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

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78 L. D. Faddeev and 0. A. YakubovskiT

freedom in a potential field has the form

H(P, 4) =

V (q).

Corresponding to this Hamiltonian function is the Schr" inger oper-

ator

+ V(Q).

For a free particle V = 0. We begin with a study of this simplest

problem in quantum mechanics. Let us find the spectrum of the

Schrodinger operator

(1)

_ P2

The equation for the eigenvectors has the form

(2)

or, in the coordinate representation,

d2,0

E .

(3)

2m dx2 -

It is convenient to use a system of units in which h = 1 and m = 1/2.

Then for E > 0 the equation (2) takes the form

(4)

+ k

2,0

=01 k

=Ej k>O.

The last equation has two linearly independent solutions

fikx

(5)

bf k (x)

21r

We see that to each value E > 0 there correspond two eigenfunctions

of the operator H. For E < 0 the equation (3) does not have so-

lutions bounded on the whole real axis. Thus, the spectrum of the

Schrodinger operator (1) is continuous and positive and has multi-

plicity two.

The functions (5) are at the same time the eigenfunctions of the

momentum operator P = -i that correspond to the eigenvalues

±k. We note that the functions (5) do not belong to L2 (R) and,

§ 15. One-dimensional problems of quantum mechanics

therefore, are not eigenfunctions in the ordinary sense. For subse-

quent formulas it is convenient to assume that -- oo < k <

k = ± VEK Then both the solutions (5) have the form

(s)

,W,/ 1 fkx

k (x - (i;) 8

The normalizing factor in (6) was chosen so that

(7)

Ok(x) 'Ok-(X) dX =a(k - k').

oo and

The solutions of (2) in the momentum representation are obtained

just as simply. For m = 1, (2) has the form

(s)

p2b=k2i',

and its solutions are the functions

(9)

Ok(P) = 8(p - k),

k =fem.

The normalization of the functions (9) is chosen the same as for the

functions (6):

Jo(-.

k) b(p - k') ap = a(k - k').

To explain the physical meaning of the eigenfunctions Ok, we

construct a solution of the nonstationary Schrodinger equation for a

free particle

dip (t)

= HO(t)

under the initial conditions

,0(0) holl = 1.

This problem is simplest to solve in the momentum representation

ft (p, 0

= p2o(p,

t),

10 (A 0) _ 0 (P)

Ji(p)l2dp

= 1.

Obviously,

0(p, t) = SP(p)

a-$Pat.

L. D. Faddeev and 0. A. Yakubovskri

In the coordinate representation the same state is described by

the function

(10)

2 -pZt) -2t

(x, t)

= (27r) JR W(P) e"

dP = JR (k) ?Gk(x) e

It is easy to verify that the normalization of the function *(x, t) does

not depend on the time:

jjp(X7 t) 12dX =

I P (P) 12dp = 1.

The formula (10) can be regarded also as a decomposition of the

solution O(x, t) of the Schrodinger equation with respect to stationary

states, that is, as a generalization of the formula (9.13) to the case of

a continuous spectrum. The role of the coefficients c is played here

by the function p(k).

The state described by the functions O(x, t) or O(p, t) has the

simplest physical meaning in the case when cp(p) is nonzero in some

small neighborhood of the point po. States of this type are generally

called wave packets. The graph of the function Icp(p)12 is pictured in

Figure 4.

We recall that I lp (p, t) 12 = I cp(p) 12 is the density of the momentum

distribution function. The fact that this density does not depend on

the time is a consequence of the law of conservation of momentum for

a free particle. If the distribution

IW(p)12 is concentrated in a small

neighborhood of the point po, then the state fi(t) can be interpreted

as a state with almost precisely given momentum.

All

10-

Figure 4

§ 15. One.dimensional problems of quantum mechanics

The function kt'(x, t)12 is the density of the coordinate distribu-

tion function. Let us trace its evolution in time. First of all it follows

from the Riemann- Lebesgue theorem 23 that

(x, t) ---* 0 as Itl --> 00

for smooth ap(p), and therefore the integral fn

(x, t) 12 dx over any

finite region f of the real axis tends to zero.

This means that as

Itt --* oo, the particle leaves any finite region; that is, the motion is

infinite.

In order to follow more closely the motion of the particle for large

Jt, we use the stationary phase method. This method can be used to

compute integrals of the form

(11)

I (N) _

F(x) esNf (x) dx

asymptotically as N -+ oo. For the case of smooth functions f (x)

and F (x) and when there is a single point i in (a, b), where

,f (x) is

stationary (f(X) = 0, f"() 0), we have the formula

(12)

IN=

F xex

iN +sn"

0 (h).

(NI;)

}

4We

rewrite the expression for 0(x, t) in the form

x t = (i-)

0(')

rJRV (p)

The upper sign in the exponential corresponds to t > 0 and the lower

to t < 0. We find the asymptotics of b(x, t) as t --* ±oo from the

formula (12). Then the stationarity is found from the condition

PP) = T 2p + Ft = 0,

from which p = 2t

. Further,

f " (p) =::F22 and as t -* ±o,c

x t =

x 1

eix(x,t)

(13)

t(, )`;P

it-

(21tj

where x is a real function whose form is not important for what

follows.

23This theorem asserts that f

°o f(x)e'

dx - 0 as N --r oo if the function f (x)

is piecewise continuous and absolutely integrable on the whole axis -oo < x < oo

L. D. Faddeev and 0. A. Yakubovskii

By assumption, ap(p) is nonzero only in a small neighborhood of

the point po, therefore, 1,0 (x, t) 12 =

I p(it)12

t +O( -It 1-3 , is nonzero

only near the point x = 2pot. From this relation it is clear that the

region with a nonzero probability of finding a particle moves along the

x-axis with constant velocity v = 2po; that is, the classical connection

p = my between the momentum and the velocity remains valid (recall

that we set m = 1/2). Further, from the asymptotic expression for

I lli(x,t)12 it is clear that the square root D,,,x of the variance of the

coordinate for large values of ItI satisfies the relation

Owx = 20wPltl

A"X A"VItj.

This means that the region in which there is a laxge probability of

finding the particle will spread out with velocity A,, v while moving

along the x-axis.

Even without computations it is not hard to see that the be-

havior of a classical free particle in a state with nonzero variance

of coordinates and momenta will be exactly the same. Thus, quan-

tum mechanics leads to almost the same results for a free particle as

does classical mechanics. The only difference is that, according to

the uncertainty relations, there are no states with zero variance of

coordinates and momenta in quantum mechanics.

We mention some more formal properties of a solution of the

Schrodinger equation for a free particle. It follows from (13) that

0 1

(j )

Itj1/2

that is, for any x

kt'(x,t)i <

jji/2'

where C is a constant. It is thus natural to expect that the vector

sli(t) tends weakly to zero as ItI --> oo.

It is simplest to prove this

assertion in the momentum representation. For an arbitrary vector

f E L2 (R),

(b(t),f) =

§ 16. The harmonic oscillator

In view of the Riemann-Lebesgue theorem, the last integral tends to

zero as ItI -> oo.

In the coordinate representation the weak convergence to zero of

ip(t) has a very simple meaning. The constant vector f is given by

a function which is markedly different from zero only in some finite

region SZ, and the region in which O(x, t) is nonzero goes to infinity

while spreading out. Therefore, (i/.'(t), f) --+ 0 as It I

--+ oo.

We verify that the asymptotic behavior of the expression (13) for

the function 0 (x, t) has the correct normalization. As ItI -+ oo, we

have

I'P(x, t)

12 dX

N f

21tj

0 (X)

2tI2

- JR

Ico(k)i2dk=l.

§ 16. The harmonic oscillator

In classical mechanics a harmonic oscillator is defined to be a system

with the Hamiltonian function

H(Pj 9') =

2m + 2 g2

The parameter w > 0 has the sense of the

freq2

uency of oscillations.

The Schrodinger operator of the corresponding quantum mechan-

ical system is

PZ w2Q2

(1)

H = 2 +

We use a system of units in which m = 1 and h = 1. Our problem is

to find the eigenvectors and eigenvalues of H. We solve this problem

by using only the Heisenberg uncertainty relations for the operators

P and Q and not by passing to a concrete representation. To this

end we introduce the operators

(2)

a =

(WQ + ip),

(WQ - ip).

Using the Heisenberg relation

(3)

[Qj P]

= ij

L. D. Faddeev and 0. A. Yakubovskii

we get that

waa*=

wl Q2 + P2+aw

( 2

(-QP+I>Q)=H+-

wa*a = -GO 2Q2 + P2) +

(QP

- PQ) =

H --

`IO

2 2

(4)

(5)

H = waa* 2

H=wa*a+

From (4) and (5) we find at once the commutation relation for

the operators a and a*:

(6)

[a,a*] = 1.

From this it is easy to verify by induction that

(7)

[n, (a*)n]=n (m

)^-1.

Finally, we need the commutation relations of the operators a and a*

with H. To compute the co mutator [H, a], it suffices to multiply

the formula (4) by a from the right, multiply (5) by a from the left,

and find the difference between the expressions obtained:

[I-I, a] = -wa.

(9)

[H, a*] = wa*.

Let, us now pass to a study of the spectrum of the operator H.

Suppose that there is at least one eigcnvalue E of H. Let 'E denote

a corresponding eigenvector. We show that the eigenvalues E are

bounded from below. By assumption, the vector

'OE satisfies the

equation H F = Ei/ , which by (5) can be rewritten in the form

wa*a +') OF = EV)E.

Multiplying this equality from the left by 'OE, we get that

wllalFll2+w

IIh/,E 112 =EII 112,

which implies at once that E > w/2, with equality possible only if

aibE = 0

From the expression (5) for H it is clear that if some

§ 16. The harmonic oscillator

vector

satisfies the condition a = 0, then it is an eigenvector of H

corresponding to the eigenvalue w/2.

We show how to construct new eigenvectors from an arbitrary

eigenvector 1E. Let us compute the expression HaE using (8):

Haz/iE= aH?/iE - wa'OE =(E - w)aViE.

It is clear from the last relation that either aE is an eigenvector

corresponding to the eigenvalue E -- w, or aIPE = 0. If a'E

0, then

either a2 ,0,F is an eigenvector with eigenvalue E -- 2w, or a2 E = 0.

Thus, from an arbitrary eigenvector bE we can construct a sequence

of eigenvectors ?PE, aE, ... , aNuE corresponding to the eigenvalues

E, E -- w, ... , E -- Nw. However, this sequence cannot be infinite,

because the eigenvalues of H are bounded from below by the number

w/2. Therefore, there exists an N > 0 such that aN E 54 0 and

aN+ 1 V)E = 0. Let the vector a N uE be denoted by 00. This vector

satisfies the equations

(10)

ado = 0,

HV)o = 2 V)o

We see that the assumption that there is at least one eigenvector

bE is equivalent to the assumption that there is a vector satisfying

(10). The vector iU describes the ground state of the oscillator, that

is, the state with the smallest energy w/2.

Let us examine how the operator a* acts on eigenvectors of H.

Using (9), we get that

(11) Ha*tE = a*HbE + wa* bE = (E + w)a* bE

We point out the fact that a*bE cannot be the zero vector. Indeed,

it is obvious from the expression (4) for H that a vector satisfying

the equation a*

= 0 is an eigenvector of H with eigenvalue --w/2,

which is impossible, since E > w12. Therefore, it follows from (15.11)

that a*uE is an eigenvector of H with eigenvalue (E + w). Similarly,

(a*)2pE is an eigenvector with eigenvalue (E + 2w). Beginning such

a construction from the vector 00, we get an infinite sequence of

eigenvectors Oo, a *o

, ... ,

(a*)?z1/.'o,... corresponding to the eigenval-

Ues w/2, w12 +w, ... , (n+ 112)w I .... Let the vector 00 be normalized:

L. D. Faddeev and 0. A. YakubovskiT

II = 1. We compute the norm of the vector (a*)flt,o:

11(a

*)nO0112

= ((a*

)noo, (a* )nOO)

(bo,a1_l(a*)nabo)

+ n(V)o,

(a*

Here we used (7) and (10).

Thus, the sequence of normalized eigenvectors of the operator H

can be given by the formula

(12)

(a*)noo,

n = 07 1127

The orthogonality of the vectors on corresponding to different eigen-

values follows from general considerations, but it can also be verified

directly:

(Ok, fin)

(0o, ak(a')"'+Po) = 0

for k # n.

The last equality is obtained from (7) and (10).

We now discuss the question of uniqueness of the vector Rio. Let

?{ be the Hilbert space spanned by the orthonormal system of vectors

On. The elements cp E 7-l have the form

n! E jCnj2 < 00.

(13)

O=i:C

n 7= jpO I

In this space we get a realization of the Heisenberg commutation

relations if, in correspondence with (2), we let

(14)

Q =

a+a*

Vrw(a - a*)

The relation (3) is then a consequence of the formula (6). The space

?[ is invariant under the action of the operators P and Q (more pre-

cisely, under the action of bounded functions f (P) and f (Q), for

example, U(u) and V(v)), and does not contain subspaces with the

same property. Therefore, the representation of Q and P acting in W

by (14) is irreducible.

If in some representation there exist two vectors 00 and 00' sat-

isfying the equation (10), then together with W we can construct a

space ?{' similarly from the vector 0o. This space will be invariant

§ 17. The oscillator in the coordinate representation

with respect to P and Q; that is, a representation in which there is

more than one solution of (10) will be reducible.

§ 17. The problem of the oscillator in the

coordinate representation

In the last section we found the spectrum of the Schr"finger op-

erator for the oscillator in a purely algebraic way, under the single

assumption that the operator H has at least one eigenvector. This is

equivalent to the existence of a solution of the equation ado = 0. We

show that in the coordinate representation there does indeed exist a

unique solution of this equation. Let us write the operators a and a*

in the coordinate representation:

72= w-

(

(2) a WX

( )

The equation ado = 0 takes the form

wx'00(x) + Rio (x) = 0.

Separating the variables, we get that

dV'o - --wx dx

and

by (x) = C'e

The constant C is found from the normalization condition

1= IC12

f--

e "s'

dx = IC12

This condition is satisfied by C = (w/ir)'/4, and the normalized eigen

function for the ground state has the form

or:a

fo(x)

= () e s