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

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L. D. Faddeev and 0. A. Yakubovskii

Of course, the relations (13) do not depend on the representation. We

note further the formulas

U(ui) U(u2) = U(ui + U2),

V(vi) V(v2) = V(vi + v2);

that is. the sets of operators U(u) and 11(v) form groups: demote

these groups by U and V

We can now give a precise formulation of von Neumann's theorem.

For simplicity we confine ourselves to a system with a single degree

of freedom

Theorem. Let (I and V be one-parameter groups of unitary operators

U(u) and V(v) acting in a Ililberl, space H and satisfyinq the. condition

(14) U(u) V (V) = v(V) U(11) P.ivuh

Then 7-l can be represented as a direct sum

(15) f=7-l, mW2O ..,

where each 7-ij is carried into itself by all the operators U(u) and V (v),

and each li can be mapped unitarily onto 112 (R) in such a way that

the operators V (v) pass into the operators i(x)

e - ivXj(x) and the

operators U(u) pass into the operators O(x)

-+(x - uh).

One can say that a representations of the relations (14) by unit ary

operators acts in the space W. If this representation is irreducible,

they i the sum (15) contains only one term.

In the conclusion of this section we prove the irreducibility of the

coordinate represent,atioii for P and Q.

Let K be an operator commuting with Q and P:

[K, Q] = 0,

[K. P] = 0.

It follows from the second equality that

KU(u) = U(u) K.

Applying the operators on both sides of the equality to an arbitrary

function cp(x), we get that

ih) dy =

K(x - uh. y) (y) dy.

§ 14. Interconnection between quantum and classical

The substitution y --- uh y on the left-hand side lets us rewrite this

equality as

K (x, y + uh),r(y) dy =

K(x - uh,J) ,?(?/) dy.

Since p is arbitrary,

K(.x,y+uh) = K(x-uh,y)

from which it is obvious that the kernel K(x. y) depends only on the

difference x -- y, that is,

K(x. y) = k(x

- y).

We now use the fact t hat, K commutes with the coordinate operator:

xk(x - y) (y) dy =

- y) y(y) dy,

-x

t:k(x

from which it, follows that

(x-y)k(x-y) =0

The solution of this equation has the. form

k(x

- y) =

c.S(x-y),

and the function on the right-hand side is the kernel of the operator

C,I.

Accordingly, we have shown 1 hat any operator commuting with

Q and P is a multiple of the identity.

§ 14. The interconnection between quantum and

classical mechanics. Passage to the limit

from quantum mechanics to classical

mechanics

We know that the behavior of a macroscopic system is described well

by classical xrrechanics, so in the passage to a macro-object, quantum

mechanics must reduce to the same results. A criterion for the "quan-

tumness" of the behavior of a system can be the relative magnitude of

Planck's constant h = 1.05.10-27 erg-sec and the system's character-

istic numerical values for observables having the dimension of action

If Planck's constant is negligibly small in comparison with the values

L. D. Faddeev and 0. A. Yakubovskii

of such observables, then the system should have a classical behavior.

The passage from quantum mechanics to classical mechanics can be

described formally as the passage to the limit as h - 0.18 Of course,

under such a passage to the limit, the methods for describing the

mechanical systems remain different, (operators acting in a Hilbert

space cannot be transformed into functions on phase space), but the

physical results of quantum mechanics as h - 0 must coincide with

the classical results.

We again consider a system with one degree of freedom. Our

presentation will be conducted according to the following scheme.

We first use a certain rule to associate self-adjoint operators A f

with real functions f (p, q) on the phase space (f - A1). Then we

find an inversion formula enabling us to recover the function f (p, q)

from the operator A f (Af -+ f) Thereby, we shall have established

a one-to-one relation between real functions on the phase space and

self-adjoint operators acting in fl, (f H A f).

They correct formula

turns out to be

1 TrA = dpdq

()

JM f (p, q)

27rh

Finally, we determine what functions on the phase space corre-

spond to the product A f o A. and to the quantum Poisson bracket

(Aj, Aq } . We shall see that these functions do not coincide

with the

product fg and to the classical Poisson bracket {f, g }, but do tend

to them in the limit as h -* 0. Thus, we shall see that the algebra of

observables in quantum mechanics is not isomorphic to the algebra of

observables in classical mechanics, but the one-to-one correspondence

f Af does become an isomorphism as h - 0.

Suppose that a quantum system with Schrodiuger operator H is

in a state M, and let A f be some observable for this system. The

one-to-one correspondence described enables us to associate with the

operators H, M, and A f, a Hamiltonian function H(p, q). a function

pi (p, q), and an observable f (p, q).

Let p(p, q) = pi (p. q)/2irh. It

18Here there is an analogy with the connection between relativistic and classical

mechanics Relativistic effects can be ignored if the velocities characteristic for the sys-

tem are much less than the velocity c of light The passage from relativistic mechanics

to classical mechanics can be regarded formally as the limit as r - no

§ 14. Interconnection between quantum and classical

follows from (1) that

(2)

ly M = f P(P, 9) dp d9' = 1,

(3)

TrMA f

h ---*O

ff(mq)P(Pq)dPdq.

.tit

The formula (2) shows that the function p(p, q) has the correct

normalization, and the formula (3) asserts that in the limit as h - 0

the mean value of an observable in quantum mechanics coincides with

the mean value of the corresponding classical observable.I9 Further,

suppose that the evolution of the quantum system is described by the

Heisenberg picture and the correspondence A f (t) f (t) has been

established for an arbitrary moment of time t.

We show that as

h - 0 the classical observable f (t) depends correctly on the time.

The operator A f (t) satisfies the equation

(4)

dA f (t)

= {H,Af(t)}h.

Ih-

It follows from the linearity of the correspondence f H A f that !YL

d, and moreover, {H, A f } H {H, f I ash --+ 0, so that as h

the classical equation

is a consequence of the quantum equation (4).

We proceed to the details. Let us consider a function f (p, q) and

denote by f (u, v) its Fourier transform:2°

(5)

f (q, p) = f (u, v)

e-`agVe-iPu

du dv,

27r W

f(u,v)

f (p, q)

dqdp.

27r

R,2

19We underscore

once more that the left-hand side of (3) does not coincide with

the right-hand side when h 0, since a function different from f (p, q)p1(p, q) corre-

sponds to the product MA f

Moreover, we note that for h 0 0 the function p(p, q) can

fail to be positive, that is, it does not correspond to a classical state, but in the limit

as h - 0 it follows from (3) that p(p, q) satisfies all the requirements of a classical

distribution function

20 We omit the stipulations which should be made in order for all the subsequent

transformations to be completely rigorous

L. D. Faddeev and 0. A. Yakubovskii

The Fourier transform of a real funct ion f (p, q) has the property that

(7)

WN ^

f(u,v) = f(-u.-v).

To construct the operator Af corresponding to the function f (p, q),

we would like to replace the variables q and p iii (5) by the operators

Q and P. However, it is riot ininiediately clear in what order we

should write the rioncommuting factors V(v) and U(u). which satisfy

the commutation relations

(8)

ihuv

U(U) V(V) = V(V) U(U) e

A natural recipe (but by no rrieans the only one) ensuring self-

adjointriess was proposed by Weyl and has the form

(

du dv.9) = 1

27r

f(u, v)V (v)U(u)e f

The appearance of t he extra factor e`h",'/2 is connected with the non-

c:onimutat ivity of V (v) and U(u) and ensures that, the operator Af is

self-adjoirit. Indeed, using (7) and (8). we have21

27r

= I

^ u v U* u V * v e- '

hte

du dv

1h III

27r

f (-u, _v) U(-u) V (-v)

e--2

du dv

0111L

f (u. v) U(u) V (v) e- du dv

27r

fJ(uv)=

V(?)) U(u) e 2 dudv = Af.

At the end of this section we verify that the operators f (Q) and f (P)

correspond to the functions f (q) and f (p) by virtue of the formula

(9), in complete agreement with the assumptions made earlier.

We now find an inversion formula. In computing the trace of an

operator k we shall use the formula

(10)

TrK = fK(xx)dx.

21 F3clow

we omit the bymbol R" in the expression for an integral over the n-

dinicrisional apace

§ 14. Interconnection between quantum and classical

where K(x, y) is the kernel of K. We find the kernel of the operator

V(v)U(u). It follows from the formula

V (v) U(u),p(x) = e-"'x p(x - uh)

that the kernel of the integral operator V(v)U(u) is the function

V (v) U(u)(x, x') = e-ivX8(x - uh - x').

By (lo),

Tr V M U u

e-ivx J -uh dx

27r6v

J-x

We verify that the inversion formula is

(12)

J(u,v) = hTrAfV(--v)U(--u)etj2+ .

To this end we compute the right-hand side of the equality, using (8)

and (11):

Tr A f V (-v) U(-u) e

i`'2"

27r

Jf(u,.v,)v(v,)u(u')e',,v(-v)u(_u)eu,du l

Tr f(u , v) V (VI) V (_V) U(u) U(-U} e 2

(v)

du dv'

2ir

Jfu',

I'41'-22t'V)

2zrv)V(v --v)U(u -u)e

du, dv

ff(u',

)b(I) --v}S(u ---n)c du dv

-- h f (u, v)

We set u=v_=0in(12):

f (07 0) = h Tr Af

UU the other hand,

/(0,0) = d d ,

f(p

q) q

27r

and we obtain the formula (1).

It now remains for us to find the functions on phase space corre-

sponding to A f o A,, and {Aj, Ay } h using (12), and to see that these

L. D. Faddeev and 0. A. Yakubovskii

functions tend to f g and {f, g} as h -* 0. We first find the func-

tion F (p, q) corresponding to the nonsymxnetric product A f A9 . Its

Fourier transform is

P(u,v) = h Tr A fA9 V (-v) U(-u) e i`'2"

f(ui,

ihut z1

= hTr

dul dvl du2 dv2

v1),q(u2,v2) V(vi) U(ui) e 2

x V (V2) U(u2) e

th 2 2 V(-v) U(-u) e th2

h Tr

(±-)

27r

fdul dv, du2 dv2 J(u1

, vi) (u2, vz)

x V (V1 + v2 - v) U(ul + uz - u)

x exp

[!(u1v1

+ u2v2 + uv + 2ulv2- 2uly - 2u2v)

Finally,

(13)

AfAg H 2 dul dvl due dvi f (ul, vl) (u2,v2)6(vi + vz - v)

x 5(u1 + U2 - u)

e2tulv2-u2v1)

The exponent in the exponential was transformed with the o-functions

under the integral sign taken into account. We recall that the Fourier

transform 4 (u, v) of the product (p, q) = f (p, q)g(p, q) of the two

functions is the convolution of the Fourier transforms of the factors:

(u,

v) =

27f f dul dvl dug dv2

x f (u1, v1) .9(u2, V2) a(v1 + V2 - v) 6(ul + U2 - u).

The function F(u, v) differs from the function 4 (u, v) by the factor

2 (ul v2

-u2 vi) under the integral sign. This factor depends on the

order of the operators A f and Ag, and therefore the operators A fA9

and A9 A f correspond to different functions on the phase space. In

the limit as h - 0 we have e 2 (uiv2-u2v1)

1 and F(p, q) - 4(p, q).

Of course, this assertion is valid also for the function Fs (p, q) corre-

sponding to the symmetrized product Af o A9.

§ 14. Interconnection between quantum and classical

We denote by G(p, q) the function corresponding to the quantum

Poisson bracket

Af, A., 1,

(Af Aq - AgAf)-

From (13) we get, that

(u,

v) 27rh

f dug dvl dul dv2 f(ui

, vi) Xu2,

v2) b(vi + V2 - v)

X b(u1 + u2 - u) a

2? (ul V2-u2v1) -

eh («2v1-u1v2)

dut A, du2 dvz .f (ul. V1) ,(U1, v2) a(v, + v2 - v)

x Oui + u2 - u) Sill' (U2V1 - ul v2),

andaash-f0

., A

G(u, v) - -

dul dv1 dul dv2 (U2V1 -- ui v2) f (U1, v1) g(u2, V2)

X Ovl + v2 - v) a(u, + u2 - u).

The integral on the right-hand side is the Fourier transform of the

classical Poisson bracket

{f,g} =

afa9

Of 'N

ap aq

since the functions -iv f and -iu f are the Fourier transforms of the

respective derivatives

and "f

Thus, we have verified all the assertions made at the beginning

of the section.

In conclusion we present some examples of computations using

the formulas in this section.

Let us find a formula for the kernel of the operator A f in the

coordinate representation. Using the formula

V (v) U (u) (x, xf) = eivx6(x - uh - x'),

L. D. Faddeev and 0. A. Yakubovskii

we get that

A f (x, x') =

f(u.v)e_itxo(x

- uh - x') e

2 du dv.

2-r

(14)

A f(x. x') =

f x - x

, v

e- 2

dv.

27rh

We show that f (q) H f (Q). For such a function on the phase space,

f(u,v)=

27r

f (q) e ivq

eiUP dq dp =

f (V) 6(u).

Here f (v) denotes the Fourier transform of the function f (q) of a

single variable. Further, by (14),

A f (x, x') =

27rh

f(v)e'dv

;+x'

= f

6(x - X') _ f(x) b(x - X'),

and the operator with this kernel is the operator of multiplication by

the function f (x). In exactly the same way it is easy to verify that

f (p) f (P) in the momentum representation.

We shall get an explicit formriula, for finding the classical state

corresponding to the limit of a pure quantum state as h - 0.

Using the inversion formula, we find pl (q, p) corresponding to the

operator P1k, and we construct p(q, p) = p1 (q, p)/(27rh). The vector

i' is assumed to he given in the coordinate representation. From the

formula

V (-V)

U(-u) e?VX

f (x') v)(x') d r' ,4'(x + uh).

it follows that

V (-v) U(-u) 131, (x, x') = eit,r2,7(x + uh) V., (x')

and

ill «:

1(it,r) _

hTrV(-r;) U(-u) 1

27Th

zz.c

thus

e l'(x + uh) v(x) e

27r

§ 15. One-dimensional problems of quantum mechanics

If we introduce the function22

(15)

F(x, u) =

V'(x + uh) G(x),

then

p(u, v) = f ezvx F(x, u) dx

is the Fourier transform of the classical distribution function corre-

sponding to the limit of the state P.0 as h -> 0. The distribution

function p(q, p) itself is given by

(16) p(p, q) = J e-iPUF(q, u) du.

Suppose that 0(x) is a continuous function and does not depend

on h as a parameter. Then

F(x, u) =

2 I(x)I2

and

p(p, q) = 1,0(q) I26(p)

The state of a particle at rest with density

1 2 of the coor-

dinate distribution function corresponds to such a quantum state in

the limit

ipox

Let O (x) = SP (x) e

, where p

(x) is independent of h and is

continuous.

In this case in the limit as h --p 0, we arrive at the

classical state with distribution function

p(p, q) = J12a(p - po).

We see from these examples that in the limit as h - 0 a mixed

classical state can correspond to a pure state in quantum mechanics.

15. One-dimensional problems of quantum

mechanics. A free one-dimensional particle

In §§ 15-20 we consider one-dimensional problems of quantum me-

chanics. The Hamiltonian function for a particle with one degree of

22This limit is not always trivial, since in physically interesting cases i/ (x) usually

depends on h as a parameter (see the example below)