Problems in Quantum Mechanics [3 ed.]

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PROBLEMS IN

QUANTUM MECHANICS

PROBLEMS IN MECHANICS THIRD EDITION

Edited by

D. ter Haar

"{Q 3 8888 22649197 9

Dover Publications, Inc. Mineola, New York

Copyright

Copyright© 1975 by Pion Ltd. All rights re.served.

Bibliog,~aphical Note This Dover editions first pub] ished in 2014, is an una brid~d republication of the third revised d.11 d enlarged edition of the work t origi na ILy pu bli~ai by Pion Ltd., London~ in 1975. This Dover edition is published by special arrangement with Pion Ltd.) 207 Brondesbury Park, London N\\12 SJN,. England.

International Standard Book Number ISBN~l 3~· 978-0-486-78080·1 ISBN-JO~~ 0•486,.78080 ..J Manufactured in the United Stales by Courier Corporation

-78080501 2014 www~doverpub1ications.com

Contenh Problems

Preface 1 One dimensional motion 2 Tunnel effect 3 Commutation relations; Heisenberg relations; spreading of wave packets; operaton;

Solutions

3

81

10

129

15

155

4 AnguJar momentum; spin

24

216

5 Central field or force 6 Motion of particles in a magnetic field 7 Atoms 8 Molecules

34

233

38

254

42

270

53

350

58

383

10 Creation and aMihilation operators;

67

430

density matrix II Relativistic wave equations Subject index

74

443

9 Scattering

··.

Preface to second edition

This is essentially an enlarged and revised second edition of a collection of problems which consisted of a text by Gol'dman and Krivchenkov augmented by a selection from a similar text by Kogan and Galitskii. In preparing the present edition I have used the opportunity to revise some of the problems in the first edition, to change a few of the solutions, and to make the notation both uniform and conforming to English usage. Also, I have added a few problems from a collection by Irodov on atomic physics and a number of new problems which were mainly taken from Oxford University Examination papers. I should like to express my thanks to the Oxford University Press for permission to include these problems. These problems can be used either in conjunction with any modern textbook, such as those by Schiff, Kramers, Landau and Lifshitz, Messiah, or Davydov, or as advanced reading for anybody who is familiar with the basic ideas of quantum mechanics from a more elementary textbook.

Oxford, September 1963

D. ter Haar

Preface to third edition

In preparing the third edition I have dropped some of the problems, slightly rearranged the order of the problems, and added new problems to the old chapters, as well as added new sections on the density matrix and annihilation and creation operator problems and on relativistic wave equations. Otherwise the aim and scope of the book remain much as they were before, but to help readers I have added stars to more compli~ c.ated problems.

Oxford, September 1974

D. ter Haar

I

One-dimensional motion

1. Determine the energy levels and the normalised wave functions of

a particle in a "potential well". The potential energy V of the particle is:

V=oo for xa; V=O

for

0 0 (outside the metal) (see fig. 7), and determine the reflection coefficient at the metal surface for an electron with energy E > 0. V(x)

E------------- ------------

'

Fig. 7.

2". In the preceding problem it was assumed that the potential changed discontinuously at the metal surface, In a real metal this change in potential takea place continuously over a region of the dimensions of the order of the interatomic distance in the metal. Approximate the potential near the metal surface by the function V

V, = - e=ia+ 1

(see fig. 8) V(.11)

•

Pig. 8.

and determine the reflection coefficient of an electron with energy E > 0.

2.8

II

Tunnel effect

3, Determine the coefficient of transmission of a particle throush u rectangular barrier (see fig. 9). F(x)

,.

0

I

IT

},'-------------

-----""

ITT

-----------"

Fig. 9.

4. Determine the coefficient of reflection of a particle by a rectangular

barrier in the case where E .> V0 (reflection above the barrier). 5. A particle is moving along the x-axis. Find the probability for

transmission of the particle through a d~lta-function potential barrier at the origin. 6. Determine approximately the energy levels and wave functions

of a particle in the symmetrical potential given by fig. 10 for the case where E 0 in the first approximation of the adiabatic perturbation theory and discuss the conditions for the applicability of the result obtained. 56. The point of suspension of a linear oscillator in its ground state starts to move slowly at t = 0 and comes to rest again at t = T. Find the probability for the excitation of the oscillator in the adiabatic approximation (compare problem 24 of section 3) and ascertain the applicability of this approximation. 57. A particle is bound in an infinite rectangular potential well of width a. At t = 0 one of the walls of the we~! starts to move according to an arbitrarily given time dependence. Reduce this problem to a wave equation with a Hamiltonian which depends explicitly on the time and discuss the particular case of the motion of the wall for which the variables in that equation can be separated, 58*. A particle is in the nth stationary state of an infinite rectangular potential well of width a. At t = O one of the walls of the well starts to move slowly according to a given time dependence. Find in the adiabatic approximation the probability that at t > 0 the particle will be found in the mth stationary state (m =f, 11). 59. Evaluate the de Broglie wavelength of (i) an electron, (ii) a hydrogen atom, and (iii) a uranium atom, if the kinetic energy of each of them is 100 eV. 60. Find the kinetic energy of an electron and of a neutron with a de Broglie wavelength of l A. 61. Under what circumstances will a particle of mass µ. and velocity v ( ~ c) show distinctly wave properties when it is scattered by a periodic structure of linear period d? · 62. For what neutron energies may we expect especially strong diffraction effects when the neutrons are scattered by natural crystals with lattice constants between 2·5 and 6·0 A? 63. Bearing in mind the wave properties of particles, show the limits of the applicability of the classical concepts for an electron of 10 eV energy and a proton with 1 MeV energy. 64. For the study of the structure of atomic nuclei several laboratories have constructed electron accelerators with energies up to 6 GeV. What is the electron de Broglie wavelength _in that case, and what is the need for such high energies?

4

• Angular momentum; spin

1. Consider the angular momentum operators satisfy the commutation relations ••

[fx,ly}_

= itllz,

• • fx,fy,

•

and

fz

which

• • • • •

•

[ly,lzl_

= 1'hlx,

[lz,lxl_

= 1111.,,,

and introduce the non-Hermitean operators •

•

•

(i) Derive the commutation relations satisfied by J+, J_, and lz. • • • • • 2 (ii) Prove the relationsJ~J± J -J,(lz + 11. • (iii) Let 'Pa, b be the joint eigenfunction of J 2 and lz corresponding to the eigenvalues a and b,

=

P,r;a, b = a,p,,, b

• ,

fzl{)a, b

~

la. b> ,

= b-.Pa, b

;

we shall use the Dirac notation 'Pa,

b

and choose the 'Pa, b to be an orthonormal set so that (a', b' la, b > = Saa' lj bb' Prove that •

,

a> b 2 , •

•

•

2 and that l+la, b) and J_la, b) are also• joint eigenfunctions of 1 and Jz, • Find the eigenvalues corresponding to J+la, b} and l_la, b}. Hence prove that there are states such that

and such that

•

l _la, bm1n> If we put bm,x/'11

= O.

= /, prove that a= j(/+ 1)11 2

,

and that / is either an integer or half-oddMintegral. (iv) If we now put b = mh and call the la, b) state the 1/, m) state in agreement with the usual notation for angular momentum eigenstates,

Angular momentum; spin

4.8

25

find the matrix representation of Jx,ly, and lz in the I/ 1 m) representation. 2. Express the operator of rotation over a finite angle '{)o around the direction n in terms of the angolar momentum operator (of a system of N particles). 3. Obtain expressions for the operators lx, !11 , lz in spherical coordinates starting from the fact that lx, lv, lz are the operators of infinitesima} rotations. • 4. Prove that under a rotation an operator n changes to a new operator d' such that

0 = ROR- 1 1

' where R is the rotation operator (see problem 2 of section 4 ). S. Give a simple interpretation of the commutability of the operators of the components of the linear momentum and the non-commutability •

of the operators of the components of the angular momentum, starting from the kinematic meaning of these operators, which are connected with infinitesima! translations and rotations. 6. Prove the following commutation relations: (a) [li, &0,]_ = ie,u,1 Xi,

(b) [l,,p,J_ = i,,,,p,,

v,1here eikl is the antisymmetric unit tensor of third rank, the components of which change sign for any interchange of two of its indices, for instance, eu,1 = - eilk• and where e11is = 1 (1, 2, and 3 correspond to x, y, z). 7. Prove the following relations:

(a) [Ì,(P!+P;+j)))J_ = o, (b) [Ì,(x'+y'+z')]_ = O. 8. Show that if ljl~l is the eigenfunction of the jz operator corresponding to an eigenvalue m, the function

1Pm = exp(-il1 ,:p)exp(-i]11 {})Y,';::) will be thc eigenfunction of the operator:

J, = Jzsin{}cos,p+JvsinDsin,p+J,cosD-, corresponding to the same eigenvalue, that is:

f,fm

= mfm-

Hint. Use the relations (see problem 21 of section 3): exp(-iJ11 fJ)1:exp(i111 D-)

= Jzcos{}+JzsinfJ,

cxp(-il:,:p)Jzexp(iJ~rp) = Jzcos,p+J11 sinip,

4.9

Problems

26

9. Show that in a state iJJ with a well-defined value of la (le!f = mf) the average values of lx and 111 are equal to zero. Hint. Find the average value in the state i/, for the left-hand side and for the right-hand side of the commutation relations

l11 fe-/Bf1I = i(,:,

lafx-fxla = i/11 •

10. Obtain an expression far the operator of the angular momentum relative to an arbitrary axis (z') in terms of the operators lx; !11 , / 1 • 11. Show that in the state IP far which lzlf = mif, the average value of the angular momentum about an axis z' which makes an angle 0 with the z-axis is equa! to 1n cus 0. 1~his result can be visualised as follows, The angular momentum vector in the state fm is evenly "spread out" over a eone with its axis along the z-axis, its slant height equal to ✓ [l(l+ 1)], and its height egual to m. The average value of its projection on the xy-plane is equal to zero, and its component along the z' -axis is, after averaging, equal to m cose. 12. Find the matrix giving the transformation of the components of the spin function of particles of spin 1 corresponding to an arbitrary rotation of the system of coordinates. 13. In a Stern-Gerlach type of experiment the deflection of a beam of atoms with total angolar momentum J depends on the value of the component of the angular momentum in the direction of the magnetic field in the apparatus. If the particles in the beam have a well-defined value of the angular momentum relative to an axis which is not along the direction of the applied magnetic field, the beam is split into 21 + 1 components. Determine the relative intensity of these components if J = 1, and if the component of the angular momentum along an axis which makes an angle e with the direction of the applied magnetic field has the well-

defined value M( + 1, O, -1). 14. The three Hermitean operators

[Ox, O,]_= 2ib,, [b,., f1z]_

f1x, Ùy,

and

Òz

satisfy the relations

= 2i0x, [O,,Ox]_ = 2iciy

,cii+cij+Oi

= 3.

Find a representation of these so-called Pauli matrices in which (Jz is diagonal. 15. If {I is the vector operator with components cix, a,., and fJ,, prove that • • (i) if A and B commute with {;, we bave

ca.A)(a,Ji)-(A .Ji)+i(a,[A and (ii) cix ciy O,

= i.

AB]);

4.21

27

Angular momentum; spin

16. Find the transformation operator for a·spin-function (spinor fora spin-½ particle) corresponding to a rotation described by the Euler angles {}, l/1, and ,.p (see fig. 19).

e

' y'

~--y

Fig. 19.

17. Find the transformation matrix for the components afa spinar, if the system of coordinates is rotated aver an angle tb with respect to an axis with direction cosines O\~, and -y. 18. Find the eigenfunctions of the operator cx6x +_aay+-'Yòz, where a 2 +P2 +y 2 = 1, and show that the expansion coefficients of an arbitrary function

(t:) in terms of these functions determine the probability th~t

the value of the spin component in the direction characterised by the direction cosines o:, p, and y is equal to +½or -½, 19. lf the z-component of the electron spin is egual to +½, what is the probabilify that its component along a direction z' which makes an angle e with the z-axis is equal to + ½or - ½? Determine the average value of the component of the spin along this direction. 20. The most general form of the spin function of a particle of spin½ in the I!, m)-representation is

ifi2 = e'P sin IL This function describes a particle state in which the probability that the z-component of the spin is equal to +!(or-½) is equal to cos 2 8 (or sin 2 8). What would be the result of a measurement of the component of the spin along an arbitrary direction ? 21. The spin function of a spin-½ particle has the following form in the I½, m )-representation:

s),

~, \ _ (''" cos ( e1P sin S

iµJ

28

4.22

Problems

1s there a direction in space along which the spin component has the well~defined value of + ½?

If such a direction exists, find the palar angles (0, 11>) of its direction. Hint. Find 0 and tP from requiring the second component of the spin function to be egual to zero. 22. Consider a system of non•interacting identica! particles. Let their momentum be the same and their spin be egual to ½, If these particles did not possess spin we could describe the system by a "pure case" ensemble. However, we do not know whether the spins of all the particles are parallel. Is it possible to use an experiment of the Stern-Gerlach type to determine whether this beam of particles corresponds to a "pure case"

or to a "mixture" ensemble? 23. Find that wave function of a system consisting of two spin-½ particles which is an eigenfunction of each of the two commuting operators, the square and the z-component of the total spin. 24. Show that in a system consisting of two spin-½ particles the total spin S is an integra! of motion, provided the Hamiltonian is symmetric in the two spins. 25. Denote by 0 1 and 6 2 the spin operators of two particles and by r the radius vector connecting these particles. Show that any positive integrai power of either of the operators

S - 3(&,.r)(&,.r)_(& & ) lll

y2

1•2•

and any product of such powers can be written as a linear combination of A, S 12, and the unit matrix. • 26. Show that the operator S 12 of the preceding question can be expressed as follows, in terms of the total spin operator S = ½(ét1 + &2), a

_

.:112 -

6($.r)'

,,

-

,, 2 o)

and that if the tota! spin of the two particles is equa! to unity, S12 can be written in the form of the following 3 x 3 matrix: -✓ 3

Y, __ ,

-2Y20 ✓3Y,,

✓6

✓3

Y, __ ,

Y,.-, Y,,

'

29

Angular mo men tum; spin

4.32

27. Consider two spin-½ particles interacting through a magnetic dipole-dipole interaction,

V= A~~1_,b2l!~=--\~-~-·-r)~6 2 .r).

'

If the two spins are at a fixed distance d apart and if at t = O one spin is parallel to r and the other one anti parallel to r, calculate the time after \vhich the parallel spin is antiparallel and the antiparallel spin parallel. 28. Find all the quartet and doublet spin wavefunctions for a sys.tem of three spin--½ particles. 29. Show that a totally symmetric spin wavefunction for a system of n spin-½ particles is an eigenfunction of the tota! spin operator

S2 ( S =

t

jho 1) with quantum number ½n and that any other state ofthe

system has. a smaller total spin quantum number. 30. Let o1 denote the spin variable of the ith electron. This variable can take on the tv.·o Yalues + 1 and -1. Sho\V that if the operators

-i)01'&iz=o(! ·(01)'(0 1 O i ll1x=

1'

ll1v=

which referto the lth electron, act upon a function_f(a 1, r; 2 , ••• , a1, ... , a 11) of the spin variables of n electrons, the result is the follov.·ing one:

&r.rf = f( 0 1, · · ·, 0 1-1, O'ivf = - ia1f( 0-1,

-

a1, 0 )t1> · • •, an),

••. , 0'1-1, -

a1, a,.,. 1 ,

.. , ,

a,,),

31. Show that the operator of the square of the total spin moment of n electrons can be v.'ritten in the form

" = n--4 n' +}: ~ i ,

52

k, and let the state of a particle be described by a wave function t/; (l,j = l + ½, m) (see problem 36 of section 4). It is clear that the spin direction of such a particle will, generally speaking, not be the same in all points of space. Find the relation between the angles 0 and Il> and the space coordìnates of the particle. 39. A system consists of two particles, one with angular momentum 11 = 1, and the other with angular momentum /2 = l. The tota! angular momentutn J can in that case take on the values l + 1, !, and l - 1. Express the eigenfunctions of the operators j 2 and Jz in terms of the eigenfunctions of the square and the z-component of the angular momentum of the separate particles. 40. A system consists of two particles, one of spin ½ and one of spin O. Show that the orbitai angular momentum is an integrai of motion for any law of interaction between these particles. 41. Show that the normalised part of a 3D 1 state wave function which refers to the spin and angular dependence can be written in the following

Angular momentum; spin

4.43 form:

I 4,1(2'1T) S12

1

1

o o o

1 --------·'"

o o o

4-,./(ZTT) S12 I

4 ✓(2~)

s2 '

31

(J = 1, M = I, L = 2, S = 1),

(J = 1, M = O, L = 2, S = !),

(J = 1, M = -1, L = 2, S = !).

1

Hint. See problem 39 of section 4. 42. Prove the following equations: (a) []',À]_=i([ÀAJ]-[J/\À]),

(b) [J•,[J',À]_]_

= 2(i'À+À1')-4J(J.À),

= (].À)~'.f (J)JM' (C) (À)•'JM• nJM J(J+l) JM•

A satisfies the commutation rule [Ji, Akl- = ieikl A, .

The arbitrary vector quantity •

43. If T is a vector operator whose commutator with the tota! angular • momentum operator J is governed by the relation •

•

[(a .J), T]_

•

= i11[T A

a], •

where a is an arbitrary constant vector, show, by taking T to be the electric dipole-moment operator that the quantum number m in the li, m} representation (see problem 1 of section 4) changes by O or + I in an electric dipole transition. Using the result (b) of the preceding problem show that [(À;. - À/ ) 2 - 2(À1+ Xr )](i, m I fii', m'>

= -4'1-1 I'). i-ln'

'

'

The summation is aver alt Z protons, I is the nuclear spin, and n is the combination of all the other quantum numbers which characterise the state of the nucleus. 60. Evaluate the quadrupole moment of a nucleus with one proton of angular moment j outside a closed-shell spherically symmetric core, neglecting any deformation of the core.

s Centrai field of force

1. A system consists of two particles of mass µ 1 and µ 2 • Express the operators of the total orbitai angular momentum 11 + 12 and the total momentum f, 1 + fi 2 in terms, of the centre of mass coordinate

- -

R

= JL1r1+µ1r2 /J,1 + µ2

and the relative coordinate r = r 2 -r1, Show that if the potential energy of the interacting particles depends only on their distance apart, V= V(lr 2 -r1 ]), the Hamiltonian can be put in the form

li - -

n2 v2 _ n2(µ1 + /J,2) v2+ V(r) 2(µ,l + /J,2) R 2µ1 /J,2 r '

where V,k and V~ are the Laplace operators referring to Rand r. 2. A particle moves in a central field. Write the eguation far the radiai part R,,,1 of the wave function in the form of a one-dimensional SchrOdinger equation. 3. Show that in the case of a discrete spectrum in a centra! field the minimum value of the energy for a given value of the orbitai quantum number l increases with increasing l. 4. Determine the wave functions and energy levels of a threedimensional isotropie oscillator. 5, Salve the previous problem in Cartesian coordinates by separation of variables. Express the wave function for nr = O, I= 1 (see previous problem) as a linear combination of the wave functions obtained. 6. Assume that a nucleon in a light nucleus moves in an averaged potential of the form V(r) = -Vo+½µ.w 2 r 2 , to determine the number of particles of one kind (neutrons or protons) which can be accommodated in a closed shell. A shell is defined as the totality of all states with the same energy. 7. Calculate the theoretical radius of the closed shell nuclei ;He and 110 assuming the same potential as in the preceding question. The theoretical nuclear radius is defined as the distance from the centre of mass of the nucleus to the point where the "nuclear density"

p(r) - "J.:.f:(r)f,(r)

5.13

Centrai field of force

35

(the surrtn1ation is over ali nucleons) decreases most steeply, thàt is,

- o 2 r-R ( O it satisfies the differential equation of an ordinary oscillator. One sees easily that the wave functions of the oscillator for odd values of n = 2k + 1 vanish at x = O and are, in the region x;;,, O, solutions of the problem under consideration. Hence

E,= liw(2k+¾) (k p2

( 2µ-

12.

= O, 1,2, ... ).

µw2 n,2 02 )

op' a,(P) = E.a,(p),

2

Ia,(p)I' = 2• .nl ✓~~µwn) exp ( -p'/µwn)H!(✓(:wn)) · 13. An investigation of the behaviour of the solutions of the SchrOdinger equation

x)'] if,=O

n' d'~2 - [E-Vo;;;-a (" -2µdx

as x ➔ oo shows that 'P behaves asymptotically as exp ( new independent variable

½f) where g is a

' - ✓(2µV,) ' 'li,a X,

~-

As

x➔O

f is proportional to

gv1 2,

with

'=; [J(Sµ~a• + 1) +1]. We make the substitution ~

= exp ( -

½e),., u(,)

and fìnd for u(ç") the following equation: ,

,

(

,

')

,

['

µa(E+2V,)] = o.

1 ,u + ,+,-, u - z+,2n ✓(2µV,) u

(1)

One-dimensional moti on

1.14

91

Equatian (1) is the equation far the canfl.uent hypergeametric function and its general salution is af the form u(t)

= c1 F(o:,v+½, f)+c 2F{tl-v+½, :-v, f),ft-P,

where we have written cr. far the expression within square brackets in equatian (1). Fram the requirement that !p(O) must be finite, it fallows that c1 = O. Apart fram tbis we must require that the wave functian should decrease as x ➔ oo, that is, that the functian u(t') reduces to a polynomial. This can be attained by putting o: equal to -n (n = O, 1, 2, ... ) so that we find the energy levels

E.=

~J(2~) l•+t+i[J(8µ::•· +1)-)(8µ::•·)] )·

The energy spectrum is tbus the same as for an oscillator of angular frequency w = 4(8Vo/ µa 1 ) pravided the zero of the energy scale is suitably chosen. It is of interest ta note that the zero-point energy of a particle in the potential Vo(afx-x/a) 1 is always larger than the zero-point energy of the corresponding ascillator. The wave functions are of the form

ifin = c3 ~exp[-J(J~2)x2] F[-n,v+ì,J(t~)x 2] , where v = ½[✓(811-Voa 11 /! 2 +1)+1], while tbe constante"' must be found from the normalisation condition. 14*. In the Schròdinger equation

~• d'•

-2p, dx2 -

[E + coshV,,(x/a)] ·'·'I'= O 2

we make the substitution

( ·)-'' u,

V,= cosh a

The equation for u takes the following form

d 1 u 4,\ xdu 4 dx" -atanh a ;&+a3(,\ll-1el)u where

(we consider the discrete spectrum, E< O).

= o,

Answers and solutions

92

1.14

If we introduce a new independent variable Jl e=

-sinh1:!1

•

then the equation for u reduces to the hypergeometric equation d1 u

du z(I-•) dz' +[½-(1-2!.)•l;i;;-(!.'-•')u-o.

(1)

The parameters o:, /3, y, which occur in the generai form of the hyper~ geometrie equation

d2 u du z(I - z) dz' + [y-(o+ µ + l)z] dz - ,µu - O, bave in our case the following values:

r=l,

o:=K-À,

/1=-K-À.

The two solutions of equation (1) which lead respectively to the even and the odd wave functions if, are of the form

u, = F(-À+1x 8 : .,.

'l'Ill

.(,;- 1,;) exp (-lm) = -z exp

p

(i7i

. 1

p ••

d ) X

ii"d) p x .

-,. (,;+1,;)exp(¼m) exp ( --

n .,

✓P

~fhe condition that the solution is quasi-pcriodic is of the form:

-i(cr-ic;) exp ( -¼11i) = ZC1, -i(c~ +-le;) exp (¼'rri) Expressing of equations:

= zc2,

ci and e; in terms of c1 and c2 we obtain the following set

e, exp (i,) [exp ( -µ) + 4exp (µ)] + ic,exp ( -i,) [exp ( -µ)- 4exp (µ)J

= 4zcu e, exp (ia) [exp ( -#)-4 exp (µ)] + ic,exp ( -ia) [cxp ( -µ)+4exp (µ)]

= 4izc

2•

The value of z is determined from the condition that this set can be solved (that is, by putting its determinant egual to zero):

z 2 -2z[exp(/3)+¼exp(-~)] cosa:+ 1 = O, so that

z 1, 2 = [exp (/3) + ¼exp ( -,8)] cosa:+ ✓{(exp CB)+ Ì exp ( -/3)] 2 cos 2 a:-1}. (2) If the expression under the square root is positive, then z is real and different from unity. In the opposite case, z is complex and its modulus is egual to unity,

z 1, 2

= [exp (/:Ì) + l exp ( -/3)1 coso:+ i ✓ {l -

lz,.,I' -

[exp (,8) + ¼exp( -fi)] 2 cos 2 o:},

[cxp(#)+lexp(-µ)]'cos'a + {I - [exp (P) + l exp ( -P)]'cos• o} - I.

Thus the equation which determines the allo\.ved bands follows from the condition that the expression under the square root in equation (2) is negative,

(3)

Answers and solutions

126

1.39

The condition that the semi-classical approximation can be applied is that the expression f p d.-.; is large compared to Ti,, that is, a►

1, f3 P,1.

If, in accordance with this, we neglect the term with the negative exponent we can write equation (3) in the form exp (2µ) cos 2 o:~ 1

o, cos 2 o:~ exp ( - 2(3).

(3')

To a first approximation, we bave

cos2o: = O,

or

..

From (1) we thus get

i••

pdx

o:=

n'l'T+½1r (=o:n)·

~ ,~(n+½) .

This is the Bohr-Sommerfeld quantisation rule. From the solution of this equation we get a set of levels in a well which is a hole in a periodic potential with impenetrable walls:

Eo,E1,E2, ... ,En,···, We now expand the left-hand side of equation (3') near the value c.11 corresponding to the nth energy level COS 2 a.

~sin11 o:,..(a!Xn) 2 ~(ao:.i) 2 ,

From the definition of o: we bave ao: = 6.E !_

fj, dx = aE fil•~ dx =

1i, aEJ:e1

1i,

):i,,BE

~ 1i,

ft:t dx = aE T Jz1 V 1i, 2'

,vhere T is the period of motion of the particle in the corresponding well-level. If we note that exp ( - 2P) is the probability of transmission through the barrier, W(E), we can write equation (3') in the form

(Ab',)'.;~: W(E,),

•

so that we get for the band-width:

2~

AE, - 7: ✓ [W(E,)]«:~w,.

•

One-dimensional motion

1.42

127

41. lf we choose as a (normalised) trial function

f

= t'(2o/rr) exp ( -

(1)

••'),

we find that the ground state energy E0 must satisfy the inequality (2)

where V(x) < O. Substituting (1) into (2) we obtain E~E" with

E = ~;; + J(~) f exp(-2ox') V(x)dx.

(3)

0

The minimum value of Ea,,, E;;"n, is obtained by finding condition

o:

from the

'!•=O=;;+ J( 2~.) f exp(-2ax') V(x)dx -J(:;')f2x•exp(-2ox')V(x)dx.

(4)

Combining (3) and (4) we fìnd

E:"n =

J(4;;,) f exp( -2ox')(l +4ox') V(x)dx< O,

from which follows that E0 < O, as was to be shown. 42. If À is small, we should expect {i) only one bound state and (ii) a wave function corresponding to the ground state which is varying only very slowly. In the region where the potential vanishes this wave function is of the form

1,=Aexp(o:x),

xa and X< -a is clearly egual to IFl2, Solving equation (8) and assuming the Airy functions u and v to be normalised to satisfy

u'(t)v(t)-u(t)v'(t)

~ I,

we find y -

-

IFI' --

,\ ~~,---c~~~----c=~~~~~~~=

[v(µ) u'( - ,\)-u(µ)v'( -,\)]'+ A[v(µ) u(-A)-u(µ) v( -,\)]'

I x [v'(µ) u'( -,\)-u'(µ)v'(-A)]'+ .\[v'(µ)u( -,\)- u'(µ)v( - A)]'' (Il) Equations (11), (9), (10), and (1) together solve our problem. The functions u(t) and v(t) which are real for real values of t and their derìvatives u'(t) and v'(t) bave been tabulated, We shall now find the limiting expressions for the penetrability of the barrier in the semi-classical case. It is well known that one may consider the motion to be semiclassical provided the particle wavelength A-

-

~

✓{Zµ[E- V(x)]}

changes little over a distance dx~ i\. In the regions x < - a and x > a the motion is free and thus also semiclassical (;\ = 1/k0 = const, so that di\/dx = O). In the regions - a:;;; x:;;; O and O:;;;x:;;;a we have respectively from equations (1), (5), and (6):

A~ (~)' _!"" K~

i ✓{'

dA _ d~ d( _ _ I dx d{dx- 2i{i'

and ;\=

1 a"\ (

I

KVi ✓ri'

dA dAd" dx = d71 dx

The condition of semi-classicism requirement

I = 2t'rii.

J d;\/dxf

~ 1 leads thus to the

(12) We consider a range of energy E< J7a (that is, k0 < Ko, µ,>O). We denote the barrier width for an energy E by 21 and we have clearly:

k' V,-E ( 1- kj) , a-l = a~. l=a~~·--=a 2

Vo

Ko

'"

(13)

138

2.7

Answers and solutions

Condition (12) is, of course, violated near the classical turnìng points x = + l (where {=O, '1J = O). One can, however, formulate a condition of "integral" semi-classicism (semi-classicism in the important region)

which ensures a "semi-classical" penetrability of the barrier. For this one must obviously demand that the sections of non-semi-classica} behaviour Ax (corresponding to the sections b..,~1, /)."f/~1) are small compared to l and (a-1). We have (A()n,n-a,m.+z,-À-z,½;

+c (cosh x)a 2

where

2

J.. •

x (

stnh aF

-sinh2 ") a

ika ika . x) ->.+z+½,-À-z+½,-i-; stnh 2 a , (I)

,\ - }[✓(I - SµV, a'/1!')- I], k -

✓(2µE)/li.

The coefficients c1 and c2 are deter,nined from the condìtion that as x ➔ +oo the wave functìon has the asymptotic form

'P~ eikx.

140

Answers and solutions

2.8

To find the asymptotic form of expression (1) we use the relation

F(a,~,y; z)

r(y)r(~-a) ( _!) ~ r(p)r(y-a)(-z) °F •,a+l-y,a+I-µ, z r(y)r(a-fi)( )-PF(" a I a I • 1) +r(o)r(y-fi) -z e,e+ -r,e+ -a, z .

We find thus:

fx_,.-o:, ~( -1)2A [(t1 A1 - c2A2) ( -½)-lka eikx

+(c1 B 1 -c2 B 2) ( -½)ika e-ikx], if x _,. +oo ~[(c1 A1 + C2 Aa) (½ )-ika e-ikx + (c1 B1 + C2 B2) (½ )ika eikx],

(2) (3)

where for the sake of simplicity we bave introduced the notation

~ ____

A

r(½) r( -ika) _ r( -À- ika/2) r(A+ ½--i"k-a/=2)'

'

r(J) r( -ika) A, ~ Ì'( -À+ ½-ika/2) r(A + I -ika/2)'

r(½) r(ika)

B,

~ f( ~-A+ika/Z)f(A+½+ika/2)' r(f)r(ika) B, ~ l'( - A+ ½+ika/2) r(X

+1 +ika/2).

The difference in sign of the coefficient c2 in equations (2) and (3) is explained by the fact that sinh (x/a) is an odd function and the second term of expression (1) changes sign when we change from positive to negative values of x. The requirement that at + oo there is only an outgoing wave leads to the following relation between c1 and c11 :

c1 A 1 +c11 A 2

= O.

The transmission coefficient will then be of the form: D- lc1B1+c2B1d2 - [c1A1-c2A2l2'

Substituting in this expression the values of the coefficients A 1, A 2, B1 , and B 2 and performing a simple transformation we find finally:

D

=

sinh2 ?Tka

sinh 2 ?Tka+ cos 2 {½7T ✓ [l -(8µVc; a 2)/h2]}'

for

8µVoa 2 I li'
bN so that we must require that C0 = DN = O. lt is easily seen that this entails that the matrix element (AN)l1 2 is equal to zero. The condition (AN) 22 = O determines the energy spectrum of our problem. To evaluate this matrix element we consider the matrix

,s t-1Y S =e' '=I+t A+21 A ' +31 A ' + ... +Nl AN + .... t'

It can easily be shown directly that the matrix S satisfies the equation

dS -AS dt

(2)

2.14

Tunnel effect

with the initial condition S(O)

149

= !.

We write equation (2) in more detail

a (s11 s,,\ = ("

dt S 111 S 11,J

or

y

ds,, s •s dt = a: +I-'

111,

dS,, S •s dt = a: 12+,-, 22,

dS111 S , dt =y 11+0S,11

dS22 S 'S dt =y 111+u 211·

11

Since the condition which determines the energy spectrum can be written in the form

(AN'J22

= (d;t~211)1-o = 0,

it is sufficient to consider the second pair of equations. If we put S 111 = fe}J, S 22 = geAJ, we get

J>. = of+pg, g>. = yf+Sg. The value of À is determined from the equation a:-À y

f3

,\ 2 -

S->. = O,

.\(2 e''+½ e·-1') cos u + 1 = O,

which gives two roots À1 and .\11 • Since À1 À2 form where

= 1 we can write À1 ,2 in the

cosu = (e'+ ì e-~) cosu.

If À1 '# À11 , the solution which satisfies the initial conditions S 12(0) S 22(0) = 1 has the form:

= O,

s,, = P(•'"- •'•')/(>., - >.,), S211=

().1 -

o),,,, - (>., - o). I>. {(!.1 -o)>.f-(>.,-o)>.f} = O. 1-2

Answers and solutions

150

2.14

We can substitute in this expression the values: À1 l! •

= e±tu

and neglect e_,, in the equation which determines cosu (this is equivalent to assuming that the penetrability is small): cosu:;;;;;eTcosa,

Under this assumption the condition determining the energy levels becomes very simply: sin(Jl!+l)u=O. s1nu This equation has the following roots:

n, U=N+l'

with u = O, w, 21r forbidden. Thus, cosu has N different values, n,

cosu=cosN+izeTcosu

(n=l,2, ... ,N).

More explicitly,

cos

(! J::pdx) = exp (-! f~:•f PI dx) cos ; : 1

Since exp ( -

(n

= 1, 2, ... ,N).

!J,'.'lp Idx)

is a small quantity, this last equation can be rewritten in the form:

1

~ J.b p dx = 1r(m +-li-)+ exp (-! (""/p Idx) cos Nrm 1 Jb1 + a1

(m=0,1,2, ... ) (n= 1,2, ... ,N).

(3)

This is the condition which determines the energy levels in the field V(x). It is very similar to the quantisation condition for the field of a separate vrell. By considering equation (3) we can conclude that the energy spectrum in the field V(x) is, roughly speaking, the energy spectrum of a separate well, with all its levels split into N sublevels. Let us determine the value of the shift, 11Em,

1 'h,

J.''a, ✓[2µ(Et~>- V)J dx+ µ'h, J.b'a, ✓ [Zµ(EJgJdx V)] 11Em

1("')

,n

=,(m+½)+exp ( -.J,,IPldx cosN+l

(n=1,2, ... ,N);

2.15

1l 1

Tunnel effect

if we now introduce the notation

we find

dEm = 1i1rw exp (-! fb~IP Idx) cos ; : l

(n= 1,2, ... ,N).

The distance between the highest and lowest sublevel is equa] to:

21'w 1T exp

('fì) ~ -nJb,·pldx COSN+l"

15"'. In the regi on x < -b we have under the circumstances of our problem only a wave which goes to - oo, or

If we extend this solution into the region x > b, we get the following expression for the wave function:

if, -

Jp exp ({ J:pdx) [texp (-; J:IPI dx+ i;) cos (! J+:p dx) + 2exp (; J:IPI dx-i;) cos (i J+:pdx)]

The quasi~stationary level is determined by the condition that there is no wave coming from + oo. If we put the second term in the last expression equal to zero, we get cot

(! J~:P dx) - i [¼exp (-; J:IPI dx)+ 2exp (~ J:IP Idx)]_,•

Considering exp ( - 02

J.'.IPldx).

152

Answers and solutions

2.16

to be a small quantity, we find that

!J+:pdx ~o(n+½)-;exp (-~J:[p[dx), from which follows the condition to deter1nine the quasi-stationary levels E~0l and their width r,

I+•0

Il1 _ ✓ [Zµ(E,\-V)]dx~(n+½)• (n~0,1,2, ... ),

nw 2?Texp (zr•) -nJalPldx =r,

where w

~•(µ J: ✓[Zµ(;;=·v)]) _,.

We find the following value for the transmission coefficient: D(E)

~ [4exp (: J:[p[ dx) cos' (!

f+:p dx)+ sin'(! J::pdx) i-t.

For values of E coincìding with one of the quasi-levels we bave D(E,\) ~ 1. For [aE[, 1,

[t), x111_

= -n,11:xn-l'

{X, P"]_

= n111fjn-t.

Tue given relations then follow by expanding f(x) and g(p) in Taylor • senes. 4. Let us consider the operator

d(t)

= exp(,L)dexp(-,L),

where s ìs an auxiliary parameter, and let us find the differential equati on ,vhich is satisfied by d(s):

d!~)

= Lexp(,L)dexp(-,L)-exp(,l)dexp(-,l)L =

[L,a(s)J_.

We shall differentiate this equation once more,

d~:(•l

=

[L,

d!~)l

=

[l, [L, d(,)]_]_.

One sees easily that the derìvative dn d(s)/dsn is equal to n successive commutators of the operator l with the operator d(s). If we now write the operator

exp(l)dexp(-L) = d(l)

156

Answers and solutions

3.5

as a Taylor series,

da(O) I d' a(O) d(l)-d(O)+ ds +2) ds' + ... , and express the derivatives with respect to s at s = O in terms of successive commutators of the operator L with d(O) = d, we get the relation which had to be proved. 5. I.et 1(.\)=exp(.\A)exp(.\ll). We then bave

~\À) - A

0

exp (.\A) exp (.\ll) + exp (ÀÀ)1lexp(.\1l)

- (A+1l)f(.\)+[exp(.\A),1l]_exp(Àil).

(!)

For [exp(ÀÀ),ll]_ we find oo ,\n

oo

,\n

L - [A•, ll]_ - L ( I)-I KA•-> - .\K exp (.\A), 1 n-o n n.. o n (2) where we bave used the fact that [A,.S]_ is a c-number and thus commutes with A. Combining (1) and (2) we bave [exp (.\A), ll]_ -

0 ~\.\) _ (A+ll+.\K)~(.\) with the generai solution 1(.\) - exp [.\(A +ll) + ½.\' K] C,

where C is a constant, arbitrary operator. By letting ,\ ➔ O, we find that C is the unit operator, which concludes the proof. 6. The proof follows, as both the left-hand and the right-hand sides of the equation satisfy the same differential equation •

dO -... __ .t - - -Bil+[A B] e- , da. ' -

=

and both sides vanish fora O. 7. We consider first of ali the case of a discrete set of wave functions ij,i. The ave rage values of the operators A and Il in the state characterised by the function ifo (i/; = :I: a, if,1} are equal to

.•

A= ,La;A,kak,

3.8

Commutation relations, etc.

157

We consider the non-negative quantity l(À) =

f [f(Au,+i,\B ,)a,.]* [f(Ai +iÀBi )a,] ;;:O, 11

1

1

where À is a real parameter. Collecting terms of the same power in À and using the fact that A and /J are Hermitean (Aik = A[t, Bui = BZ1,), we find

+ ,\2 at Bki Bilai] = A2 - ,\e+ À2 .S2. Here

é is thc

Hcrmitcan operator

1 C - ; (All- llA).

' The quadratic form J(À) is non-negative and thus 4A

2

l1 2 ;,:(C)2 • If we note that the operators b.A = A -A and l!J.É = 11- É satisfy the same commutation relations as A and !J,

Il.A ll.Jl-ll.Jl Il.A - iC, \VC

get

J[(ll..4)' (ll.1ll'l ~

L;l.

The proof of this relation for a continuum set can proceed along similar lines. The expression

J(,\) - Jc(A + i,\Jl) ,',]' [(A+ i,1/l) ,',J d, with real ,\ is non-negative and can be transformed as follows: J(,\) -

J[(A,',)' -i,1(/l,fo)'] [A,',+ i,1/l,',J d,

-f[,','A•,',+

i,\,',*(All- llA) ,', + ,1• ,',' Jl• ,',] d,,

where we use the fact that, as the operator

f(A,',)•~dT -

f•· A~d,.

The rest of the proof proceeds as before.

s.

A is Hermitean,

-

(i.lq)'(AF)'>~ ~: '.

158

Answers and solutions

3,9

9, The energy of the oscillator in a stationary state is

' s1nce

P ~ 1r means that rp is completely undetermined and thus essential l!,,.rp ;>,, 1r corresponds to Arp = oo. This is a physical argument, but it does not completely remove the difficulty. This difficulty is removed when we bear in mind that (1) is derived by assuming that "5fz and I{' are Hermitean operators (see problem 1 of section 3). However, "5fz is nota Hermitean operator, since the relation

only holds for f(rp) and g(rp) which are periodic functions of rp, but not for all functions f(rp) and g(çi). For a more detailed discussion of this problem we refer to short notes by Judge and by Judge and Lewis (Physics Letters 5, 189 and 190, 1963).

Commutation relations, etc.

3.18

163

17. The wave function 1/J(x, t) of a free particle depends on 1/i(X, O)

as follows: where

5+•

( I 5+• - (2;,n)I

I a(p) = (Z1r1l)' -,;;:, if1(x, O) exp

,pfi,x)

-i

dx

0 -p)x) (;(p -oo ~(•) exp n dx,

The function a(p) differs appreciably from zero only for those values of p ,vhich satisfy the condition

IPo;Pls;;;I. If x varies Vi'ithin the interval

(A)

+s,

-◊ ~ - µw' i(t) dt

(3) ·

Because these equations are linear they can be' solved as any other set of equations for ordinary, c-number quantities. We get then

X(t) P(t)

= é'1 coswt+fasinwt, ) = é'9 µ,wcoswt-é'1 µ.wsinwt.

(4)

The constants of integration è1 and t 2 are only constants as far as the time is concerned, They can be determined from the initial conditions. Indeed, from the definition of the Heisenberg operator it follows that

L(t)[,_,

~ exp(ibt/n)Lexp (-ibt/n)[,., ~L.

Hence we bave

X(t)l,=o =X= Ci, Jl(t)[1. , ~j) ~ e,µw. We bave thus finally

.i!(t) = Xcoswt+Lsinwt, µw

P(t)

= pcoswt- µwXsinwt.

These equations are valid in any Heisenberg representation,

(5) In

particular, for the coordinate representation ,ve get

Xx(t)

= xcoswt+. n, sinwt ò, tµ,w

0X

0 p.,,(t) = ~coswta -µ,wxsinwt. l

. él\llint 31.111 òt

=

~ (H1)1nt\llint,

(6)

X

"

ihL1at

=

~ ~ [Lini,Ho]_.

33. The operator corresponding to the time derivative of a quantity A which does not explicitly depend on the time is of the form

(!)

3.34

183

Commutation relations, etc. •

The average value of A in a stationary state of the discrete spectrum with a wave function 'f..., is equal to the corresponding diagonal elemcnt of the matrix of the operator (1), À

=

f

f!ÀfndT=Ànn

= ~(HA-AH),.

11 •

(2)

By virtue of the fact that the Hamiltonian is diagonal in the energy representation and that the matrix elements of A are finite in the case of the discrete spectrum, the right-hand side of equation (2) will tend to zero identically, which proves the required relation A= O. (The • classica! analogue of this property of A is the fact that the time average of any bounded quantity A vanishes.) A concrete example of this equation can be found in problem 16 of section 5. 34. The virial theorem in classica! mechanics states that if the potential energy of a mechanical system is a homogeneous function of its (Cartesian) coordinates, and if the motion of the system is bounded to a finite region of space, tbe time averages of tbe kinetic energy T and the potential energy V are connected as follows:

nV = 2T.

(1)

Tbe bars indicate bere time averages and n is tbe degree of homogeneity of the function V(x 1, x2 , .•. ,xi, ... , x3N), wbere N is the number of point particles in tbe system, so tbat we have from Euler's tbeorem on bomogeneous functions (2)

Tbeorem (1) remains valid also in quantum mecbanics if we understand by averaging tbe quantum mecbanical averages (over tbe wave functions}; we sball denote tbem here by ( ). It is convenient fora proof to use matrix metbods. (We note tbat a proof using tbe Scbrtldinger equation in tbe coordinate representation is appreciably more cumbersome, tbe same applies to the proof of tbe sum rute in problem 83 of section 7 .) We bave

(Pl av)

' ' 2T-nV ~ a.V ~ ~-&;"- . i=1

Using tbe operator relations

/J-i

oxi

,Pi=-av Ox;

(3)

184

Answers and solutions

3.35

we write equation (3) in the following form: $,V,

•

2T-nV ~ I; (x,p,+&,p,). A

A

,_,

(3')

This operator relation is, of course, valid in any representation. We now go over to the energy representation, and use as a base the orthonormal set of eigenfunctions ifim of the Hamiltonian fl = T + V. Since the motion is supposed to be bounded these functions refer to a discrete energy spectrum so that we are dealing with matrices all of whose matrix elements are finite-this is essential for our proof. Let us write down the diagonal matrix element (m, m) of the operator {3') or, in other words, the average value of the operator (3') with respect to the wave function of the stationary state i/Jm• From the rules for multiplication and addition of matrices we get 8N

,._= (2T- nV)mm

~(-11 I; I; [(x,)m1(P,),m + (x,),,,(p,J,ml·

(4)

We have also the well-known matrix relations

where E,,, and E1 are energy levels of the system. The e.xpression under the double sum sign in equation (4) is thus equa! to zero and we have found the required quantum mechanical generalisation of relation (1),

n = 2(T>.


-e'

i e1c,i, +.,.,) cos 2 ½"

-

;; 1/1~,

v'2

; . - e 1.., sin{}

;

y 2 e-1

cos1'

y2

-e-i(o/1-,p) sin 2

-e1 sinl~cosj(~- o/), ~sinjo!> = sin!asinl(~-;), 1sinjo!> = cosjo!>sinl(~+ o/),

and Il>

222

Answers and solutions

4.19

Hence, using the results of the previous problem we find

T = cos½ ,,I>+ i(tl!{i;" + (30y + 'Yiìz) sin½ IJ>. 19. W(+½)

=

0 2 COS

z,

The average value of the spin component is equa! to

½cos B.

20. We use the matrix of the transformation of the components of the

spin function under a rotation of the coordinate axes. Tlùs rnatrix was gìven in the solution to problem 16 of section 4. Using this matrix we -fmd the spin function in the new system of coordinates

if,l,

= exp [½i(g;, + lf) + ia:] cos ~. cos 8 + i exp (½i'( il,, -

ip)+ i,8] sin!. sin 13,

v,; = i exp [ ½i(~-if,)+ù~'] sin! .cos 13 +exp [-½i(~+if,)+ i,8J cos

!

,sin 13,

We find the probability that the spin is directed along the z' ~axis,

= 'Pi."" if,l, = cos2

w1

!.

!

cos 2 /3 + sin 2 .sin2 13 +½sin 0 .sin28 .sin (ip+ o:- (3).

From this formula it follows that the probability for the value of a spin component along the orbital direction depends only on the difference lX - /3 and not on cx and /3 separately. 21. The spin direction is determined by the angles

0 = 28,

=

¼w+,8-,.

22. It is possible. In the case of a mixed ensemble for every direction of the inhomogeneous magnetic field one will always get a splitting into two beams. In the case of a pure ensemble by a suitable alignment of the instrument one can obtain the disappearance of one of the beams. 23. The operator of the square of the total spin is given by the equation À

2

À

À

s = Si+S1+ 2(S l. S2) = ;11 i 112+ ½11 2

'

2

{ Ò1xÒ2x

+ 61yÒ2y + Ò1z Ozz 1

=!n'(b ~),(b ~),+in•[(~ b),(~ b),+(~ ~;),(~ --;), 0 + (b -1), (b -~) ,] · where the indices I and 2 distinguish the two particles and the operators

221

Angular momentum; spin

4.28

with index 1 (2) operate only on the spinar of the fir9t (second) partici,, while 1 is the unit operator

(b ~).

We now introduce the unit spinors Cl! and {3:

The general spinar of a system of two spin-½ particles can then be written' in the form •

•

•

and one finds that '11T is an eigenfunction of Sz; = S 1.a-+ S 2 z = }h(0 1 z + 17:iz ), provided b e or b = -c. Solving the eigenvalue problem S 2 '11 = X'V we finally find the following joint eigenfunctions:

=

'112

s2 = 211 2 , S' = 2n' , S' = o ,

= 0:1/32+/310!2,

'l/'3=/31/32, '114

Sz = 11 ; Sz = O; S' =-h., Sz = O.

s2 = 211 2 ,

W'1=0:10:2,

= IX1/32-f310::2,

27. The possible wave functions for the two~particle system are i/i1,1

= IX1ct2,

!ft,-1

= f1tf3i,

I

'Pt,O

= ✓2(0:1/J2+/J1ix2};

I

~"-" = ✓2(,,P,-fi,,,). The first three wave functions correspond to the triplet state and the last one to the singlet state. The dipole-dipole interaction energies corresponding to these wave functions are

(V),,,= (17 ),_, = -ZA/d'; The wave function at t

f

(V),,0 = 4A/d 3 ;

= O is ✓ (2)[1/1 1 , 0 +!f 0, 0]

(V)0,0 = O.

so that at time t we bave

= ✓ (2) [f,,o oxp ( -4iAt/d'n) +f,,J,

and we see that after a time 'li,d 3 /4trA the wave function will be 1P = - ✓(2) [tJ,1,0 -!fo,o], corresponding to a spin-flop. 28. Proceeding as in problem 23 of section 4 we find four quartet state spinors: ir1 i"2 \Jr 3 \Jr4

S' = 'f·n• , = 0!10!1ct3, = a,a2,l33+0!1./32ct3+,/31a20:3, S'='fh', = a, .132.133 + /3, 0:2 /33 + 131 .132 0:3 , s2 = 1:1-112 • S'='fh', = f31f32f33,

s, = ;n;

s' = 0' n ·,

S,=-½h;

s, = -;11 ;

224

Answers and solutions

4.31 and two, degenerate, doublet state spinors (). and µ are arbitrary constants) 'I' s = a:1 azf33 + Xa1/3 2a3- (1 + X)/3 1 a: 2 a 3 , '1'6=0i.1/l2/33+/JIJ1CXz/33-(l+µ)f31/320:3,

S2 = ¾11 2 , s2 =¾12 2 ,

s, = ½11 ; S;:

=-½11.

31. By writing out explicitly the corresponding 4 x 4 matrices one finds that

(I) where Jkl is the 4 x 4 unit matrix. Using (1) and the results obtained in the preceding problem, the proof foUows in a straightforward way. 32. Between the components of a spinar function and the components of a symmetric spinar there exists the following relation: .,,, 'f'

-

.,. 'f'll

.,.,, -

.,.,, -

'f'

'f'

-

-

- I .,. ~2 't'O•

.,.•• - .,. 1 'I' 'f'- '

(I)

where ij; 1 , l/) 0 , and 1/1- 1 are the components of the spin~l spinar (cf.problem 12 of section 4). A spinar of the second rank transforms as the product of two spinors of the first rank, that is ,P'll

= rx2 ,p11 + 2c.{J,P12 +{J',P2 == = -½

5.4

flft{9 +:; l(l;l)}if, dT, 0

1

As far as the integrai J1.!:..(l+2 l)if;t-+1 1f1+1 d-r is concerned it is always µ '

larger than zero. Hence statement given above.

Ef11ti < BM_n,

that is, we bave- proved the

4. The potential energy is V(r) = ½µw 2 r 2 . The radia! part R of the wave function satisfies the equation

If we substitute x

= Rr and

use the notation

k - )(2µE)

-

n,

'

E - ,\ n,

•

we have

, + /k' - ,\',r - --'---:,---C 1(1+1)) ,, X = O·

X

(1)

If we consider the asymptotic behaviour of X as r ➔ O and as r...;,.oo, we can put the solution for x in the form X= r1+ 1 exp(-½Àr 2)u(r),

(2)

If we substitute expression (2) into equation (1) we get an equation to determine the function u(r):

1 1 u"+z( : -,\,)u'-[2,\(/+J)-k'Ju- O. By introducing a new independent variable f

= Àr2

(3)

equation (3) goes

over into the following differential equation:

d2 u

du f df' +[(1+!)-fldg+[i(l+!)-t,]u- O, where

The solution of this equation is the confluent hypergeometric function

u-F[½(l+j-,),1+!; ,].

5.8

Centrai field of force

235

Rcquiring that R vanishes as r ➔ co wc get

~(l+1-s)=-11r (nr=0,1,2, ... ) and hence we find that the energy levels are egual to E,,,1 = 1iw(l + 2n,. + ½), and the wave functions gìven by

1/1,,,im

= r1 exp ( -

½,\r 2) F { -

n,., l + ,J, Àr 2} 1;"111 ('!?, 1) ).

5. The wave functions are where

••••••(x.y.z) - ? •. (•)?••(Y)?••(z).

1

1 (

o). exp(-½.\x2).

1l,,(x)= .J(2n,\n➔ nJ)~7T Àx-ax

The corresponding energy levels are (see problem 5 of section 1) E 11 ,,,,n,

= 1ìw(n1+n2+n3+}).

The connection bet1,veen fn,~m and (l)n,n,n, for n,. = O, l

form:

fo11

= .J~ (a,

oo

r O will be of the form

e,(~) exp [ -i(k+. r)] + e,(~) exp [i(k;. r)], where C+ and C,i, are the expansion coefficients of the initial spin state in terms of the states ·(~) and (~). A simple estimate shows that even

6.17

Motion of particles in a magnetic field

263

for X~ 104 Oe an appreciable reflection can only take piace for very slow (thermal) neutrons (~~ 1 A) and for angles of incidence q; which differ from 1r/2 only by a fraction of a degree. 17. The Schr6dinger equation for the spin function in the z-representation,

(;j, is of the form

where µ, is the magnetic moment of the particle. We shall introduce the notation ~Xcosi9-=a,

~Jf'sinD=h.

In this notation the equations which determine the components s and s 1 2 have the following form:

~: = ias1 + ib exp ( -t'wt) s2, d,, dt

= i"b exp ("iwt) s1 -

. • zas 2

The solution of this set of equations is s1

= Aexp(zj)1 t)+Bexp(ip 2 t),

s2

= exp(iwt)[-ab+Pi Aexp(ip1 t)+ -ab+p2 Bexp(tf2

t)],

where

J(~ +a +b +wa)-~, P = - J(~ +a +b +wa)-;. 2

2

p1 =

2

2

2

2

2

The quantities A and B are determined from the initial conditions and the normalisation condition ]s1 ]2 +/s 2 !2 = 1. After some simple calculations we get for the transition probability the following value:

P(½, -½)

= l +q :in;t'}, sin 2 [zt w(l-2qcosD+q 2)1], - qcos 0

where q is the ratio of the Larmor precession frequency to the frequency w of the rotating magnetic field,

q--

2µ.JF liw

WL

- -

w

264

Answers and solutions

6.18

The quantity q is positive if the magnetic field rotates in the direction of the precession and negative if the rotation is in the apposite direction. If the angle f} is small, that is, ✓(Jlt'!+J'f;,)/.Jl;I'' 2 - 2µ, Oz 2 + Op 2 + p Op+ p2 Or; 2 - 2µc ~Or;+ 8µc 2 P f

= Eif,,

which far a particle which is constrained to move along the circle p = a reduces to

6.27

Motion of particles in a rnagnetic field

269

with thc solution

,f; = exp(imq;),

m =O,+ 1, +2, ....

The corresponding energy eigenvalues are

E -

_I_ [""'

,a.Yt'] '

mzp,a+Zc'

and we see that the ground state energy as function of :tris given by the equation 1

Egrst

= 2p,

(mn eaJlf)' a+ Zc '

ea 2 .7t" - Znc -t~m~

-

ea 2 .Yt' 21ic +t.

This mcans that Egrst is a periodic function of JR with periodicity Znc/ea 2 and with discontinuities in slope at .Yt' = (2m+ 1)1ic/ea2 .

7

Atoms

· 1. From the given inequality we get

f! 11,f,l'd, +Z f(171+1'- 17,)d,+Z' f(V, .17r)/,f,/'d,;,O. Integratingthe second term by parts and usìng the fact that (Vr. Vr) and 'v' 2 r"" 2/r, we bave

!f!l7,f,j'd,- ffJ,t,j'd,;,

= l,

-f fi+/'d,.

The Ieft-hand side of the inequality is the average value of the Hamiltonian operator {in units e = 1i, = p, = 1) 11 = - ½~-Z/r for the state !p. The lower limit of the energy, -Z 9/2, is reached for the state with wave function 1/,0 , which satisfies the first-order differential equation

Vl/,0 + Zifi0 Vr -= 01 from which it follows that

,f,,~exp(-Zr). 4. We evaluate .ilrst of all the wave function in ·the momentum representation from the general formula

fff

~(p) = 2(2:li)'

exp [ -i(p. r)f7i],t;(,)d,.

For the ls-state we find

1 (2a)'

~,.(p) = w

I

1' [(P'•'I•')+ 1j;•

Siroilarly we find for the 2s-state

Atoms

7.7

271

For the 2p-state there are three eigenfunctions (m. == -1, O,

+ 1)

'''() = .!(")' Pza n A[(pla2/1ì.2)+¼P'

g;~ p

-t 7f

(")t (P~±iPv)a p - - ✓(2)• n n[(p'a'/•')+ìJ''

1±11( )·-

~.p

i

From these expressions we can find the normalised momentum distribution w(p) - lf(P)I'· s•. The Hamiltonian function in relativistic mechanics is of the fonn

H~

✓(l'','+p'c')- l'''+ V(r)~~+ V(,)+H,,

where

H1=-

p•

s2•

8I' e

We shall now take p to be the operator p = -ih'v, and H1 will be considered ·to be a small perturbation. We then get in the approximation used by us in the Schr6dinger equation [pa/2µ,+ V(r))Y, = Eifr, and the required correction for the energy in a state with n, l, m is

!:.E,~ - ,,~,, f,f,•p•àf,d, 8

3E'

~- f

(µe'/1!")'

z!,,

[ 3

f,f,'[E- V(r)]',f,d,

1

lli'~ ,,.. ('')'·

~2,,,,-n,,,,,(21+1)~ Sn'-(21+1)•'

7. The tritium nucleus goes aver into the 3 He nucleus during /3-decay. The influence of j3-deeay on the atomic electron is essentially contained in the fact that during a short time interVal, t ( ~'li-3/JJ-l'), the potential energy of the electron in the atom changes: it becomes equal to V== -e2/r instead of V= -2e2/r. The time t can be estimated as the time of flight of the µ-particle through the atom,

t ,..,:!]

•'

wbere a0 = 1i2/µ.e 2, vis the velocity of the fl-particle. Since the energy of the p~particle is of the order of severa! keV, we find ,,..,Q,11t3/µ.e'. The wave function of the electron does not have tìme to change during t, since we get from the Scbri:>dinger equation ,, t

6,f,~- .• ,f,4.,f,,

r '"

272

Answers and solutions

7.7

Let us expand the wave function of the electron !p in terms of the eigenfunctions of the electron in the field with Z = 2,

i/i =

f Cn ifin + fe,. if,li d k. 3

The expansion coefficients

e,.=

f

#i.d-r

determine the probability for excitation Wn

and ionisation

Wion

= ~]cnla

= J1c,./2d3k,

Since t/i is spherically symmetric, c.. and e,. will be different from zero only if the states n and k are s-states (I= O}. Since

(Z)'

(

2Zr) ,

R~1 =2,i exp(-Zr/n)F-n+1,2, n we find -

rooRIZIRIZ'I

Cn - Jo

no

10

Ifwe put Z= 2, Z'

'd -

8 r T-(Z'+Z/n)a

(ZZ')'F(I n n+ '

= 1, we get for n = C1=

3

2Z )

2 ' 'nZ'+Z,

1:

!6J2 27 •

that is, the probability for the 3He-ion to be in the-ground state is equal to w1 = lc1 [2 = (8/9)3 = 0·70. The total probability of excitation and ionisation will thus be egual to 1-w1 = 0·30. For

n = 2, c3 =

-½,

w3 =

0·25.

Using the equation

F( cx., /3, y, x) we find

= (1- x)1-a-p F(y- cx., y- /3, y, x)

7.8

273

Atoms

We find for the value of the probability for excitation of tbc firat few levels, using this formula,

=

W3

zo 30 SlO :::::

1•3%,

W4

= 2'' :::::0•39%, 612

8. Let the electron be in a stationary state with well-defined value m of the z-component of the angolar momentum. The wave function of such a state is equal to:

= Rm(r) P/ml (cos .i-) exp (i'm11),

lfnim(r, fJ, 'P)

The electrical current density in the state coordinates, r, {}, .>.

+ :!:

M

i>k


),

we find that the expression within braces in equation (3) is equal to

Z✓U~ + I)] [(j + m1) (j- m1 + 1)-(j + m1 + 1) (j- m1)] - ✓[i(J~ I)].

7.47

Answers and solutions

302

lf we then integrate over·r, we get

-!•./[•'-(i+½)']. We thus :find finally for the perturbation matrix elements (3):

-

-

.J[•'-(i+l:)'] V.,-V,,--!n iU+l) m,elf. We find the required energy correction by solving the secular equation -

li

JJ;l

l'i2 -- o,

-

+ ,,

E=_f'lll•

fi

'= ±¾./[•'-(i+½)'lJ(i:11)""' Fora given value of n the term withj = n-½ is not split in the electrical field since it is not degenerate with respect to the quantum number l (I has a fixed value, l =j-} = n-1). Ali the other terms of the· fine structure are split into 2j+ 1 equidistant levels (mj = -j, ... , +j). 47. « = jai, where a is the Bohr radius. 8

,//, (j+½)'

48.

µo= Zµc j(j+l)mJ• 49*. In the case under consideration the spin-orbit interaction Vi., the relativistic correction due to the change in mass J'2, and the energy of the electron in the external homogeneous electrical field Vs = - etfz are all ~f the same order of magnitude. We shall therefore consider their sum as a small perturbatìon of the originai syatem. For our calculations we shall start from a state where the orbitai angular momentum I~, its z-component m1, and the z-component of the spin ffi.e (the electrical fièld is along the z-axis) have well-defirted values. If we evaluate the matrix elements of Vi_, J!;,, and V,. we have (in atomic units) .1

2

mi(m-m1)

8

•" •'l(l+½)(l+l)" ('')l'm( "1 1,111 -

lo

, ✓ [(1+½)'-m'] •'1(1+½)(1+ 1) 811·

fo, fo,

m1 = m- ½or the other way round, O

otherv.,ìse,

1 V.)•·~•=_!:( . ( 2rm, zns !+½ -~)a,,a 4n mimi•

7 ,49

Atoms

303

If 11 = 2, the energy of the st,1te ,vith quantutn numbers l=l,

(}=])

1n=m1+111s=+~

,vili not change in the electrical field. 'fhe shift of this leve! due to Vi_ and li; is egual to C< 2/128 at un (see problem 23 of scction 11 ). The splitting of the leve! with quantum numbers n = 2, m = + ~-, can be found from the solution of the secular equation

8J2

o

-fS-Ertl

-3eo'

= O,

-3e6' where 38 = ù 2/32 at un· is the fine structure splitting of the leve! n = 2 when there is no external field. If wc introduce in that cquation .. v.1hich is connected with EUl by the relation E!l)

= E-J:l·ò

[ - 1,/·ò

is tbc cnergy of tbc ccntre of gra\'ity of the thrcc energy lcvcls: (Efl)+Eil)+E~1 l)/3 = _J_/S], we get

-,

o =0,

-3eiff

O

-3eC

-S-€

or

\Ve shall solve this equation both for weak fields (ed - E{01 I bet,vccn thc ground state and the first excited state. If wc solve this condition for tJ, ,ve get

li' ,g «s,'(3l;i1· As we should bave expected, this inequality also means that the perturbation energy ( ~t!d) is small compared to the distance between consecutive levels ( ~1l 2//). We use the relation betv.,een the normalised spherical harmonics

J[(z~:i;;;:~)] Yi+sm J(~ -;"'J Yi-,.m, l m~7i,

cos 0 Yi,m

~

+

1

and we find for the polarisation in the lm~state 01

2/µ,2 [

(/+ 1)2-m2 z2-m2 (Zl+3)(Zl+l)(l+l)-(2l+l)(Zl-l)l'

where the value for l = O is found by putting m = O and letting l ➔ O. I t is rather paradoxical that a rigid dipole is "polarisable". Consider the case when l = 1. If m = O, the dipole behaves diamagnetically, and if m = ± 1, it behaves paramagnetically (or "norma!"). The degeneracy is only partly lifted, as the energy shift depends

only on Imi, Finally we note that +I

L

CX1,n

m--l

53, (a)

EnJ =

= o.

. i~; eà((2·+t)2Jf' 8J(J+l)µc J 3

+ J\(Zj + l)'Jf''+ ~:' [n'- (j + ½)'] ,g, ))

(see problem 46 and 49 of section 7).

(b)

308

Answers and solutions

where

1:

7.54

follows from the solution of a cubie equation 1,3

+Z#t:2 -

E'.(382 + 9F 2 -,BZ) ± 28"' ,B- zaa = o,

where

p--2JAC' !E_ Jlt' Fora strong field (6

F

,i( t•)

=e',.W2

'

< F, 6 ,( II) we bave (see problem 49 of section 7)

54. Let the z•axis be along a magnetic field direction and the x•axis along the electrical ditection. 1~he potential energy operator of an electron in these :fields will then be given by

We shall consider this expression to be a small perturbation, and we shaU characterise the unperturbed stationary states by the quantum numbers n, l, m, a (m and o are the Z•components of the orbitai angular momentum and the spìn). The non•vanishing matrix elc1ncnts of x are of the form: 2 v-,.,,, _ ( .u) Vl •+(I+ Au)(V ,{ft ,f,,). V,u)+ A,J,t,p(V,u. V,u). (3) If we integrate this identìty over the whole of configuration space we get

j(+t ,f,,(1 + Au) Vl u + (1 + ,lu)(V,(ft f,). V, u)) d, = -).f,J,tf(V,u.V,u)d,.

(4)

Substituting expression (4) into equation (2) we get ~

T=-

z1•1r, ,~, 2 J[,J,:(I + Au)' 'l!,j,,+,J,,(1 +Au)' V!ft] d,

i 2

+).~

i=l

(ifitiJ; 0(V,u.Vtu)dr.

j'.

rfhe Hamiltonian operator 1ì is equal to 1J = 110 + u = T + fr +u, Sin ce , V commutes with (1 + >.u), expression (1) can be written as follows:

~

H

½r(I +>.u)'(,J,t 1i,J,,+,J,,1ift)dT+ÌÀ' i r,i,: ,f,o('i,u. v,u)d,

= Eo+ J'

Jc1

1-1J' + ).u)',f,3 ,j,0 d,

Since 8 0 if,0 = Eo'Po, we bave

_

f,i,3u(l+Au)',f,,d,+tA•i f,t,3,f,,(V,u.'i,u)d,

ii= Eo+ J:

i--1

J'.

J(I + ).u)',f,t ,f,, d,

•

315

Atoms

7.63

or

(5) where (u) 00

=

fft

uif,0 d'T, (u 2) 00 =

f1Pt u2 i/io d'T

and so on.

We shall

expand the second term in equation (5) in a power series, neglecting terms like (u 3) 00, (u)~o• .... For the energy correction /),,_E we find an approximate expression of the form

•

t:.Es (u) 00 + 2,\(u')00 - 2,\(u)&, + ½,\' L [(V, u. V, u)] 00•

,_,

(6)

The variational parameter À is determined from the condition

dc.E d,\

•

~ 2(u')00 -2(u)!o + ,\,_, L [(V,u. V,u)J 00 ~ O,

so that we find for À the expression (u)~o -(u2)oo

L• [(V,u.V,u)J.,

•

i..,l

Substituting this value into equation (6) we get the required relation

LìEz(u)oo-2 l(u)Ho-(u2)oo]2 .

L [(V,u.V,u)J ,_,

(7)

00

63. If the atom is in a unifonn electrical field of strength &, along the z~axis the perturbation operator is

•

u = -é"Lzt ,_, = -é"z, whose matrix element (u) 00 is equa! to zero (we use atomic units). From equation (7) of the preceding problem we have t:.Es - 2.ff' [(z'),,l ,

n

where n is the number of electrons. I t follows thus that the polarisibility is egual to:

a= 4[(z2)ool~. n

7.64

Answers and solutions

316

It is necessary to note that this equation which is obtained by using one vati.ational parameter À is a reasonable appYoximation only for hydrogen and helium. For the case of atoms with several electron shelts the deformation of the shells will not be the same. To obtain a better result from the variational method we must thus introduce for each electron shell its ov.•n variational parameter. For the hydrogen atom we bave 14n J.w exp (-Zr) r' dr= 1 (z2 ) 00 = "5(r2) 00 == 3-;0

and thus a:=4atun.

In the cgs system we bave a=

4(~,·I!') cm

3

,

For the helium atom we put the ground state \\'ave function in the form (see problem S9 of section 7) 1/10

= z:ttexp[ -Z.,u{r1 +r2)], n

The final result is a=

From this value of normal conditions

0·98 at un ix

or

a

where Z,,a =

~i"-

A•)' cm'. (~•'

= 0-98 - ·

we get for the dielectric constant of helium under

,

~

I ·00049,

while its experimental value is 1·00074. The relatively large difference between the theoretical and experimental values is mainly due to the fact that \ve have used a rather rough approximation for the unperturbed wave function. 64. The potential of a point charge is the same as the potentiat of a sphere charged on the surface outside that sphere, if the total charge is the same in both cases. Inside this sphere the difference between the two potentials is given by

Aq; =

-ze(~-1).

The change in the potential energy of the atomìc electrons is equal to:

AV=-Ze'Z

-1-1

(!-!),(,,), a ,1

7 .65

Atoms

317

wbere we bave introduced the function

1:(r)

=(1

r a.

According to first•order perturbation theory tbe energy shift is

L\E

(-1- a

~ -z,, jlfirl',tL .. 1 r,

1 ) ,(r,)d,, ... d,N.

If we integrate over all variables except one, we get

fir(,, ... rN)J'd,, ... d,N

~ ~p(r),

where p(r) is tbe electron density. In this way we get

L\E

~ -z,•fp(r) (; -

~) ,(r)d,.

We can use the fact tbat p(r) changes little in tbe region r < a. If we take this quantity from under the integra! sign and put it equal to its value at tbe origin we get

(1) 65. We use for the shift the equation (compare preceding problem)

(1) where 6. V is the difference between the actual potential energy and the Coulomb potential and if,0 tbe ground state wave function. If p(r) is the nuclear charge density ,vhicb is normalised so that

Jp(r) d 3 r = Ze,

(2)

wbile its mean square radius (r 2) satisfies the equation

Jp(r) r2 d 3 r = Ze(r 2 ),

(3)

we bave for .6.V the equation

'V()_ r -

~

J' (

Ze 1 p r ')d' r, --+rro·

f.~ r

p(r')d'r' . r'

(4)

318

Answers and solutions

7.66

Substituting (4) into (1), integrating by parts, using. the fact that Y,0 changes little over the nuclear charge density, v,e find, using (2) and (3), tbat

(5) If the charge distribution is a spherical shell as in the preceding problem, (S) reduces to equation (1) of that problem. 66*. The wave function of an s-electron is

1 x,(r) ,t,,(r) = ✓(4n-) - , , where Xn satisfies the equation

x;+ T,[E.-V(,)Jx. = o

fo X~, dr = 1. 10

and the normalising condition

In the semi-classical approximation the solution of the equation for X,t is of the form

(!) where

P.

~ ,f\Zµ[E. -

V(r)]).

This solution is1 however, useless for small values of r. lndeed, if r is small (r~'/i.2/21 µ,ci}, then first of all we can neglect the screening of the field of the nucleus and put V(r} = -Ze2 /r; secondly, we canalsoneg1ect En compared tO V(r). If we put p = ~(2p.Ze 2 /r) into the condition. of applicability of the semi-classical approximation

d(•!P.1~ 1 dr

we get

'

n•

'~z,,.e2· To obtain an expression for·x» which can be used for small values of r, we start again from the originai equation and substitute for V(r)

-Ze~/rand also neglect En:

7.66

Atoms

319

1'his equation has the follov.'ing solution: (2)

To find the connection hetv.'een the constants C 11 and A,,, wc note that the region of applicability of the se1ni-classical solution (1),

Il'

r'p----

Zµe2'

and the region of applicability of solution (2), where the screening of the nuclear field has been neglected,

1i2 r~ Zt µe?.' overlap for large values of Z so that solutions (1) and (2) must coincide in the region

We shall show that solutions (1) and (2) are the same in the entire region of applicabi!ity. To do that we put in solution (1) p = ✓(2µZe 2 /r). We then get

Xn

=

[2 ✓(2eZe'r) ] {l(ZµZe 2) cos "li + (t). 72. Instead of flqùations (l) to (4) of the preceding problern we now bave &(() W

-

10 -

PJ( 2µwn ),-,,, .'

p

= -e-t/T • ~IO = V,-OI = _ r er ' ~;;.-~ P'

2-,iJUl.w

II.

e O

iwr-t/T dt

I'

- 2µ1ìw(lP' + w2-r2)

•

73*, The generai equation for the probability of a transition per unit time from a state of the discrete spectrum into a state corresponding to an infinitesima\ energy interval of the continuous spectrum through the action of a periodic perturbation of frequency w is of the form

(I) ln this expression n is a set of quantum numbers characterising the state of the discrete spectrum, v a set of quantities for the states of the continuoùs spectrum, dv the corresponding infinitesima! interval, E~01 and Ev are the unperturbed energy levels of the discrete spectrum and the continuous spectrum, while F,,11 is the matrix element of the perturbation operator for the transition under considcration, which will be defi.ned more exactly in a moment. The wave functions of the discrete spectrum are normalised to unity and those of the continuous spectrum to ~(v-v'). I,et r and - e be the radius vector and charge of the atomic electron and let 8 0 be the amplitude of the intensity of the ltniform electrical field. In our case the perturbation operator Vis of the forro

V~ ,(C(t).r) =

,(C,.r)sinwt = Fexp(-iwt)+F*exp(iwt),

,vhere

(2)

•

P = ;,(.t,.r).

(2')

The 8-function which occurs in equation (1) shows the res.onance character of possible transitions; the energy can make a transition to a state of the continuous spoctrum only if (strictly speaking, equation (3) is only approximately taken into account as we shall discuss at the end of this problem) (3) E = EW 1+1iw

"

"

7.73

Atoms

329

(absorption of a "quantum" of frequency w). Since for the hydrogen atom in the ground state (E~-E~ 1)mtn = µe 1/21i, 2, it follows that the minimum frequency which is necessary for ionisation is given by the equation

We shall now evaluate the matrix element

Rvn

-f-1·*

'f'v ft.1.(0) 'l'n

dT

•

(4)

We shall substitute in this expression equation (2') and also ip~> = i/;100 = (1Ta3)-i exp ( - r/a), VJ,, ~ i/Jk = (2n )-1 exp [i(k. r)], where we bave used the normalisation mentioned earlier. We shall choose for our set of quantum numbers v, the wave vector of the free electron k; it is clear that the use of plane waves for i/1" is, strictly speaking, only correct if w ► w 0 , that is, \Vhen the electron is moving fast. We get in this way from equation (4)

F,,

J

= ~ (2,)-1 (na')-1 exp [ -i(k. r)- r/a] (8 0 • r) d' r.

(4')

To evaluate the integrai we introduce a spherical system of coordinates (r,t?,1) with the polar axis along k and we denote by 0 the angle between k and tf 0. We get then

(80 . r)

= tf0r[cos 0 cosi?+ sin 0 sin Ucos (1-10)],

where p 0 is the azimuth of 8 0 in this system of coordinates. Substituting this last expression into equation (4') the second term will clearly give zero, when integrated over p, and we get ( cos t'J-::: x):

F,,n

= 2-l~ai 21rCocos 0

~

rl

[Joct,exp( -ikrx- r/a)r 3 dr] xdx

i'eC0 cos0J1

n(2a)I

3!xdx _1 (1/a+ikx)'

Substituting this expression for

d, = d 3 k

~-

16ka5 n(2a)t (l+k'a')''

e40cos0

l~nl 2 into equation (1) and writing

~k'dkdQ, ~k' dEkr11,2 dk dQ dE = µk df1,dEv

(we use the equation Er

= h k /2µ,), 2

2

'

we get

26 µa 7 e2 C2 k 3 cos 2 0 dw,w=-;;· 'h,3 ·cl+k2a2)6 O(Ev-E~ 1-hw)dOkdEP,

330

Answers and solutions

7.73 where d0.1, is an element of solid angle with its axis along lt. If wc finally integrate dw1w over E", we find the probability of ionisation while the electron has a final wave ve'] r21+adr, and hence for the first sum from (A), using (3),

S = _ 2 l + ~ [ (w,21+4 p(n'> F(n) dr] [ ( 1 3 2! + 1 ~ Jo 1+1 1 • Jo 1

00

r21+a p(n') 1+1

p:il:2(Jl1l+1iì:)-

0~ 2]fl>e1dr,

li' I B = 2M p2· The quantity Bis called the rotational constant. 4. The basic physical fact in the quantum mechanics of molecules is the very large value of the ratio M/µ, where M and p, are respectively the nuclear and the electronic masses. Indeed, the presence of this large dimensionless parameter, of the order of magnitude of 1000 to 10,000, causes a considerable difference between the orders of magnitude of the quantities mentioned in the present problem. Let a be of the order of magnitude of the linear dimension of a diatomic molecule. The distance between the nuclei will clearly be of the same arder of magnitude. Indeed, it cannot be larger than a because of the meaning of this quantity but it can neither be considerably smaller than a because of the mutuai electrostatic repulsion of the nuclei. (We can easily arrive similarly to the conclusion that a must be of the order of magnitude of the linear dimensiona of an atom, This fact about the

354

&.4

Answers and solutions

order of magnitude of a does not play any rote in the estimates in which we are interested.) 1. We shall first of all estimate the arder of magnitude of the energy of the valence electrons and at the same time the order of magnitude of the intervals between electroniç levels. Since the valence electrons move, in contra-distinction to the electrons of the closed inner ehells at each of the nuclei, in a regi on of space of linear dimensions a, so that the u.ncertainty in momentum ò.p wi11 be of the order of magnitude 1i,fa,

~

the zero-point energy of the electron, or the difference in energy of successive electronic levels, will be of the order of magnitude E81 : (!) We shalt now consider the vibrations o( the nuclei in the molecule,

We can use as a model, at any rate far the ground state and the lower excited levets, the moti on of a harmonic oscillator with a mass of the order of M (or, rather, with the reduced mass of the nuclei) and a stiffness coefficient K. We can estìmate the latter from the fact that a change of the distance between the nuclei of ordet a must correspond to a change of arder unity in the electronic wave function, that lS, it must be connected to a change in energy of the arder of magnitude

Ka2 ~Eel• so that we get i11 the usual way for the frcquency of the vibrations of the nuclei in the molecule w~J(K/M) and using equation (1) \\·e get for the interva1s between vibrational levels (2)

Finally tbe rotatlonal levels of the molecole can ctearly be treated as the 1evels of a rotator with a moment of inertia J ~ Ma 2, so that we get for the intervals between rotational levels ;,,,

.,,,,

µ,

Er~t~7~ Ma2""1k[Ee1•

(3)

It is clear from equations (1) 1 (2), and (3) that the quantities Eeh Evlb• and Erot forma geometrie progression with the factor (µ,/M)~ .... 10-11• 2. Let b be the amplitude of the zero point vibrations of the nuclei i.n the moleçule. It will be of the order of magnitude of ✓(1i/Mw) [we can obtain this estimate in different ways, for instance, by equating the arder of magnitude of the oscillator energy ( ~ 1iw) and the potentW

8.5

Molecules

355

energy for a displacement over a distance b ( ~ Mw 2 b~) which fron1 the expression (2) for w gives us

(4) The ratio of the vibrational a1nplitude b to the equilibrium distance a of the nuclei in the 1nolecule is thus of the order of rnagnitude of (µ/M) 1 ~ 1, v\'e can consider this quantity to be a small expansion para~ meter in the theory of n1oleculcs. According to the results obtained earlier Eci, Ev; 1., and Erot are quantities of respectively the zeroth, second, and fourth order in this small paran1etcr, 3. 1'he periods of the electronic moti on and of the nuclear vibrations in the molecule are of the fullowi.ng arder of n1agnitude: 1

7i

1

(µM) 1 a'

Wvill

Il

µa 2

7;,~-·-~E--~,-, wol 'el ri

(5)

Tvih~·--~-i-- •

(6)

The corresponding characteristic velocities are clearly equa! to

v,,~;,;

[o, -J(;1)]-:a'

b [ or. vvih ~ Tvili

~

J(E,-M-

0 ,)]

li

~ µ.t Ml a.

(7) (8) '

From equations (5) to (8) it follows that 1

Tvlb~(M\ ► 1, Vvib~(}'_)i , l..lu• LiH molecule (Li atom in 2 S 0 state, H atom in 2 Su state): 1I:+, 3I:+. HBr molecule (Br atom in 2Pu state): 1I:+, 3I:+, 1 TI, 3 TI. CN molecule (C atom in 3~ state, N atom in 4 Su state): 2:E+, 4 I:+, _6I;+, 2n, "Il, 6IJ. (The number which stands before the term symbols indicates the multiplicity of the terms.)

356

Answers and solutions

8.6

6. The helium atom in its ground state is characterised by the fact that both its electrons are in the lowest state (parahelium). The total eigenfunction of the ground state of the helium atom can be written

approximately in the form

1

Jz f

0

(1) f 0 (2) ["+(u,) "-(o,) - "+(o,)"-(o,)J,

where !fa is a hydrogen function, The hydrogen atom has the eigenfunctions f 0(3J,+(o,)

o,

f,(3),_(u,).

If both atoms are at a large distance apart the wave functions of the system can be written as a product

1 ✓Z y,.(l)f.(2)f,(3) ["•(•,)"_(o,)- ,+(u,)"_(u,)J "+(o,). Taking exchange of electrons into account the wave functions of the system must be anti-symmetric with respect to an interchange of any

two electrons. There is only one anti-symmetric eigenfunction which is also the eigenfunction of our system in zeroth approximation 1 'l' - ✓[ 6 ( l _ S)J ( f "(I) ,f,0 (2) f,(3) ["+(I) s-(2) - ,+(2) ,_(I)] "+(3)

+f.(3)f.(1) f,(2) ["+(3) ,_(1)- "·(1)"_(3)] "+(2) +f,(2)f.(3) f,(1) ["+(2) ,_(3J-"+(3)"_(2)J"+(I)). In expression (1) ✓(

6 (l~ S)] is a normalising factor and

f -f -f

S -

f.(1) f,(2) f,(3) f,(2) y,,(3) f,(1) dr, dr, dr,

f.(1) f.(2) ,f,,(3) f,(1) y,0 (3) f,(2) dr, dr, dr,

f.(1) f,(3) f,(2) f,(2) f,(3) f,(1) dr, dr, dr,.

If we apply the usual perturbation theory we have

(1)

8.7

Molecules

357

where lY is the eigenfunction in zeroth approximation and fl the perturbation energy, ,vhile the summation is over the spin variables, We must take into account that f1 is different for different parts of lfr, and indeed for f,,(l)f,,(2)1f'b(3) the perturbation energy is equa! to 2

f1 = e

(!-

r~

3

-

r~~- r~

,~j,

2

+ r~ 3 +

3

+ r~ 2 + r~ 2 )

and for f,,(l)f,,(3)ij.,b(2) it is equal to

fJ =

2

e

(!-,:

2

-

r~

1

-;~

•

If we take into account the fact that the integrals which differ only in the numbcring of the elcctrons are the same, ,ve bave

K-A i;=

1-S'

(2)

The integrals K, A, S are in generai of the same kind as the corresponding integrals in the hydrogen molecule problem, An evaluation of the intcgrals shows that equation (2) corresponds to a repulsion. This is correct not only for helium but for ali inert gases. 7. If we split off the motion of the centre of mass we find for the wave function of the relative motion of thc nuclei the following equation:

v',f,+ ZM [E+2v(!- ..!_)],~O, p

11,2

2p2

where M is the reduced mass of the nuclei. If \ve separate the variables in spherical coordinates and put

,f,

~ x~)YK,,,(0,r),

,ve find for X the differential cquation

d2x+[-À2+2y2_y2+K(K+1)]' x=D dp2

P

P~

358

Answers and solutions

If we make the substitution x(p) s

=p

8

8.7

exp ( -Àp) u(p), where

= ½+ ✓[r2+(K+½)2],

the last equation goes over into the hypergeometric equation

pu" +(2'-2!.p) u' + ( -2,,\ + 2y2) u = O. The solution of that equation which is finite for

p

form:

= O is of the following

2

u = cF(s-';. ,2.r,2,\p). In the states of the discrete spectrum the wave functions X must tend to zero as r ➔ co. This means that the expression for u must become a polynomial,

s-xr' = -v, where v is a non-negative integer. From this condition we get the energy levels y4

ll,2

E,K

= -2Ma (v+½+ ✓[r'+(K+½)'])'" 2

The dimensionless parameter y 2 is proportional to the reduced mass of the nuclei M, so that y 2 ~ 1. If v and K are not very large, v

where we understand by v J JM., the expression VJJM_,(x)

= c(l-x)J-.11t.,(l+x)J+.i111,

(7)

From equation (7) and the recurrence relation (6) it follows that JJ-k

vkJM1

= e dxJ-k [(1- x)J-M,(1 +x)J+.ill-'],

Hence

4>,,J,l1.,(B, if,, q;) = e exp (ik,p) exp (iMJ if,) (1 - cos 8)-Hk-.1t1) (1 + cos B)-i(k+M,1) Ò )J-k x (- [(l-cos0)''-.'11,(l+cos0)J+,lIJJ• Ocose '

for MJ = O these generalised spherical functions go aver, as we should

have expected, into the ordìnary spherical functions and are the wave functìons of the rotator

(i,f,) (cos 1+i cos 9 sin ip)

o

(1 /,r] ✓C-t) cos ,p sin 8

-1

E,

C( (J) k•I, m -

.

4'kd-l, ,~)

f,1(3)/411] elCp (-i,{,) sin O

(1 /11"1 ,/( ¼) exp (-i,/,) (CO!! q,-i C01l Bsin rp)

1

[1 /,.] ,J( ¼) ex.p (i,f,) (cos Ocos q.i +i sin ip)

o

(1/u} J{i) sin Bs!n q,

.-.1 [1f.r] J( ¼) e,cp { - i,fJ} ( -cos Il cos rp+i sin tp) ~

25. The' splitting_ of the terms is detennined by the spin-spin interaction. To determine this splitting it is necessary to average the operator of the spin-spin interaction, o.(S.n)~, over a rotational state.

8.26

Molecules

367

:For a given value of K the quantum number J takes on the values

J-K+l,K,K-1 . • The matrix clements of ( n. S} which are different from zero have the form

S)K J-K-J(K+I) ( n. S.)K-1,,-K-( K ,r=.K- ''· K-1,,=K2K+1.' ,>-K ( n. S"JK+l x J=I'

. = ( n · S)KK+t J-!( J=E. =

J(2k.-+K )

t '

J=1•+1-(n · S)K+i.r~x+i-J(K+Z) (n .,>")K K+lJ=K+1K ,T=K+l2K+·j ' a . {n . .,)f-1:~ .. f=l

J(K-1) zK_:-i ·

1:f:J:l = • = (1i.S)~-

Using these Iast relations we find for the splitting of the components of the triplet

K-1 AEJ~K-t = 2K -1

a.

26*. We find first of aU the off-diagonal elements ofthe operator w, (.,) nAO:vJ nA0.'11J'•

One sees easily that the Only matrix elements which arè different from zero correspond to O.' = O ± 1, J = J'. Since - 1'(NlQ)•'" (/(l{ )llAQ nA'1±1 = + nA'1±1• we bave

]òOJ (!2 M~)nAOt'

/i' [ {) i [j (w)~±'{.,J=M ±~0 +. 0 ;--+(0+1)cot8 11 SlO 11'{,

l),.Q±lJ

p

nAO,tlti

,

To evaluate the rnatrix elements

]""J

& +. ; ,+(r.l±l)cot9 a . ±ii[ 0 sin 0 11~

~n:1:1J

,

we note that if we put in the operator 1,+i11 (see problem 13 of section 8) r, = O and -i(O/i}p) ➔ lvlr,, we have

(,',±•J',),-,

.

-tlo/?iol -+.H,

~+i[±t-8 + _i 8 ;'_+(r.l±l)cot8]. Il

Sffi

vy,

Answers and solutions

368

8.26

If we evaluate this last equation we bave

a

· a

· ]"'' -±i✓[(J+fl)(J±fl+l)].

[ ±, 0+,i~B~+(O+!)cotB """''

In the case of small vibrations near the equHibrium position the matrix element

(-;p 10',)''"'

MO±l11

can be put approximately equal to

I -(J\1)•AO ~ ,; MO±l• where p 0 is the equilibrium distance between the nuclei, Since •

(l, + S,)::1&,, - (S,)~8,., - ± ; ✓[S(S + 1)-:E(:E ± I)],

we bave fina1ly: (w);Afll:{,, - B, ✓ [S(S + 1)-:E(:E ± I)] ✓ [J(J + 1)-(A + :E) (A+ i;±!)]. Here B 0 = fi 2/2Mpi is the value of -the rotational constant in the f:quilibrium state which corresponds to p = p0 • In the generai case the doublet splitting may be of the same order of magnitude as the matrix elements we bave evaluated, If we consider, therefore, the displaced levels of the doub)et term we can apply pert:urbation theory in a somewhat different form. Instead of the functions ifsi,.M+if!J,

'Pnt.A-nJ

we use as zeroth approximation a linear comhination of them

i/J = C11Pw.M+}1lJ + Ca 1Pn.M..-i:11J• If we substitute this expression in the perturbed equation and use standard methods we find the secular equation (O) -E E nM+i1JJ

wnM+i•' n/1.A- T/.J

wnM-j•' nAA+ 'f!J

- o.

EJi1.A-ivJ -E From the solution of the secular equation it follows that where

E - ½E'''±½J(AE''' +4B![(J+½)'-A']}, .6.E(OJ

= E~A+h..r-E~_,,,J,

(1)

Molecules

8.28

369

In the a case when the multiplet splitting is large compared to the distance between rotational levels, it follows approximately from equation (!) that

_ E'"'nAA+ì.,J + Bj[(J+½)'-A~ fl.ElO) ,

El -

E - E'"'

nAA-h,J

2 -

- Bj[(J+½)'-A'] fl.EW)

'

In the b case we get from equation (1)

6.ECO) E,.,~ ½E'"'± B0 [(J +½)'-A~+ BB,[(J +½)'"'-~A""']. 27, K = O, 2, 4, ... , if the total spin S = 2, or S = O; K if the total spin S = 1, 28. The magnetic moment of the molecule is equal to ,li

2µe

=

l, 3, 5, ... ,

(A+Z~)n,

where n is a unit vector in the direction of the molecular axis. To determine the splitting of the energy it is necessary to average the quantity ,li -

2 µe

(A+Z~)(n.J!")

aver a rotational state, that is to determine the matrix element

(n):fMJ•

Since J is a vector quantity which is conserved, it is clear that the matrix elements of the vector n must be proportional to the matrix • elements of J, that is, •

(n):fMf ~I"+ µ,:Yt', V~

A

A'

l:! ~ (MK- l)ZK(K+ l) A-(MK-1) K(K+ I) µ,K- µ,:Yt',

Vllf:C,!1- ½A K(:+l) ✓[(K-M+l)(K+M)] VM:~!i*-½A(n,-in,)M:-'

. ½A(n,+in,JM:_,,

~ ½A K(:+l) ✓[(K-M+l)(K+M)].

Molecules

8.34

371

Substituting those expressions into the secular equation and solving it we get E(1l A2 (M ') .>r' AA

'·'-K(K+1)

-4K(K+l)

K-,µ,

1

+ ZK(K + l) ✓{[AA(M -½)-A' µ 0 .>r' + Zµ, .>r'(K + 1) K]' +A' A'(K+MK)(K-MK+ 1)), We consider two limiting cases, If p, 0 Jf' ► A, we get for E1~J the expression E(IJ

A2

1,2=-K(K+l)

(M

l - 1) K-lf+g

P..o

,r'-

,r'

+P,o',

which coincides with the equation obtained in the preceding problem. For A ► p, 0 .#' we get

l

E'" _ AA (M ') [ A' 1 .>r' ' -Z(K+l)- K-, (K+l)(K+jj+K+½ µ, '

l

E'"_ AA (MK-lf') [ K(K+~)-K+} A' 1 P.o .>r'• 2 --2KThe second terms of these equations, that is, the terms which are linear in .!/f', agree with the corresponding expressions which we obtain if we substitute in the equation found in problem 28 of section 8

J - K + ½,

S - ½, and J\f1 - MK-1• 32. Because of the axial symmetry the dipole moment of the molecole is directed along the line connecting the nuclei, that is p-pn. If we proceed as in the solution of problem 27 of section 8, we find

n

l>EM, - -Cp J(J+ 1) M;,

33.

t,E --C J\iAJ(J+1)-S(S+1)__+K(K_±_1) ''' p 1 ZK(K+1)J(J+1) .

34. The potential energy V(x 1 , x 2 ) of the system is to lowest order

in R-1

372

Answers and solutions

8.35

where x1 and x 2 are the displacements of the two electrons from their respective centres. The Schr5dinger equation for the two electrons involved.can thus be reduced to one for two linear harmonic osci1lators by bringing the kinetic and the potential energies simultaneously in the form of squares only, and the cigenvalues of the energy are ·

2jl' [(2n, + I) J(k +:f.)+ (2n, + I) J(k-~:)] , n,, ~ O, 1, .... In the ground state we bave n1 = n2 = O, and expancling the radicals, we find for the dispersion energy of attraction, again to lowest order in R-1

'

3S•. The interaction energy consists of two parts. First of aU, there ìs a contribution due to the polarisability a of the hydrogen atom (see problem 47 ofsection 7). The field & ofthe H~ ion at the position ofthe hydrogen atom is by the relation

e/R 2 so that that part of the interaction energy is given •

(I) The second contribution comes from the symtnetry properties of the problem (compare problems 6 and 12 ofsection 2). Let lp 0 (x,y,z) be the wavefunction of the electron corresponding to it moving in the vicinity of proton I which is situated at x ½R ,Y z O (see fig. 43a),

=

= =

,, -tR

•

,, I

o

tR

Fig. 43a.

where we have chosen the x-axis along the line connecting the two protons and the origin midway between the two nuclei. In that case ,P 0 will be of the form iJ;·_A{x,y,z)ex 0 (2) y'1rai( p as ' -

(-.:.!)

with A(x,y, z) a slowly varying function which satisfies the boundary

condition A(x,y,z) ➔ 1

as

x,y,z ➔ ½R,0,0.

(3)

8,35

Molecules

373

The function 1/1 0 must satisfy the equation h' -2µV',j, 0 +(U-E0),j, 0 =O,

(4)

where

Uc.::.

e2

e2

e2

----+r1 Y2 R '

(5)

while E 0 is the binding energy of the hydrogen atom. If we denote by i/I 00 the function VI O with A(x, y, z) 1, l/) 00 satisfies the• equation

=

h' 2 \IJ + ( -,;;.-Eo e' ) Vloo =O, -2µV 00

(6)

so that we get far A(x, y, z) the equatian

òA

A

A

"ax-R+ ½R+x

=o'

(7)

where we have neglected all second derivatives and all derivatives with respect to y and z af A and where we bave put r2 = ½R + x, The solution af (7) satisfying the boundary condition (3) is

R

A= ½R+xexp

x-½R R

.

(8)

There is, af cowse. also a solution of (4) whìch is centred around proton 2. It will be 1/, 0 (-x,y,z) (which in accordance with theRsSump-, tiom of our problem can be assumed not to overlap with l/) 0 (x, y, z)) and also canesponds to an energy E 0 • The true states of the system will be the symmetric and antisymmetric combinations of VJ 0(x, y, z) and !/J 0 (-x,y,z); (9)

and these combinations also satisfy equation (4) with energies E,, •. If we multiply equatian (4) far l/, 0 (x,y,z) by Y/ 5' a and equatian . (4) far ..;,,,a by if; 0 (x, y, z) and integrate the difference of the two ·resulting equations aver the half~space x ;;.,: O, we find, using the fact that far x = O we have ,;, =y'2,j, 0 , ,i,,= O, a,i,,/3x = O, ò,j,,/òx =.j2o,J, 0 /òx, that E 1,~-E0

h'ff

= f/J

where the ìntegration is aver the piane x

a,;,

1/1 0 òx dydz,

= O,

(I O)

Answers and solutions

374

8.36

Using expressions (2) and (8) far 1/1 0 we then find E

•• a

R ( R) (a•)' ] -! . ea a R

n' [ 2 -E O = ~ +--exp - µali

8

(11)

8

We note that this result is valid, provided R/a 8 ► l. The symmetric state has the lowest energy. The exact E. has a minimum for R "'= 2a 8 , egual to -16·3 eV (compare next problem), while the minimum far Ea-which follows from (11)-occurs atR ~ 13a 8 and equals-0·002 eV. 36"'. We start from equation (4) of the preceding problem and use as a tria! wavefunction (1)

where

(2) so that we bave two parameters, o: and R, in our problem. In contrast to the discussion in the preceding problem we now must take into account the overlap integrai S

=

f

\}1 1 IJ; 2 d 3 r.

(3)

From symmetry considerations it follows that the lowest energy will correspond to the case where a= b = {2(1 +S)}-½, where we have used the requirement that VI must be normalized. The energy of the system is now found from the equation E= 1

:s{f

-.J, 1 Ii,Jl 1 d

3

r+ f

-.J, 1 Iiiµ 1 d 3

r},

(4)

where

(5) Hence we find _

[-i, E-1+sµai C(+ I

h'

1

a(a- l)-C+(a-2)/ ..1.J l+S +R,

(6)

where C is the dimensionless Coulomb interaction integrai

.

a'Jexp(-2ar 1 ) 3 e= aa-~-~~a r,

,,

(7)

and / the dimensionless exchange integrai,

a'Jexp(....,ar 1 -a:r) 1

l=aa w

r2

d 3r

•

(8)

8.36

Molecules

375

Integrai (7) can be evaluated by expanding r1. 1 in terms of r 1 /R (if r 1 or R/r 1 (if r 1 > R) (see fig. 43a)

I

,,

• ('; )'P1 (cos1'), RI ,io

ifr 1

-I I" (R)' - P, (cosD), '11~0

< R)

a).

The phase ·80 and the coefficient A follow froro the condition that both the wave fuuction and ìts derivative are continuous at r = a. We get in this way

8= arctan (!,tank'a)-ka. 0

The partial cros~section for I= O is thus

a0

= ~ sin2 80 = .; sin2 [ arctan (i,tan k'a )-ka] ·

(1)

Far small velocities of the incident particles (k ➔ O) 80 will be pro,, portional to k S k' = 2µV, (Z) o,;:,, ka ,o n,2·

ka(tank,a - l)

Because of the factor 1/k

2

' . the cross-section o-

0

will be

Answers and solutions

9.1

a~4wa11 ( tank koa0 a -1)' (for small k).

(3)

384

We consider the cross-section o-0 as a function of the well depth, whic_h determines k0• If the well ìs shallow (k 0 a41), we bave

uo

= 47ra2

}4a4 9

16,r a6

= 9-

nn,! µ,~ .

We note that we get from perturbation theory

f(if)~

1 Zµf

-fuli'

V(r)dT -

2µ,

a3

li'V,T

and thus

-41T IJ(·")I'av - l6ffa'J'lµ' 9 7i4 • The cross~section increases with increasing. Vo and diverges for k0 a = ,,, /2. The condition k0 a == 11/2 ìs the condition far the appearance of the :first level in the well, If we deepen the we11 even further, the cross-section starts to decrease again and tends to zero for tank 0 a = k 0 a, When Va is further increased the cross--section continues to oscillate between O aild oo, becoming infinite whenever a new level appears in the well. The sharp oscillation of the cross•section for the scattering of slow particles explains why the cross-section for the scattering of slow electrons by an atom can di-ffer appreciably from the geometrie cross~section. We note thnt if k 0a is near to an integer times ½11, we must alter equations (2) and (3). lndeed, in that case tank'a is a large number and we cannot use the expansion which leads from equation (1) to equation (2). In that case we can still neglect the term with kada - k' ~ (l+l)sìnS1 sin6,,.1 cos(ll,+1 -8J, l.,.41

4,,i

(•3cos•O-ld /(1+1)(21+1). •S 81 J, 2 a~ k' 1.,(21+1)(21+3) n 1

+

12,.

~

kB l;O

(l+ 1)(1+2) . 21 +3

.

sin 81s1n 81+2 cos (-81+s - 8,).

4*. The radiai function satisfies the equation

, + [k'-1(1+1)_ 2~

Xi

1 11

-O

,i"iyiij X -

and the boundary conditions x1(0) = O and solution satisfying these conditions is

x = finite as

r ➔ OJ.

x, -.J(r)J,(kr), where

A-

J

[(I+½)'+

2

:;;4].

From the asymptotic behaviour of l;,..(kr) we get the phase shifts,

s, - - ;c1.-1- ,> - -i(J [ci+ ½l'+

2

:;;4] -cz+tl).

The

386

Answers and solutions

9,5

The independence of 81 on k means that we get for the scattering

amplitude

where f 0({}) is independent of the energy of the scattered particles. The scattering cross-section

I do~ k,lf,(D)l'd!l is inversely proportional to the energy and is characterised by a universal angular distribution, Since the sum

I

~

f

1-0

f(O)- Z'k :1: (21+ l)P,(cos#)[exp(2i81) - J],

which determines the scattering amplitude diverges for t't ➔ O, it is clear that large values of l are essential far the evaluation of f({}) at small values of ff. For large values of l we bave

(I) so that

I ~ f(D)~k :1: (21+1)P,(cosD)81

,_,

If 8rAAI

n,a ""' '

expression (1) for 61 is valid for all ! and thus for all values of {}

1 /(#)~- kli' Zsin½D' 1rµA

7Ta µA2

da= 21i,2 E cot½Ddb. S. The Schr6dinger equation of our problem is of the form -

li' 2

µ "l',f,+V,f,-E,f,.

Scattering

9.6

387

Introducing, as usual, k2 = 2µE/1i 2 we get this equation in the form

2,/: V,f,.

V',f,+k',f, -

(1)

We shall consider the right-hand side as an inhomogeneous term. We must thus find a solution of an inhomogeneous equation, satisfying a given boundary condition, . exp (ikr) ,f, o exp[t(k0 .r)J+J~~ • t ➔ OO

f

Using the expression for the Green function of equation (1),

G( _ ')- _ _l__exp(ik!r-r'!) TT l r-r 'I' we easily find the required solution

''-4

.

µ

f

,f,- exp[t(k,.r)]-zwk' V(r')if,(r')

exp(iklr-r'I) lr-r'I d'r'.

For large values of r we bave

exp(ik[r-r'!) exp(ikr) [ "(k ')] 'I a .exp -1 .r r-r r where

I

(r ➔ oo),

r k - k-.

r

Hence,

f

,f, o exp [i(k0 • r)] - :,, exp ;ikr) d' r' exp [ - i(k. r')J V(r') ,f,(r'), 2

so that

/(k) -

-Z:., fa• r exp [ -i(k .r)J V(r),f,(r).

This equation is convenient for various approximate calculations. For instance, by substituting if,(r)~ exp [i'(k0 • r)] we get the scattering amplitude in the first Born approximation. 6. The scattering amplitude is in the Born approximation given by the equation

/nom(#)- -

µ ,r,, 2

I .

••q,

exp[,(q.r)] V(r)dr - - nrA

where

q=k'-k,

q=2ksin½D.

388

Answers and solutions

9.7

Hence

a•••,.= lf(D)l'dQ =

w 3 µ,A 2

2J!'E cot½DdD.

In classical m.echanìcs we bave the following connection between the angle of scattering and the impact parameter p:

f.

00

µ,v p dr ,, r' ✓[2e(E- V)-(!'vp/r)']

1r -

=

,O.

2 '

where r0 is the zero of the expression under the square root sign. If we integrate, we get A 1 (~-11)' Pt = EO 2nr-lJ. ' and thus

If

we can apply the Born approximation far ali angles (see problem 4 of section 9).

In the apposite limiting case, where 8µA/1i, 1 >, 1 the classical result holds for not tao srnall angles,

•' •'

D ;e Bl'A'

while far smaller angles,

D;SBl'A'

the Born approximation result i.s valid. 7*. We have the following equation for the radiai function,

x" +j$[E+ V.exp(-r/a)]x= O. Using the notation ka = Zp.Ef'Ji. 2, K 2 = 2µ,VofA 2 and the independent variable f = exp ( - r/2a), we must solve the equation

x"+ ~x'+4aa(~+1/ [exp (ikr)- ( trl=Ol

✓•

''

lY exp ( -ihr)]

00

- ~k :1: :1: C/;• '1>/,[✓(I + I) Yt + ✓(/) Y,] exp (ikr)/r t

i=O I

-

✓k• S :1: C/••'1>) [,/(/+ !) Yt + ✓(I) Y,] ( t

!=OI

'

l)'exp ( -ikr)/r.

(4)

'

34*. We shall use the expansion of the incident wave in terms of the eigenfunctions of the operators which are conserved (see equation (4) of the problem 32 of section 9). Each term of this sum, which corresponds to well-defined values of I, J, and /, wìll be scattered independently of the others. The number of particles corresponding to given values of l, J, and I v,,jll thus not be changed during the elastic scattering process, so that the influence of the scattering centre will be to introduce some phase factor exp (2i8[ ±)· The quantity 8[± = 8l .J-l+l depends on l, J, and I but it is independent of I, because of the hypothesis of isotopie invariance. We must note that the scattering centre does not influence the incoming wave but only the outgoing one. The wave function of the system can thus be written, if we take scattering into account, in the form

,f,

= ,j__k• :1: :l: Cj;•J,[✓(I+ I) Ytexp (2;81,) + ✓(/) y- exp (2i8/_)] exp (;k,) i

r

11.. 0

- ✓,.k":1: :l: Cj;•f,[✓(l+l)Yt+ ✓(l)Y,]exp(;;k,)_

(!)

l I-O

At large distances from the scattering centre we can write it in the form

.,. _ 'f'lnc .,. +

'f' -

,xp(ikr) • 1~~~ r

The quantity f is the scattering amplitude. ~f we take from equation (1) the expression far 1/l;nc [see equation (4) of the problem 32 of section 9] we get

f

= ✓k• :1: :l: Cf'•f,{✓(I+ 1) Yt[exp (2;8/+)-1] t 11-0 • + ✓(/)

Y1 [exp(2i8/_)-1]).

(2)

414

Answers and solutions

4>f.

The functions the operator I,,

9.34

can be expanded in terms of the eigenfunctions of

(3) 1) 8(n- 7,,) + Cf;--r• S(1T- 1rk) O(n + Ts), where the Clebsch-Gordan coefficients CI;• are given in table (2) of the preceding problem and where wk = 1r1 +2T~, Substituting equation (3) into equation (2) we get

f. = CJ;• 8(1T -

f where

✓w

1r

~

= ik :;: 1~ 0 [GJ;: 8(1r- 1r1) S(n-Tz)+ GJ:,~ 8(1T-11k) 8(n+ri.)] 1•

x {✓(I+ 1) Yt[exp (ZiSf+J-1] + ✓(/) Y1 [exp (ZiSf_)-1]),

T:

and denotes the final state of the nucleon. If we now put

f we get

= J;; 8(1T -ni) 8(n- r +J:: 8(1T-7Tk) ò(n +T 2), 2)

✓w

~

J;•,' - ~k:i: :i: Gf•; ,{✓ (I+ 1) Yt[exp (2iSf+)-1] l ll"'O •• + ✓(l)Y-[exp(ZiS/_)-1]),

(4)

where GJ;.;, is given in the following tahle for all reactions necessary for our problem: Reaction p+-+ p+

p-----►

p-

p--+ no

GiT,

I

!

✓(2)/3

GÌT1 l,T,'

o

!

-.J(2)/3

[,T '

If we substitute these values of Gf;r; into equation (4) we get finally ✓w ro

f(p+,p+) - -,k L l

{✓ (I+

1) Yt[exp(ZiS/+)-1]

1=0

+ ✓(/)

f(p-,p-) -

Y1 [exp (2iS/_)-1]),

3✓~k L {✓(I+ I) Yt[exp (2iS/+l + 2exp (ZiSI,_)- 3] t l=O

+ ✓(I) Y1 [exp (2iS/_) +2 exp (2iS/_)- 3]),

f(p-, n,) -

✓(3~Zkw) L {✓(/+ 1) Yt[exp (2iSt)- exp (2iS/+ll l 1-0 + ✓(I) Y1 [exp (2iS/_ - exp 2iS/_)]),

(5)

415

Scattering·

9.35

3S*. The table of the coefficients GJ;~,' for ali possible reactions of the pions with nucleons is of the form No,

Reaction

I,

Gi,-,

1

p+-p+

I

I

cJ••, ••• o

2

pO ➔ pO

I

i

I

3

pO ➔ n+

!

✓(2)/3

-,/(2)/3

4

-5

p- ➔ nO

-!

,/(2)/3

-✓(2)/3

p--,.p-

-!

I

I

6

n+-+ n+

!

l

I

7

no ➔

-!

.j(2)/3

-,/(2)/3

8

n+--,.po

!

✓(2)/3

-,/(2)/3

9

710-..110

-!

I

!

10

n--,,.n-

-t

1

o

--

p-

I,r,'

(1)

Since the phases are independent of J. because of our hypothesis of isotopie invariance it follows immediately from equation (4) of the preceding problem and table (1) that (1) f(p+,p+) = f(n-,n-),

(2) f(p-,p-) = f(n+, n+),

(3) f(P', n+) = f(n+,p 0) = f(p-, n') = f(n°,p-),

(4) f(P',P")

= f(n°, n').

The expressions for the first three amplitudes were given in the preceding problem. From table (1) one sees easily that

f(P',P')

= ½[f(p+,p+)+ f(p-,p-)].

Using table (1) we get

= j(n-,n-) = JI, f(p-,p-) = j(n+, n+) = ¼(/' +2/1),

(1) f(p+,p+)

(2)

(3) f(P", n+)

= J(n+,p•) = f(p-, •') = f(n•,p-) =

(4) f(p 0 ,p0 ) = f(n°,n°) = ¼(2/1 +/1),

f (/

1 - /1),

416

Answers and solutions

9.36

36*. The differential scattering cross•section is egual to

i~ I/I',

where dO.

~

sin BdBd~.

The total scattering cross-section is equal to

J:"l l/lll sin 8 d(} drp. 11

a

=

If we substitute here the scattering amplitudes for the reactions considered, which are given in problem 34 of section 9, and use the orthonormality of the Pauli functions

f f (Yf)+(Yt)dO.

~f f (Y,)+(Y,)dO. ~ 8w,

ff(Yt)+(Y,.)dO.

~f f(Y1)+(Yt)d0. ~0,

we get

a(p+, p+) =

!; :i:

[(

l + 1) sin 2 OJt + / sin 2 Of_J,

1 0

4 a(p-,p-) = 3k~ {(l+l)(sin2 Sf++2sin 2 Sf+-isin 2 (0f+-8it)] /,. o + l [ sint O/~ + 2 sin2 Ot_ - f sìn 2( S/_ - Sf_))}, 8 a(p-, n°) = k1T2 + 1) sin 2( Oft - 8f+) + l sin 2( Of_ - Sl_)]. 9

i:

,_,i [(/

37*. We shall give the detailed solution for the reaction (p+,p+). From problem 34 of section 9, we get far the scattering amplitude for S- and P-waves

/(p+,p+)

,,

~°},;[o, Yt+ ✓(2),, Yt+fi, Y;-],

where o: 0 =exp(2iSt)-1;

o: 1 =exp(2i8f+)-1;

fJ1 =exp(2iOj_)-1.

Using the explicit expression for the Pauli functions of problem 32 of section 9, we get for the differential scattering cross-section for S- and P-waves

! 2

;~

= IYool 2 lo:ol 2 + ✓~IYoollYiol[Z(o:00:f+o:to:1)+(0:o,Bf+aj',81}]

+!I Y,,I' [41 a, I'+ lfi,I' +2(,,fif +•t fi,)J +il r;, I' [I •,l'+I fi,l'-1 ~

1~&·>1 ~ ( 2mc p )'

7, In this case we have the Klein-Gordon equatian far a charged particle in an electramagnetic field described by the patentials

Ze

A~O,

(Il

,;~--.

r

As we are laaking far a stationary solutian, we put

= l/)(.r)e-iEt/11, with E > O. The

i/J(r, t)

and we look far solutions becomes

(2)

equatian far ip(r) naw

(3)

Changing ta spherical polars and writing (4)

the radiai function R 1 (r) satisfies the eq uation

d2 /(!+I)-Z2a2 2ZaE m2c4-E2} - - ,2 + 1!cr - 122c2 R(r)-O {d,2 ' '

(5)

where a is the fine-structure constant,

e'

a=-~

hc

(6)

Answers and solutions

446

< mc 2 •

lf we are looking for bound solutions, E

= l(l+

k(k+ 1)

'Y (.j

=

11.8

J)-Z2 a: 2

If we write (7)

,

2ZaE "lìc(3 '

(8)

2

= h