Faddeev L.D., Yakubovskii O.A. Lectures on Quantum Mechanics for Mathematics Students

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L. D. Faddeev and 0. A. Yakubovskii

A square-integrable solution 0 will exist only for values of E such

that

F1(E) = F2(E).

The roots of this algebraic equation, if they exist, are the eigenvalues

of the operator H. From the foregoing it is natural to expect the

presence of a simple point spectrum for E < 0.

2)0<E< VO.

We write the equations (2) and (3) in the form

"+k21p =0, x< --a, k 2=E>0, k>0;

7P"-xj,0=0,

x>a, 4=-(E-V0)>O, x1>0.

The functions e±i

for x < -a and efx1x for x > a are linearly inde-

pendent solutions. It is immediately obvious that there are no square-

integrable solutions, but a bounded solution can be constructed if we

choose C1 /C2 so that 0 has the form C1 e-xx for x > a. Therefore,

in the interval 0 < E < VO the spectrum is simple and continuous.

3) E > Vo.

In this case both equations (2) and (3) have oscillatory solutions

±ikx for

x < -a and

e+ 1x for

x > a, k1 = E - VO), hence any so-

lution of (1) is bounded, and there are no square-integrable solutions.

For E > VO the spectrum of H is continuous and has multiplicity two.

In Figure 6 the eigenvalues of H are represented by horizontal

lines, the ordinary shading shows the region of the simple continuous

spectrum, and the double shading shows the region of the spectrum

of multiplicity two.

We discuss the physical meaning of the solutions of (1).

The

square-integrable solutions describe the stationary states with en-

ergy equal to an eigenvalue. These functions decay exponentially

as lxi -* oo, and thus the probability of observing the particle out-

side some finite region is close to zero.

It is clear that such states

correspond to a finite motion of the particle. The eigenfunctions of

the continuous spectrum do not have an immediate physical mean-

ing, since they do not belong to the state space. However, with their

help we can construct states of wave packet type which we considered

for a free particle.

These states can be interpreted as states with

§ 20. The general case of one-dimensional motion

almost given energy. A study of the evolution of such states shows

that they describe a particle that goes to infinity (infinite motion) as

It) --+ oo We shall return to this question when we study the theory

of scattering.

In classical mechanics, as in quantum mechanics, the motion is

finite for E < 0 and infiinite for E > 0. For 0 < E < V0 the particle

can go to infinity in one direction, and for E, > VO in two directions.

We direct attention to the fact that the multiplicity of the continuous

spectrum coincides with the number of directions in which the particle

can go to infinity.

In the example of a particle in a one-dimensional potential well

we consider the question of the classical limit of quantum stationary

states For the computation of the limit (14.15) it is convenient to

use the asymptotic form of the solution of the Schrodinger equation

as h -* 0. Methods for constructing asymptotic solutions of the

Schrodinger equation as h --+ 0 are called quasi-classical methods.

We use one such method: the Wentzel Kraxners--Brillouin (WKB)

method.

We write the Schrodinger equation in the form

(s)

E --- V(x)

- 0.

Here, as earlier, we use a system of units in which m = 1/2. Setting

(x} = exp h fem. g(x) dx], we get an equation for g(x):

(6)

ihgf -

+ E - V = 0.

The solution of this equation will be sought in the form of a power

series in h/i:

(7)

k=l

Substituting (7) in (5), we have

oo n

-E -7 9n -I EE --

9k9n -

n=1

(t)

n=1 k=O

9(X) = E (2 ) A (X)-

100

L. D. Faddeev and 0. A. Yakubovskii

Equating the coefficients of (h/i )', we get a system of recursive

equations for the functions 9k (X):

(8)

g02= E - V.

(y)

9n-1 =

>29k9n-k.

k=0

From (8) we find go(x):

go W V -(X) = ±P(X) -

1 sere p(x) with E > V (x) is the classical expression for the abso-

lutes value of the momentum of a particle with energy E in the field

V (x).

For E < V (x) the function p(x) becomes purely imaginary.

Setting n = 1, we get from (9) that

' = --2 =

-1 g-

_ -1 d to

x .

9'a .90.91, g3

2 9()

2 dx g

1p(

) I

Confining ourselves to these terms of the expansion (7), we get the

asymptotic form as h --40 for the. two linearly independent solutions

of the Schrodinger equation:

r x

(10)

*i.z =

lp(x) I

exp I f Z

p(x)

+ O(h)

h o

with p(x)

The functions (10) are sometimes called WKB-

solutions of the Schrodinger equation.

The precise theory of the WKB method is fairly complicated. It

is known that in the general case the series (10.7) diverges and is

an asymptotic series. A finite number of terms of this series enables

us to construct a good approximation for the function V) if Planck's

constant h can be regarded as sufficiently small under the conditions

of the specific problem.

In what follows we assume that V (x) = 0 for I x i > a and that

E < 0 We suppose that for min,, V (x) < E < 0 there are two points

x 1 and x2 (-a < x i < X2,< a) satisfying the condition E --- V (x) = 0.

These are so-called turning points, at which according to classical

mechanics the particle reverses direction. It is not hard to see that

in the classically forbidden region (x < x 1 or x > x2) one of the

WKB-solutions increases exponentially while the other decays upon

§ 20. The general case of one-dimensional motion

101

going farther from the turning point into the depth of the forbidden

region. For j > a, the WKB-solutions coincide with the exact so-

lutions and have the form efxx, where E = --x`2. We recall that

the eigenfunctioris of the discrete spectrum of H decrease exponen-

tially as x --- ±oo. As h - 0, an eigcnfunction must coincide in

the allowed region with some linear combination Cl ip1 + C22 of the

WKB-solutions. The construction of such a linear combination is a

fairly complicated problem since the WKB-solutions become mean-

ingless at the turning points. It can be shown that the conditions for

decrease of the function O(x) hold as x --+ --oo if

(11)

O(X) _

C sin

p(x) dx +

+ O(h).

p(x)

Similarly, it follows from the conditions of decrease as x --> +oo that

+ O(h).

(12) fi(x) = C

sin (hf4)

These two expressions for O(x) coincide if

(13)

xdx=-rh

n+1 n=012...

This condition determines the energy eigenvalues in the quasi-classical

approximation and corresponds to the Bohr Sommerfeld quantization

rule in the old quantum theory.

We proceed to the computation of the classical limit of a quantum

state. The limit as h --+ 0 can be found under various conditions.

For example, we can consider a state corresponding to the energy

eigenvalue En for a fixed value of n in the condition (13). It is easy

to see that then En --+ Va = minx V (x) as h -- 0, and the state of a

particle at rest on the bottom of the potential well is obtained in the

limit. Let us analyze a more interesting case: h -. 0, n -+ oo, and

the energy E remains constant. (We remark that in the given case

the integral on the left-hand side of (13) does not depend on h, and

h -+ 0, running through some sequence of values.) Substituting the

asymptotic expression (11) for O (x) in the formula (14.15), we get

102

that

L. D. Faddeev and 0. A. YakubovskiI

F(x'

27r

n m G(x +

ufa) (X)

(1'x+h

(x + uh) p(er)

sin ji

p(x) dx +

+uh

x C

1cosx sin I p(x)

dx + )

p(x)

him

Lh J

p(x) dx

- cos

J ztuh

p(x)

+ h f

p(x)

+ 2 J

We denote all normalization factors by C. The limit of the second

term in the generalized function sense is zero, therefore,

F(x, u)

coti(p(x)u).

(X)

Using (14.16), we find the distribution function for the limiting clas-

sical state:

p(q,p) =

p(q)

f__oo

e-sue

Cos(P(9)u) du

Finally, we get that

C °°

e-ipu(eip(q)u

e-ip(4')u) du.

p(q)

(14)

P(R 4)

PC )

[a(p -

P(4)) + b(P + p(4'))]

The state described by the function p(q, p) has a very simple

meaning. In this state the density of the coordinate distribution func-

tion is inversely proportional to the classical velocity of the particle,

and the momentum of the particle at the point q can take the two

values ±p(q) with equal probability. The formula (14) was obtained

for the allowed region. In the same way it is not hard to verify that

p(q, p) = 0 in the forbidden region.

§ 21. Three-dimensional quantum problems

103

§ 21. Three-dimensional problems in quantum

mechanics. A three-dimensional free

particle

The Schrodinger operator for a free particle in the coordinate repre-

sentation has the form

The equation for the eigenfunctions (for h = 1 and m = 1/2) is

(2)

-AO = k 201 E=k

2 > 01

and it has the solutions

(3)

fi(x) =

eikx

The normalization constant is chosen so that

fi(x) Oki (x) dx = a(k -- k').

We see that the spectrum of H is positive and continuous, and

has infinite multiplicity. To each direction of the vector k there corre-

sponds the eigenfiunction (3) with the eigenvalue k2. Therefore, there

are as many eigenfunctions as there are points on the unit sphere.

As in the one-dimensional case, the solution of the Cauchy prob-

Iem

=HAG,

(0)=

for the nonstationary Schrodinger equation is most easily obtained in

the momentum representation

80(P, t)

= P2O(P, t),

'0 (P, 0) = O(P)

It is obvious that

i(P,t) =

w(P)e_

2t.

Passing to the coordinate representation, we get that

2 Z

x t)

eitpX_P t) d

(k)

k x

eik t

104 L. D. Faddeev and 0. A. YakubovskiI

Just as in the one-dimensional case, the functions p(x, t) or b(p, t)

describe an infinite motion of the particle with a momentum distri-

bution function independent of the time. Using the stationary phase

method, we can show that a region in which there is a large probabil-

ity of observing the particle moves with the classical velocity v = 2p{

(we assume that the support of the function p(p) is concentrated in

a neighborhood of the point p}) . We have the estimate

kL'(x,t)I <

It13/2

where C is a constant. Finally, as in the one-dimensional case,

lim

b, 00

for any

E 1I.

22. A three-dimensional particle in a potential

field

In the coordinate representation, the Schrodinger operator for a par-

ticle in a potential field has the form

A+V(x).

The importance of the problem of the motion of a particle in a po-

tential field is explained by the fact that the problem of the motion

of two bodies reduces to it (as in classical mechanics). We show how

this is done in quantum mechanics. Let, us consider a system of two

particles with masses m1 and m2 with mutual interaction described

by the potential V (x 1 - X2)- We write the Schrodinger operator of

this system in the coordinate representation:

h2 h2

H =

- 2m

+ V (XI - X2)-

2m,

Here 01 and A2 are the Laplace operators with respect to the coor-

dinates of the first and second particles, respectively.

We introduce the new variables

mlxl + m2x2

x=x1 -x2;

m1+m2

§ 22. A three-dimensional particle in a potential field

105

X is the coordinate of the center of inertia of the system, and x gives

the relative coordinates. Simple computations lead to the expression

for H in the new variables:

H -- --

x --

A, + V (x)

Here M = rnl + m2 is the total mass of the system, and ji

rn 1 m2 / (m 1 + m2) is the so-called reduced mass. The first term in

H can be interpreted as the kinetic energy operator of the center of

inertia of the system, and the operator

A+V(x)

is the Schrodinger operator for the relative motion. In the equation

H4' =ET,

the variables are separated, and the solutions of such an equation can

be found in the form

P (Xj X) = 7P (X)?Pl (X) -

The function.5 O(X) and 01(x) satisfy the equations (h = 1)

- ZM OlV (X) _ 160 (X)

Alp, (x + V(X)'Oj x1 = 61 01 x1

2/u

with E = e + 61. The first of these equations has the solutions

eixx

_ .

27r

The problem reduces to the solution of the second equation, which

coincides in form with the Schrodinger equation for a particle with

mass It in the potential field V (x). We note that the spectrum of the

operator H is always continuous, since the spectrum of the operator

-- 2 'A is continuous.

The most important case of the problem of the motion of a parti-

cle in a potential field is the problem of motion in a central force field.

In this case the potential V (x) = V (jx1) depends only on lx} = r. The

problem of two particles reduces to the problem of a central field if

the interaction potential depends only on the distance between the

106

L. D. Faddeev and 0. A. Yakubovskii

particles. Before proceeding to this problem, we study properties of

the angular momentum and some questions in the theory of represen-

tations of the rotation group. This will enable us to explicitly take

into account the spherical symmetry of the problem.

§ 23. Angular momentum

In quantum mechanics the operators of the projections of the angular

momentum are defined by the formulas

L 1 = Q2P3 - Q3 P2,

(1)

L2 = Q3P1 - Q1P3,

L3 = Q1 P2

- Q2P1

We introduce one more observable, called the square of the an-

gular momentum:

(Z)

L2 = L, + L2 + L3.

Let us find the commutation relations for the operators L1, L2 ,

L3, and L2. Using the Heisenberg relations [Qj, Pk] = iSjk and the

properties of the commutators, we get that

[L1, L2] = [Q2P3 - Q3P2, Q3P1 Q1P3]

= Q2P1 [P37 Q3] + Q1P2[Q3, P3]

= i(Q1P2 -- Q2P1)

= iL3.

Thus, the operators L1, L2, L3 satisfy the following commutation

relations:

(3)

[L1,L2] = iL3,

[L2, L3] = iL1

[L3, L1] = iL2.

It is not hard to verify that all the operators L1, L2, L3 commute

with L2:

(4)

[LjI L

2] = 01

1,2,3.

§ 23. Angular momentum

107

Indeed,

[Li, L 21 = [L1, Li + Lz + L3]

= [Lil L2] + [L1, Ls)

[L1, L2] L2 + L2[L1, L2] + [L1, L3]L3 + L3[L1, L3]

iL3L2 + iL2L3 -- iL2L3 -- iL3L2 = 0.

The properties of the commutators and the formulas (3) were used in

the computation.

It follows from the commutation relations (3) and (4) that the

projections of the angular momentum are not simultaneously mea-

surable quantities. The square of the angular momentum and one of

its projections can be simultaneously measured.

We write out the operators of the projections of the angular mo-

mentum in the coordinate representation:

C l

- xl

C l

-Z xl

ax -x3

C a l

L3 - i x2

ax,

-XI

It is not hard to see that the operators L1, L2, L3 act only on

the angle variables of the function O(x). Indeed, if the function

depends only on r, then

L3'0(r) = i x2 axl - x,

a2 (xi + x2 + x3)

= io'(xi + xi + x3)(x2 2x, - xl 2x2) = 0.

For what follows, it turns out to be useful to have the formula

(5)