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Solutions to Problems in Merzbacher,

Quantum Mechanics, Third Edition

Homer Reid

March 8, 1999

Chapter 3

Problem 3.1

If the state (r) is a superposition,

(r) = c11(r) +c22(r)

where 1(r) and 2(r) are related to one another by time reversal,show that the probability current density can be expressed withoutan interference term involving 1 and 2.

I found this to be a pretty cool problem! First of all, we have the probabilityconservation equation:

d

dt = J.

To show that Jcontains no cross terms, it suffices to show that its divergencehas no cross terms, and to show this it suffices (by probability conservation) toshow thatd/dthas no cross terms. We have

=

= [c1

1+c

2

2] [c11+c22]

= |c1||1|+|c2||2|+c1c

21

2+c

1c2

12 (1)

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Problem 3.2

For a free particle in one dimension, calculate the variance at timet, (x)2

t

(x x

t)2t=

x2tx2

twithout explicit use of the

wave function by applying (3.44) repeatedly. Show that

(x)2t

= (x)20+ 2m

12

xpx+pxx0 x0px

t+(px)2

m2 t2

and(px)

2

t = (px)

2

0= (px).

2

I find it easiest to use a slightly different notation: w(t) (x)2t

. (Thewreminds me of width.) Then

w(t) =w(0) +tdw

dt t=0

+ 1

2t2

d2w

dt2 t=0

+ (2)

We have

dw

dt =

d

dt

x2

x2

= d

dt

x2

2 x d

dtx (3)

d2w

dt2 =

d2

dt2

x2

2

d

dtx

22 x

d2

dt2x (4)

We need to compute the time derivatives of x and

x2

. The relevantequation is

ddt

F= 1ih

F HHF+

Ft

for any operator F. For a free particle, the Hamiltonian is H = p2/2m, andthe all-important commutation relation is px = xp ih. We can use this tocalculate the time derivatives:

d

dtx =

1

ih[x, H]

= 1

2imh

xp2 p2x

=

1

2imh xp2 p(xpih)

= 12imh

xp2 pxp+ihp

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= 1

2imh

xp2 (xpih)p+ihp

=

1

2imh2ihp

= p

m (5)

d2

dt2x = 1

mddt

p= 0 (6)

d

dt

x2

= 1

ih

[x2, H]

=

1

2imh

x2p2 p2x2

=

1

2imh

x2p2 p(xpih)x

=

1

2imh

x2p2 pxpx+ihpx

=

1

2imh

x2p2 (xpih)2 +ih(xpih)

=

1

2imh

x2

p2

xpxp+ 2ihxp+ h2

+ h2

+ihxp

= 1

2imh

x2p2 x(xpih)p+ 3ihxp+ 2h2

=

1

2imh

2h2 + 4ihxp

=

ih

m+

2

mxp (7)

d2

dt2

x2

= 2

m

d

dtxp

= 2

ihm[xp,H]

= 1

ihm2 xp3 p2xp

= 1

ihm2

xp3 p(xpih)p

= 1

ihm2

xp3 pxp2 +ihp2

= 1

ihm2

xp3 (xpih)p2 +ihp2

= 1

ihm2

2ihp2

= 2

m2p2

(8)

d3

dt3 x2

= 2

ihm2 [p2, H]= 0 (9)

Now that weve computed all time derivatives ofx and

x2

, its time to

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plug them into (3) and (4) to compute the time derivatives ofw.

dw

dt =

d

dt

x2

2 x d

dtx

= ih

m

+ 2

m

xp 2

m

x p

= 2

m

ih

2 +xp

2

mx p

= 2

m

pxxp

2 +xp

2

mx p

= 2

m

px+xp

2

2

mx p (10)

d2w

dt2 =

d2

dt2

x2

2

d

dtx

22 x

d2

dt2x

= 2

m2p2

2

m2p2 =

2

m2(p)2 (11)

Finally, we plug these into the original equation (2) to find

w(t) =w(0) + 2

m

1

2px+xp x p

t+

(p)2

m2 t.2

The other portion of this problem, the constancy of (p)2, is trivial, since(p)2 contains expectation values ofp and p2, which both commute with H.

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Problem 3.3

Consider a linear harmonic oscillator with Hamiltonian

H=T+ V = p2

2m+

1

2m2x2.

(a) Derive the equation of motion for the expectation value xt,

and show that it oscillates, similarly to the classical oscillator,as

xt= x

0cos t+

p0

m sin t.

(b) Derive a second-order differential equation of motion for theexpectation valueT V

tby repeated application of (3.44)

and use of the virial theorem. Integrate this equation and,remembering conservation of energy, calculate

x2t.

(c) Show that

(x)2t

x2t x2

t = (x)20cos

2 t+(p)2

0

m22 sin2 t

+

1

2xp+px

0 x

0p

0

sin 2t

m

Verify that this reduces to the result of Problem 2 in the limit 0.

(d) Work out the corresponding formula for the variance (p)2t

.

(a)Again I like to use slightly different notation: e(t) = xt. Then

ddt

e(t) = 1ih

xH Hx

= 1

2ihm

xp2 p2x

=

1

2ihm

xp2 p(xpih)

=

1

2ihm

xp2 (xpih)p+ihp

=

1

2ihm2ihp

= p

m.

d2

dt2 e(t) =

d

dt

p

m

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= 1

ihmpHHp

= 2

2ih

px2 x2p

=

2

2ih (xpih)xx2p

= 2

2ih

x(xpih)ihxx2p

=

2

2ih2ihx

= 2 x .

So we haved2

dt2e(t) =2e(t)

with general solutione(t) = A cos t + B sin t. The coefficients are determinedby the boundary conditions:

e(0) =x0

A= x0

e(0) = p0m

B= p0m

.

(b)Lets define v(t) = T Vt. Then

d

dtv(t) =

1

ih(T V)HH(TV)

= 1

ih(T V)(T+V)(T+ V)(T V)

= 2

ihT V V T

= 2

2ih p2x2 x2p2

.

We already worked out this commutator in Problem 2:p2x2 x2p2

=

4ihxp+ 2h2

so

d

dtv(t) = 22 xp+ih2.

= 22 xp+2 xppx

= 2 xp+px (12)

Next,

d2

dt2v(t) =

22

ih xpH Hxp

= 22

ih

12m

xp3 p2xp

+ m

2

2

xpx2 x3p

(13)

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The bracketed expressions arexp3 p2xp

=

xp3 p(xpih)p

=

xp3 (xpih)p2 +ihp2

= 2ihp

2

xpx

2

x3

p

=

x(xpih)xx3

p

=

x2(xpih)ihx2 x3p

=

2ihx2

and plugging these back into (13) gives

d2

dt2v(t) = =42

p2

m

m2

2

x2

= 42v(t)

with solutionv(t) = A cos2t +Bsin 2t. (14)

Evaluating att = 0 givesA= T

0 V

0.

Also, we can use (12) evaluated at t = 0 to determine B :

2 xp+px0

+ih2 = 2B

so

B = xp+px

0

2 .

The next task is to compute

x2t:

x2t = 2m

2V

t

= 1

m2H (TV)

t

= 1

m2[H

tv(t)] .

SinceHdoes not depend explicitly on time, H is constant in time. Forv(t)we can use (14):

x2t

= 1

m2

T

0+V

0[T

0 V

0]cos2t+

xp+px0

2 sin2t

= 1

m2

2 T

0sin2 t+ 2 V

0cos2 t+

xp+px0

2 sin2t

=p2

0

m22sin2 t+

x20

cos2 t+12

xp+px0

sin2tm

. (15)

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(c)Earlier we found that

xt

= x0cos t+

p0

m sin t

x2t

= x2

0cos2 t+

p20

m22sin2 t+x

0p

0sin 2t.

Subtracting from (15) gives

(x)2t

=

x2

x2 =

x20

x20

cos2 t

+ 1

m22

p2

0

p20

+

1

2xp+px

0 x

0p

0

sin2t

= (x)20cos2 t+

(p)20m22

sin2 t+

1

2xp+px

0 x

0p

0

sin2t

m .

As 0, cos2 t 1, (sin2 t/2) 1, and (sin2t/) 2, as needed

to ensure matchup with the result of Problem 2.

Problem 3.4

Prove that the probability density and the probability current den-sity at positionr0 can be expressed in terms of the operatorsr andp as expectation values of the operators

(r0)(rr0) j(r0) 1

2m[p(rr0) +(rr0)p] .

Derive expressions for these densities in the momentum represen-tation.

The first one is trivial:

(rr0)=

(r)(rr0)(r)dr=

(r0)(r0) = (r0).

For the second one,

1

2mp(rr0) +(rr0)p=

ih

2m

[(rr0) +

(rr0)] dr

The gradient operator in the first term operates on everything to its right:

= ih2m

[(rr0) + 2(rr0)] dr.

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