region only one of the two linearly independent eigen-functions is bounded (in fact, it has a decreasing exponential). In the other asymptotic region, where E - Vex) is positive, this eigenfunction oscillates indefinitely between finite bounds and in consequence it corresponds to an unbound state. The energy spectrum is continuous and non-degenerate. In the third case the difference E - V(x) is negative in both asymptotic regions. The bound state solution, if it exists, must approach zero exponentially at both the limits x -+ 00. It can be shown that such a solution exists only for discrete values of E. The case in which V + :> V_is similar to the one in which V + -< V _, and does not add anything new to the above conclusions. If V + = V _ = VO' the eigenfunctions correspond to a bound state ifVo :> E, and to an unbound state in a continuous and doubly degenerate energy spectrum if E :> Vo.

3. For the sake of argument let us suppose that the opposite is true. Let "PI(X) and "P2(X) then be two linearly independent eigenfunctions with the same energy eigenvalue E. From the equations

,,2m( \, .. 0 "PI + 7z2 E - V"f'l = , " 2m ( \, .. "P2 +7z2 E-V''Y2 = 0,

we obtain

I.e. " " (')' (' )' 0 "PI "P2-"P2 "PI = "Pl'1fJ2 - "P2"PI = .

33


After integrating this equation we find that

Since, at infinity, "PI = "P2 = 0 (bound states), we must have the constant = 0 and hence

Integrating once more we have In "PI = In "P2 + In c, i.e. "PI = C1p2, which contradicts the assumed linear independence of the two functions.

4. Consider two eigenfunctions "Pix) and "Pn+1(x) with eigenvalues En -< En+l . Since the spectrum is supposed to be non-degenerate, all eigenfunctions can be taken to be real, by a suitable choice of phase factors. From the equations

after some simple calculation we obtain {J

["P~"Pn+1 -"P~+I"Pn I = !~ (En+1 -En) f "Pn"Pn+1 dx. IX

Let us take ce and fJ to be two consecutive zeros of"Pn (n ;;.. 3). Then {J

[ "P~"PIl+I I = !~ (EIl +I -En) f "Pn"Pn+1 dx. IX

Now, in the interval (ce, fJ), "Pn does not change sign; suppose, without loss of generality, that "Pn :> O. This means that "P~(ce) :> 0 and "P~(f3) -< O. It follows that "Pn+1(x) must change sign in the interval (ce, fJ), since otherwise the right-hand side of the equality would have the same sign as "Pn+1' while the left-hand side would have the opposite sign. Hence, between two consecutive zeros of "Pn' "Pn+1 has at least one zero. In this connection note also that eigenfunctions which are even (odd) with respect to the reflection x -+ -x have an even (odd) number of zeros, and that the ground state is always even.

5. Since H(x) = H( -x), we have H"P(x) = E"P(x), H"P( -x) = E"P( -x), (5.1)

i.e. "P(x) and "P( -x) are eigenfunctions of H with the same eigenvalue E. We distinguish two cases:

34


(1) The level E is non-degenerate. In this case tp(x) = Ctp( -x), and hence tp(x) = tp( -x), that is, the eigenfunctions

corresponding to non-degenerate energy levels are either even or odd. Suppose now that all the energy levels are non-degenerate. Then, if we write the energy eigenvalues in increas-ing order of magnitude, E1 -< E2 -< E3 ... , the corresponding eigenfunctions will occur in increasing order of the number of their zeros, the function corresponding to En having n-l zeros (see problem 4). Since the even (odd) functions have an even (odd) number of zeros, it follows that the eigenfunctions will be alternately even and odd, the ground state wavefunction being always even.

(2) The level E is degenerate. The degeneracy being twofold, the general solution of equation (5.1) can be written as

C1tp(X)+C2tp( -x) = A [tp(x)+ tp( -x)]+B[tp(x)-tp( -x)] = Atpe(x)+Btpo(x) where A + B = C h A - B = C 2. Thus the two eigenfunctions having the same eigenvalue can be written in the form of a linear combination of two functions of well-defined parity, which are themselves eigenfunctions with the same eigenvalue.

6. Consider a potential V(x) having a finite discontinuity at x = Xo (Fig. 11.16). In the interval (xo -d, xo+d), by replacing V(x) by the line segment shown in the diagram,

v(x) VI (x)

FIG. II.16.

one obtains a continuous potential V1(X). The Schrodinger equation then becomes

whence

(V'i) ... +r( ~i) ... -, ~ -: T' (V,(x) -E) 'I'(x) dx. xo-d

35


As d -- Owe have VI (x) -+ V(x), tpl(X) -- tp(x), wheretp(x) is the solution which corresponds to the discontinuous potential V(x) and the same eigenvalue E. Because the integrand is bounded we find that (tp')xo+o = (tp')xo-o in the limit d -- O. Note that at a point where the potential V(x) has an infinite discontinuity, the derivative wavefunction does not have the above property. The condition of continuity of tp(x) and of its first derivative is equivalent to the condition of continuity of the "logarithmic derivative"

d dx In tp(x) = tp'(x)/tp(x) of tp(x).

7. For I x I >- a, the solutions must be identically zero. Inside the well the particle states are bound and the energy spectrum is non-degenerate. Because V(x) = V( -x) the solu-tions of the equation tp"(x)+k2tp(x) = 0, where k2 = (2m/Iil-)E, will have well-determined parities, tp(e)(x) = A cos kx, tp(O)(x) = B sin kx. The continuity conditions tp(e)(a) = 0 and tp(O)(a) = 0 will be satisfied if ak = nn/2, where n = 1,3,5, ... , and n = 2,4,6, ... , respectively. Hence

{

Acos ~ x = tp~), for n odd tpn(x) =

. nn (0) c B sm Tax = tpn, lor n even.

The possible values of the energy are

En = ;;, k2 = ~:;;2, n = 1, 2, 3, ... In this example, the results of problems 4 and 5 can be verified directly. Thus, the eigen-functions which correspond to the levelsE l


The Schrodinger equation (11.4) can be decomposed into one-dimensional equations (see problem 1), whose solutions are

tp1(X) = A sin k1x+ B cos k1x, 0 < X < a tp.J..y) = C sin k2)'+ D cos k2)', 0 < Y < b tpa(z) = F sin kaz+G cos kaz, 0 < z < C,

h k2 k2 2 2m f) were 1+ 2+ka = /FE and tp1(X)tp2\Y tpa(z) = tp(x, y, z). Because at the walls tp = 0, the eigenfunctions normalized to unity are given by

The energy of the particle is thus quantized, the possible values being given by

If the ratio of any two sides is an irrational number, all the energy levels are non-degen-erate. Otherwise the energy spectrum is in general degenerate. Thus, e.g., if a = b = c, the level for which ni +,; + ni = 6 is threefold degenerate, since three linearly independ-

. f . h h . 1 6'Jr,2/j2 ent elgen unctIons ave t e same elgenva ue: E121 = E112 = E211 = -2 2. The ground

ma

state E111 is, however, always non-degenerate. 9. If bound states exist, the particle's total energy E in such states will be in the range

- Vo < E < O. The eigenvalue equations are then

112m I ) tp1 (x)-h2IE tp1(X = 0,

tp;'(X)+ !r: (Yo-lEI) tp2(X) = 0, tp~' (x) - !r: I E I tpix) = 0,

if x >- a

if Ixl


For'IjJ to describe bound states, we have to take A = G = (the vanishing condition at in-finity). We can then see a typical quantum-mechanical effect. Since / 'ljJl/ 2 = JJ2e-2a.x .,,= 0, and / 'ljJa/2 = Pe2a.x .,,= 0, the particle may be found outside the potential well, with a probability which decreases exponentially with distance (Fig. 11.17).

-0 o

FIG. 1I.17.

From the continuity conditions at x = a we obtain

2C sin pa = (B-F)e-a.a } 2pC cos pa = -IX(B-F)e-a.a 2D cos f3a = (B+F)e-a.a }

2pD sin pa = IX(B+ F) e-a.a

IfC .,,= 0, and thus B .,,= F, we obtain from (9.1)

p cot f3a = -IX. If D .,,= 0, and thus B .,,= -F, we obtain from (9.2)

fJ tan fJa = IX.

x

(9.1)

(9.2)

(9.3)

(9.4)

The relations (9.3) and (9.4) cannot both be satisfied at the same time. We have therefore to distinguish two classes of solution:

(1) C = 0, B = F and p tan pa = IX, (2) D = 0, B = -F and p cot pa = -IX.

In the/first class the eigenfunctions are even and in the second class they are odd, a result which follows directly from problem 5.

The corresponding energy values are given by the solutions of the transcendental equa-tions (9.3) and (9.4), which can be solved graphically as follows. Put X = pa, Y = lXa; the

38


energy levels E = -~ Y2 are then obtained from the intersections (if any) of the curves 2ma2 2r.na2 or

XtanX=Y I X2+Y2 = --w:- Vo

XcotX= -Y

X2+y2 = 2ma2_ Vo /12

in the region X >- 0 and Y >- 0 (Fig. 11.18). y

FIG. 11.18.

I respectively

We can see from Fig. 11.18 that the number of bound states increases as the product a2Vo (the "well parameter") increases, and is finite if a2Vo is finite. It can also be seen that if N (N+l) (2m )1/2 Tn ~ R < -2- n, where R = a2Vo and N = 0, 1,2, ... , then the number of bound states is N + 1 (for N = 0, the condition becomes 0 < R < nI2).

10. The energies of the bound states wil1lie in the interval - Vo < E < 0, and the eigen-functions will have the form

\

tp1 = Be-a.x, tp(x) = tp2 = C sin (:Jx+D cos (:Jx,

tpa = 0,

if x >- a if O


x

FIG. 11.19.

From the continuity conditions at x = 0 and at x = a one obtains D = 0, and hence

/3 cot /3a = -~. (10.1) Since this transcendental equation is the same as (9.3), the possible energy values in the present problem will coincide with a part of those found in the previous problem. However, the wavefunctions are now neither even nor odd, since V(x) ?"" V( -x).

We observe that, if R = c; voa2f'2 < ; , i.e. if a2Vo < 1hc2/8m, then the potential well cannot bind the particle; the condition for the existence of at least one bound state is

1hc2 a2Vo ;'= 8m . (10.2)

11. From the behaviour of the wavefunction in the asymptotic regions x-- 00, it follows that the energy spectrum of the particle is either discrete, or continuous and non-degenerate, or continuous and doubly degenerate, according to whether 0 < E < V!, or VI < E < V2, or E >- V2 respectively (see problem 2). The energy of the particle is thus quantized in the range 0 < E < VI and the corresponding eigenfunctions will have the form

where

[2m ]1/2 /31,2= 7t2(Vl,2-E) ,

x>-a

O


Eliminating ifJ and introducing the notation

q = (2m~1)1/2, ~ = : = V ~, cos y = V~: the following transcendental equation is obtained:

nn-aq~ = arc sin ~+arc sin (~cos y) (11.1) (here n = 1,2,3, ... , and the values of arc sin are to be taken in the interval (0,n/2). The roots of equation (11.1) give the energy levels E = V1~~' When E increases from 0 to Vl' ~ increases from 0 to 1, the right-hand side of equation (11.1) increases from 0 to (n/2)+ arc sin (cosy) = n -y, and the left-hand side decreases fromnn to nn- aq. Equation (11.1) can be solved graphically, by determining the abscissae of the intersections of the curve given by 0


energy levels in this potential well is finite. Observe that any bound particle can be found outside the well with a non-vanishing probability. In the particular case VI = V2, we have cos y = 1, Y = 0, and the condition (11.2) is certainly satisfied, at least for n = 1, which means that there always exists a bound state in this case (see also problem 9).

12. The solution of the one-dimensional Schrodinger equation satisfying the boundary conditions tp(O) = tp(c) = 0 can be written in the form

I Al sin klX, if 0 < x < a tp(x) = A2 sin k 2x+ B2 cos k2X, if a < x < b Aa (sin kax-tan kac cos kax), if b < x < c, (12.1) where

[2m ]1/2 k; = /i2(E-V;) , i = 1,2,3. (12.2)

The continuity conditions at x = a and at x = b give the system of homogeneous equations:

Al sin kla-A2 sin k2a-B2 cos k 2a = 0 } A1k1 cos kla-A2k2 cos k2a+B2k2 sin k 2a = 0 (12.3)

A2 sin k2b+B2 cos k2b-Aa (sin kab-tan kac cos kab) = 0 A2k2 cos k2b-B2k2 sin k2b-Aaka (cos kab+tan kac sin kab) = O.

Non-trivial solutions for the variables AI, A 2, Aa, B2 exist only if the determinant of the coefficients of these variables in (12.3) vanishes, which gives

ka cos ka(c-b) [k2 sin k1a cos k 2(b-a)+k1 cos k1a sin k2(b-a)] = k2 sin ka(c -b) [k2 sin k1a sin k2(b -a) -kl cos k1a cos k(b -a)]. (12.4)

From (12.2) and (12.4) the possible values of the energy of the particle can be found. We mention that for E - V; < 0, i.e. for imaginary k j , the trigonometric functions of (12.4) become the corresponding hyperbolic functions.

13. Because this is the particular case b = C, V2 = Va of problem 12, the energy eigen-value equation (12.4) becomes

kl cot k 1a+k2 cot k2(b-a) = 0, (13.1) and the corresponding eigenfunctions satisfying the boundary conditions tp(O) = tp(b) = 0 are

I tpl = A sin klX, tp(x) = 2 = A sin k1a tp sin k2(b -a) if O


the following expression for the amplitude A is obtained

(13.3)

The probabilities WI and W2 of finding the particle in the intervals 0 -< x -< a and a -< x -< b respectively are given by

and W2 = 1 - WI = 1 -~ (1 _ sin 2kIa) A2. 2 2kIa

In the particular case ; (V 2 - V 1)a2 = (kIa)2 -(k2a)2 = 1, b -a = a (i.e. the two intervals have equal widths), equation (13.1) becomes

(13.4) and then

The transcendental equation (13.4) can be solved numerically, the smallest eigenvalue being obtained for k 2a = 1388, kia = 1710. In this particular case WI = 055 and W2 = 045.

Note that if VI -< E -< V2, then the eigenfunctions are no longer periodic in the region a -< x -< b. Substituting k2 = ik, we find from (13.1)-(13.3) that

ki cot kIa+ k coth k(b -a) = 0,

I tpi = A sin kIx, tp(x) = sin kia . tp2 = A sinh k(b-a) smh k(b-x), if O-


will also be periodic, in the sense that tp(x+d) = Ctp(x) (14.2)

where Cis a constant factor. From (14.2) it follows thattp(x+nd) = Cntp(x). If lei - 1, then along one direction of the x axis, tp(x) increases (decreases) without bound. Solutions having a physical meaning can be obtained only if C is a phase factor, e.g. C = ei9, where ifJ is real. Then Ie! = 1 and

tp(x+d) = ei9tp(x). In the period -b


We have now to distinguish two cases:

(1) O


The energy spectrum will also, in this case, have a band structure. The relation rJ.a + f3b = nn, n = 0, 1, 2, ... , which would lead to a discrete spectrum does not satisfy the inequality

(14.7). Indeed, by substituting arJ. = n ; -fjj and bf3 = n ; +fjj into (14.8), we find that

I(E! ~ { which is contrary to (14.7).

1 + (rJ.-f3): sin2{]> >- 1 2rJ.f3

_ 1 _ (rJ. -(3)2 cos2 fjj -< -1 2rJ.f3

for n even,

for n odd,

The results obtained above for rectangular potentials can be generalized for any periodic bounded potentials, such as occur, for instance, in ideal crystalline lattices.

15. By writing the positive constant Vo in the form

fz2 Vo = 2m rJ.2).().-1), ).:> 1, (15.1)

the Schr6dinger equation becomes

(15.2) where k2 = 2mE/fz2

Changing the independent variable x in (15.2) to y given by y = sin2 rJ.x (15.3)

we obtain the equation

y(l-y)-+ --y -+- -- tp = O. d~ (1 ) dtp 1 [k2 ).().-1)] dy2 2 dy 4 rJ.2 1 - y (15.4) The possibility of introducing the new variable y requires discussion, since (15.3) does not establish a one-to-one correspondence between the variables x and y. Indeed, it trans-forms each point of the complex plane of x into points on a Riemann surface having an infinity of sheets in the plane of y. For Vo :> 0 (the upper half of Fig. II.22) the periodic potential (l5a) consists of a series of valleys separated by walls of finite width and infinite height. Since these walls present impenetrable obstacles to the particle, we can restrict our-

selves, in the study of the eigenvalue problem, to the range - ; -< rJ.x -< + ; , and use the

boundary conditions tp( ;) = O. Since the range ( - ; , +;) of rJ.X corresponds to a single sheet of the Riemann surface of y, the transformation (15.3) will, with this restriction (warranted by the above physical considerations), be one-to-one. This argument is not

46


trivial, since for E:> 0 and 0 -< }. -< 1, i.e. Vo -< 0 (the lower half of Fig. 11.22), the particle can move over the whole of the plane - 0:> -< :{x -< + 0:>, and the above argument is then invalid.

V(x)

.".

2"

FIG. 11.22.

Introducing a new function f(y) through the relation 1p = (1_y)A/2 fey), (15.5)

we obtain from (15.4) the equation

y(l-y)-+ --(}.+l)y -+- __ }.2 f= O. d2f [1 ] df 1 (k2 )

dy2 2 dy 4 !X2 (15.6)

The general solution of this equation can be written in the form (A.53), f = C1F(a, b, c; y)+C2y1-c F(a+ 1-c, b+ 1-c, 2 -c; y), (15.7)

where 1

c=2' (15.8)

According to (A.54), both F functions have a singularity at y = 1 of the form (l_yy-a-b = (1_y)1/2-,'. Thus the function 1p, which differs from f(y) by a factor (1_y)A/2, diverges like (1_y)(1-A)/2 near y = 1, i.e. !Xx = n/2. At these points, however, we must have 1p = O. But from (A.54) and (A.51) it follows that near y = 1 the function 1p behaves like

1 )(1-,1)/2 [ C t1(c) C 2T(2 -c) ] T( ) (-y T(a)T(b)+T(a+1-c)T(b+1-c) a+b-c. (15.9)

47


Hence the coefficient of (1_y)O-J.)/2 must vanish. Note that all the above expressions are un-changed if IX -+ -IX, with the exception of (15.8), in which a = b. The results which will be obtained below for a chosen value of b can thus be transformed, using the transformation IX - -IX, which does not change the physical problem, into the results which would be ob-tained for a chosen value of a. Now, since the function F(z) has simple poles at z = 0, -1, -2, -3, ... , the quantity (15.9) will vanish identically either if b = -n (n = 0,1,2,3, ... ), and C2 = 0, or if b+ l-c = -n and Cl = o. In the first case we obtain

(15.10) where

(15.11)

and in the second case

VJ=(1_Y)A/2yl/2F[~ (A+l+:), ~ (A+l-:), ~;Yl (15.12) where

(15.13)

Thus the corresponding eigenvalues and eigenfunctions are

E2n = -A(;~1)(A+2n)2, E2n+1 = A(;~1)(A+2n+l)2, (15.14) VJ2n = cosA IXx.F(A+n, -n, f; sin2 IXx),

VJ2n+l = cos' IXX. sin IXX .F(A+ 1 + n, -n, t; sin2 IXX). (15.15) 16. Arguments similar to those used in the preceding problem give VJ = 0 for x = 0 and

for IXX = n12. Using the notation

1.:2~2 1.:9 __ 2 2 f/ ~ 1 ) rt-rL 1 1 I ) k2 = ~ E VI = 2m 1')(1')- , V2 = 2m 1'.(1'.- , 112' (16.1 )

where 'rI, A >- 1, the corresponding Schrodinger equation with the new variable

y = sin2 IXx (16.2) becomes

y(1 _ y) _cfLtp_ + (~_ y) _dVJ_ + ~ [_k2 _'rI --=--=(-,-'rI -_1-,-) dy2 2 dy 4 1X2 Y

A(A-l)] = 0 1 VJ -y (16.3)

Let us change now to a new functionf(y) defined by the relation VJ = y'1/2(1_y)Af2 fey). (16.4)

48


Forf(y) the following equation is obtained

y(l-y) ~; + [ (rJ+ ~) -Y(rJ+ A+ 1)] X + ! [ :: -(rJ+ A)2]f = 0, (16.5) whose general solution can be written in the form (15.7) with

17+A k a=-2-+~' b = 17+A_~ 2 21X ' (16.6)

Near y = 1, i.e. IXX = n12, both F functions behave like (1-y)1/2-", so that "P '" (1- yyt-J.)/2. Near y = 0, i.e.lXx = 0, we have f = C1 +C2Y1- C = C1 +C2Y1/2- I1, and hence "P = ClY I1/2 + C2Y(1-I1)/2. Since rJ >- 1, the boundary condition '1p(0) = requires that C2 = 0. Thus both boundary conditions will be satisfied if we choose C2 = 0, and b = (rJ+A)/2-kI21X =-n (n = 0, 1,2, ., .). The energy eigenvalues will thus be given by

n21X2 En = 2m (17+ A+ 2n)2 ,

and the corresponding eigenfunctions by

"Pn = yl1/2(I-y)A/2F(rJ+ A+ n, -n,17+t;y) = sin l1 lXx cos" IXX F(17+ A+ n, -n, rJ+ t; sin2IXx).

17. Introducing the dimensionless quantities

~ _ x A = 2E - Xo ' nro '

where Xo = (h/mro)1/2, the Schr6dinger equation for the oscillator

becomes

(17.1)

Energy levels are obtained only for those values of the parameter A for which the solutions of equation (17.1) vanish at infinity. We seek these solutions in the form

(17.2) where "Poo(~) = exp( -i~) satisfies (17.1) in the asymptotic regions ~ --+ =, in which (17.1) reduces to

and tends toward zero as ~ --+ =.

49


By substituting (17.2) into (17.1), the following diffetential equation for u(~) is obtained d2u du d~2 -2~ d~ +()'-l)u = O. (17.3)

Looking for a solution of the form

(17.4)

the following recurrence relation for the coefficients C k is found:

(17.5)

Now, for large values of k, Ck +2 ~ (2/k)Ck The latter recurrence relation holds also for the coefficients of the series

1 exp (~2) = I _k_~k.

k=O, 2, 4, .. ( 2 ) ! (17.6)

Hence, for the same range of values of k, it follows that as ~ --+ 00, u(~) behaves like exp(~2) and the wavefunction 'If' diverges like exp( ~2). To avoid this divergence the series (17.4) has to reduce to a polynomial. If C n ~ 0, we can have C k +2 = for all k """ n only if), = 2n + I, from which follows the well-known quantization of the energy of the linear oscillator:

Er. = hro(n+t), n = 0,1,2, .. , (17.7) To calculate the corresponding eigenfunctions, the coefficients Co and C 1 must be deter-mined. Since, for the harmonic oscillator, V(x) = V( -x), the eigenfunctions corresponding to bound states of energy En will be either even or odd, which corresponds either to C 1 = 0, Co ~ 0, or to C 1 ~ 0, Co = 0, respectively. Taking Cn = 2n (which means in effect specify-ing Co or C 1), the recurrence relation (17.5) with), = 2n+ 1 gives for the solution u the Her-mite polynomial

Hnm = (2~)n-n(nl~l) (2~)n-2+ n(n-1)(;~2)(n-3) (2~)n-4_ (17.8)

The polynomial Hin satisfies equation (17.3) with), = 2n+ 1, i.e.

d2Hn t dHn d~2 -2" d~ +2nHn = O. Hi~) can also be written in the form

50

dne-"" Hn(~) = ( -l)ne"" (ffI'

(17.9)

(17.10)


Consequently the time-independent wave functions of the oscillator are

(17.11)

where the constant Nn is determined from the normalization condition

Integrating by parts n times, we obtain

But and

so that, with the exception of a possible phase factor,

and consequently

(17.l2)

The orthonormalization condition then holds for these wavefunctions, i.e.

+00 f "Pin"Pn dx = (jm,,' (17.13) 18. Since the general solution of the equation of motion of a classical oscillator, x + ro2 x

= 0, is of the form x = C sin (rot+ cP), the total energy

of such an oscillator is given by E1 = mm2c2/2. Since T""" 0, we have E1 """ V, which means that, classically, the particle can be found

only in the range -a",," x ",,"+a. At the ends of this interval, where E1 = V, its kinetic energy vanishes; the points x = a are called "turning points". Accordingly, C2 = a2 = 2E1/mm2 = 3fz/mro. The classical probability of finding the particle in the interval (x, x+dx)

51


is proportional to the time dt which it takes to pass through this interval. If the period of oscillation is T = 2n/ro, then

dt ro dx OJ dx 1 ( X2)-1/2 Wc\(x) dx = 21' = 'on T = Jt am cos (rot+c/J) = na 1- a2 dx,

which is the required expression. wlx)

x

FIG. II.23.

It can be seen that this probability is greatest at the turning points x = a (Fig. 1I.23). According to quantum mechanics the probability of finding the particle in the interval (x, x+dx) is

W (x) dx = 2n-1I2 x-3X2 exp (_ X2) dx qu 0 x~'

It should be noted that Wqu(x) has maxima near the classical turning points (a = V3h/mro, a' = V h/mw), but, in contrast with the classical case, it does not vanish beyond these points. This phenomenon, of the penetration of a particle into regions with "negative kinetic energy" (I x I :> a), does not lead to any contradiction because the equality E = T + V in quantum mechanics is not a simple relation between numbers, but between operators; the kinetic and the potential energies cannot in fact be determined simultaneously.

For higher levels, it is found that the curve 2Ui c\(x) becomes the envelope of the peaks of Wqu(x) in the classical limit n --+00 (cf. Fig. 1I.24, which represents WquCx) = /"PIO(X) /2, a = V21h/mw).

19. Since the particle cannot penetrate into the range x -< 0, the eigenfunctions of the corresponding Schrodinger equation have to vanish for x = 0. On the other hand, in the range x> 0, these eigenfunctions are the same as those of the harmonic oscillator. Hence

52

Ch. II

WcL ,-1---

J

I n I I

I

I I I

: \ t-

,

I

I

I

It: [I I I

I

I

-a

:'

:"'0

wlx)

~-

o FIG. 11.24.


III I I I

I I

\

i I I I

I

I I I: I

i" I I ~Wq I I

---. I

I

I I I 1\

a x

the odd wavefunctions of the oscillator, with 11 = 2k+ 1, which vanish at x = 0, are the solutions of this problem. Therefore

Ek = nco(2k+t), k = 0,1,2, ...

20. Since V(x,y, z) = V1(x) + V2(y) + Vs(z), this problem reduces to the problem of three independent harmonic oscillators of frequencies COl. CO2, coa, along the axes x, y, z respectively (see problem 1). Therefore

where

( mro1 )1/2 ~1= -h- x,

and Ill. 112, I1s = 0, 1,2, ..

( mco2 )1/2 ~2= -h- y,

(20.1)

(20.2)

( mros )1/2 ~s = -h- z

53


If the ratios of the eigenfrequencies are irrational, the energy levels are non-degenerate, otherwise they may be degenerate. The ground state Eooo is always non-degenerate.

For the isotropic harmonic oscillator

(20.3) In this case all the energy levels with the exception of Eo are degenerate. To calculate the degeneracy of the level of energy En' consider for the moment a particular value of the quantum number nl. n2 can then have any of the values 0, 1, ... , n - nl, and the sum n = nl +n2+nsforgivennandnlcan be obtainedinn-nl + 1 ways. Since nl = 0,1,2, ... , n, the degeneracy of En will be

n I (n-nl+l) = t(n+l)(n+2). (20.4) n,=O

21. Since, as x -+ - =, the SchrOdinger equation

(21.1)

has, for any value of E, a single bounded solution (the one which decreases exponentially as x -+ - =), which, as x -+ + =, oscillates endlessly, it follows that the energy spectrum of the particle in a homogeneous field is continuous and non-degenerate. In other words, to each energy value, in the range - = -< E -< + =, there corresponds a single solution. and this describes a motion of the particle which is limited in the negative direction of x and unlimited in the positive direction.

Introducing the dimensionless variable

( E) (2mk)1/S y= x+7Z ~ , (21.2) equation (21.1) becomes

(21.3)

This equation does not contain the energy as a parameter. Therefore, after obtaining its correctly bounded solution, we can readily find the eigenfunction corresponding to any arbitrary value of the energy. The solution of (2l.3), finite for any x, has the form

1p = NA(-y), (2l.4) where

~

A(y) = -In f cos (~ +Uy) du (2l.5) o

54


is Airy's function, and N = (2mn-3/2k-1/2fz-2)1/3 is a normalization factor (see problem 18, Chapter V). Thus, the wavefunction of a particle with energy E will be

-~Et 'ljJE(X, t) = NA( -y)e Ii ,

where y is given by (21.2) above. 22. By making the substitution

in the Schrodinger equation

-~Et 'IjJ(x, t) = VJ(x) e Ii

. O'IjJ(x, t) [ fz2 02 ] Ifz at = - 2m ox2 + V(x) 'IjJ(x, t),

we obtain, for x -< 0,

and, for x > 0,

The general solutions of these equations are

i i -qx --qx

'ljJ1 = Ae li +Be Ii , q = [2m(E+ Vo)]1/2, i i -px --px ~ /--

'ljJ2 = Ce li +De Ii , P = 'v2mE, x >0.

(21.6)

(22.1)

x -< 0,

According to classical mechanics, if E > 0, the electron has sufficient energy to overcome the potential barrier at the surface of separation and hence it will leave the metal. In the quantum-mechanical treatment the answer is not so simple.

The electron wavefunction is

1 ~ (qx-Et) -~ (qx+Et)

Ae li + Be Ii ,if x -< 0 'IjJ(x, t) = .

!.... (px-Et) -~ (px+Et) Ce li +De Ii if x>O.

The term with the coefficient A represents a plane wave which arrives at the surface from the left (incident particle), the term with B represents the reflected wave, the term with C represents the transmitted wave and that with D represents a wave arriving at the surface from the right. Since such a wave does not exist under the conditions of this problem we put D = O. The continuity conditions at x = 0 then yield the equations

A+B = C, q(A-B) = pC, (22.2)

55


whence B=q-PA,

q+p C=~A.

q+p

It can be seen that the certainty of transmission (which would correspond to a total lack of reflection, i.e. to B = 0) prescribed by classical mechanics, occurs only in the trivial case q = p, i.e. Vo = O. Since

we obtain in fact

T = Jkl = 1 1'2 !!.- = 4qp = 4VE(E+ Vo) IiAI A q (q+p)2 (VE+Vo+VE?'

(22.3)

IjBI I B 12 (q_P)2 ~ R=IiJ=IA = q+p = (VE+Vo+VE)4' (22.4) T+R = 1.

Note that, according to quantum mechanics, reflection occurs with a probability different from zero even if E >- O. However, if E Vo the reflection probability decreases rapidly with increasing energy:

On the other hand, if 0

Ch. II

whence

B A

2i l+71 qd 2i ' 1-71 qd


2i d C 7l q

-A- = -2-----;;2.-:-i-I-T qd

It follows from the expression for 1jJ2 that jc = 0, and hence T = 0 and R = IB/A 12 = 1. Thus, as in classical theory, an electron having a total energy smaller than the potential barrier height will be reflected with certainty. A new result, however, is that the probability of finding the electron outside the metal (x:> 0) is different from zero, since

This phenomenon is similar to that of the "total internal reflection" which occurs when the passage of light from a denser medium to a less dense one is impossible because the angle of incidence exceeds a certain critical angle. The theory of wave optics shows that, in the less dense medium, there is then a wave whose amplitude decreases exponentially, in analogy with the exponentially decreasing electron wave considered above.

23. Since the electrons, with momentum p = V 2mE, encounter a potential drop - V 0 at the metal surface, they will all enter the metal, according to classical mechanics, and, because of the law of conservation of energy, they will acquire a final momentum q = [2m(E+ Vo)F/2 after doing so.

According to quantum mechanics, on the other hand, some of the electrons may be reflected by the metal surface. Using the notation of the preceding problem, D exp ( -i/l'zpx) now represents the incident wave, C exp (i/l'zpx) the reflected wave, and B exp ( -i/I'zqx) the transmitted wave inside the metal. In this case, A = O. The continuity conditions at the point x = 0 give C+D = B,p(C-D) = -qB, whence

and T=liBl=I~12!L= 4pq. IjD I D P (p+q)2

Thus, if E = 01 e V and V 0 = 8 e V, the reflection probability is

I-V~2 = c~r = 0'64,

1+ V1+ ~ R = (VE- -YE+Vo)2 =

VE+VE+Vo

PQMS S7


which is greater than the probability (0,36) of entering the metal. The higher the energy of the incident electrons, however, the less probable is their reflection. For example, the prob-ability of reflection of an electron bombarding an anticathode in the usual Rontgen X-ray tube (Vo """ 10 eV, E = 105 eV) is approximately 62X 10-10

For greater familiarity with the phenomena of electron reflection at, and penetration into, metals, we suggest that the reader make a plot of the quantities Rand T as functions ofE.

24. By introducing the new variable

the Schrodinger equation d~ 2m ( VO)_ dx2 +712 E+ 1 +&/a 'IjJ - 0

becomes ~ d'IjJ (rJ2 1.2 ) y(y-l) dy2 +(1-2y) dy + y(l-y) Y 'IjJ = 0,

with the following notation:

2mVo 2 _ '2 ~a -Fl..

(24.1 )

(24.2)

(24.3)

(24.4)

Equation (24.3) has non-essential singularities at y = 0, 1 and 00, and we accordingly introduce a new functionf(y) through the relation

'IjJ = y(I-y)" f(y). If we impose on v and fl the conditions

v2 = 1.2 _rJ2 and fl2 = _rJ2 , the following differential equation for f is obtained

y(l-y)f" + [(2v+ 1)-(2v+2fl+2)y]f' -(fl+ v)(fl+ V+ l)f= o.

A particular solution of this equation is the hypergeometric function

f = F(fl+ v, fl+ V+ 1, 2v+ 1; y). Now, for

'IjJ = yV(I-yf F(fl+ v, fl+ V+ 1, 2v+ 1; y)

(24.5)

(24.6)

(24.7)

(24.8)

(24.9)

to be a physically acceptable solution of equation (24.2), it has to satisfy the appropriate conditions as x -- 00. This can be achieved by establishing correctly the signs of v and fl, which are not specified by (24.6).

58


If -Vo - 0. If E >- 0, we have }.2


In the limiting case a -- 0 we obtain once again the result of problem 22, viz.,

R= (k-K)2 = (VE-tVo-VE)2 k+K VE+Vo+VE

25. (l)E > Vo. The general solution of the Schrodinger equation is

i i -px --px

Ae" +Be" ; x

Ch.1I Simple Quantum Systems

Note that reflection occurs with a non-vanishing probability. If the energy of the particle equals the height of the potential barrier, i.e. if E = Vo, we have

( ma2Vo )-1

To = 1 + 2112 and ( 2112 )-1

Ro = 1 + ma2Vo .

The barrier becomes completely transparent (R = 0, T = 1) if sin qa//j = 0, i.e. if qa//j = Tm, n = 1, 2, 3, ....

This happens when stationary waves are formed inside the barrier, i.e. when

h A. a = n 2q = n"2' n = 1,2,3, ....

Thus the passage of particles through rectangular barriers leads to resonance phenomena of a type unknown in classical physics (Fig. 11.25).

To

o

(2) 0


Substituting q = i(n/2d) in (25.3) and (25.4), we obtain for Rand T the following expres. sions

4E(Vo-E) T=------'----'----V5 sinh2 ~ +4E(Vo-E)

2 h2 a Vo SIn 2d R= ---------

V5 sinh2 ~+4E(Vo-E)

(25.5)

(25.6)

Since T 7"'- 0, the particle has a certain probability of passing through the barrier even if, classically, its energy is not sufficient for it to do so. This phenomenon is called the "tunnel effect". It is important only when a/2d '" 1. If d a, the transmission probability, and thus the tunnel effect, decreases rapidly according to the approximate formula

T ~ 16 E(Vo-E) e-a/d V5

(25.7)

In the classical limit fz -- 0, Ttends towards zero. Thus the tunnel effect is a purely quantum-mechanical phenomenon.

26. Since the system is conservative, the particle passes through the barrier with constant total energy. The transmission probabilities Tj through the hatched regions will therefore be approximately independent, which means that the transmission coefficient through the whole barrier will be given by the product

(26.1)

Increasing the subdivision of the interval XI. X2 as far as is allowed by the condition dj iJxj , the approximate formula

(26.2)

is obtained; this expression is known as "

Constantinescu--Problems in Quantum Mechanic

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