Physics 215 Solution Set 2 Winter 2018

1. At time t = 0, the wave function of a free particle moving in a one-dimension is givenby,

ψ(x, 0) = N

∫

+∞

−∞e−|k|/k0eikx dk , (1)

where N and k0 are real positive constants.

Before attacking this problem, we shall first evaluate the integral that appears in eq. (1)and determine the value of the constant N . Since k0 > 0 the integral is convergent, and itfollows that

ψ(x, 0) = N

{

∫

0

−∞dk exp

{

k

(

1

k0+ ix

)}

+N

∫ ∞

0

dk exp

{

−k(

1

k0− ix

)}

}

= N

[

(

1

k0+ ix

)−1+

(

1

k0− ix

)−1]

=2Nk0

1 + k20x2.

The normalization constant N is determined by demanding that∫ ∞

−∞|ψ(x, 0)|2 dx = 1 . (2)

Hence,

4|N |2k20∫ ∞

−∞

dx

(1 + k20x2)2

= 1 . (3)

The integral in eq. (3) is easily computed if you remember that

I ≡ k0∫ ∞

−∞

dx

(1 + k20x2)

= tan−1(k0x)

∣

∣

∣

∣

∞

−∞= π . (4)

If we multiply eq. (4) by k0 and define q ≡ 1/k20, then eq. (4) becomes,

k0I =

∫ ∞

−∞

dx

q + x2=

π√q.

Finally, taking the derivative of the above result with respect to q yields

d

dq(k0I) = −

∫ ∞

−∞

dx

(q + x2)2= −k40

∫ ∞

−∞

dx

(1 + k20x2)2

= − π2q3/2

= −12πk30 ,

after making use of q1/2 = 1/k0. We therefore end up with,∫ ∞

−∞

dx

(1 + k20x2)2

=π

2k0.

1

Inserting this last result into eq. (3) yields,

2π|N |2k0 = 1 . (5)

Since N is a real positive constant by assumption (one is always free to rephase the wavefunction such that N > 0), it follows that

ψ(x, 0) =

√

2k0π

1

1 + k20x2. (6)

REMARKS:

1. A good check of the results above is to make use of dimensional analysis. If x hasdimensions of length, then k0 has dimensions of inverse length. We can indicate this bywriting [x] = L and [k] = L−1. Then, eq. (3) implies that [N ] = L1/2 and [ψ(x, 0)] = L−1/2.In addition, the parameter q = 1/k20 has dimension [q] = L

2. Using these results, one canverify that all the equations above are dimensionally correct. Such a check helps immenselyin avoiding algebraic mistakes.

2. Actually, there is a much quicker way to derive eq. (5). Using the results of problem3(b) of Problem Set 1, it follows that if f(x) = 1√

2π

∫

+∞−∞ a(k)e

ikx dk , then

∫

+∞

−∞|f(x)|2 dx =

∫

+∞

−∞|a(k)|2 dk .

Applying this result to the evaluation of eq. (2), we can identify a(k) = N√2π e−|k|/k0 and

f(x) = ψ(x, 0). Hence,

2π|N |2∫ ∞

−∞e−2|k|/k0 dk = 1 .

The above integral is elementary, and we immediately recover eq. (5).

(a) What is the probability that a measurement of the momentum performed at timet = 0 will yield a result between −p1 and p1? How does the probability change if themeasurement is performed instead at time t? Explain your result.

Applying the postulates of quantum mechanics, the probability that a measurement of themomentum performed at time t = 0 will yield a result between −p1 and p1, denoted byP (p1, 0) below, is given by

P (p1, 0) =

∫ p1

−p1|〈p|ψ〉|2 dp =

∫ p1

−p1|φ(p, 0)|2 dp , (7)

where the momentum-space wave function at t = 0 is the Fourier transform of ψ(x, 0),

φ(p, 0) =1√2π~

∫ ∞

−∞e−ipx/~ ψ(x, 0) dx ,

2

Inverting the Fourier transform yields,

ψ(x, 0) =1√2π~

∫ ∞

−∞eipx/~ φ(p, 0) dx .

If we write p = ~k and compare with eq. (1), we can identify

φ(p, 0) =

√

2π

~Ne−|p|/p0 ,

where we have defined p0 ≡ ~k0. Hence, using N = (2πp0/~)1/2 [cf. eq. (5)], it follows that

φ(p, 0) =1√p0e−|p|/p0 . (8)

Finally, inserting this result into eq. (7) yields

P (p1, 0) =1

p0

∫ p1

−p1e−2|p|/p0 dp =

2

p0

∫ p1

0

e−2p/p0 dp = 1− e−2p1/p0 .

Next, we evaluate P (p1, t). The wave function evolves according to,

|ψ(t)〉 = U(t) |ψ(0)〉 ,

where the time evolution operator is given by U(t) = exp(

−iHt/~)

. For a free particlemoving in one dimension, the Hamiltonian is H = P 2/(2m). Hence,

U(t) = exp

{

− iP2t

2m~

}

. (9)

We proceed to evaluate the momentum space wave function,

φ(p, t) = 〈p|ψ(t)〉 =∫ ∞

−∞〈p|U(t) |p′〉 〈p′|ψ(0)〉 dp′ =

∫ ∞

−∞〈p|U(t) |p′〉φ(p′, 0) dp′ , (10)

after inserting a complete set of momentum eigenstates. Using eq. (9) and P |p〉 = p |p〉,where 〈p|p′〉 = δ(p− p′),

〈p|U(t) |p′〉 = 〈p| exp{

− iP2t

2m~

}

|p′〉 = exp{

− i~p2

2m~

}

δ(p− p′) .

Plugging this result into eq. (10), the integration over p′ is now trivial due to the deltafunction. Hence,

φ(p, t) = exp

{

− i~p2

2m~

}

φ(p, 0) . (11)

It therefore follows that |φ(p, t)|2 = |φ(p, 0)|2. That is, |φ(p, t)|2 is independent of time.This result immediately implies that the probability P (p, t) is also independent of time.

3

This result is expected since for a free particle, the energy eigenstates are also momentumeigenstates. Since the Hamiltonian, H = P 2/(2m) is time-independent, it follows that theenergy (and consequently the momentum) eigenstates are stationary states. In particular,the probability of finding a particle with a given range of momenta must also be time-independent.

(b) What is the form of the (position space) wave packet at time t = 0? Calculate theproduct ∆X∆P at time t = 0. Describe qualitatively the subsequent evolution of the wavepacket.

The form of the wave packet was explicitly obtained in eq. (6). Using this result as wellas the corresponding result for the momentum space wave function obtained in eq. (8), wecompute the following expectation values.

〈X〉 =∫ ∞

−∞x|ψ(x, 0)|2 dx = 2k0

π

∫ ∞

−∞

x dx

(1 + k20x2)2

= 0 ,

〈P 〉 =∫ ∞

−∞p|φ(p, 0)|2 dp = 1

p0

∫ ∞

−∞pe−2|p|/p0 dp = 0 ,

〈

X2〉

=

∫ ∞

−∞x2|ψ(x, 0)|2 dx = 2k0

π

∫ ∞

−∞

x2 dx

(1 + k20x2)2

=1

k20,

〈

P 2〉

=

∫ ∞

−∞p2|φ(p, 0)|2 dp = 1

p0

∫ ∞

−∞p2e−2|p|/p0 dp = 1

2p20 ,

(12)

where p0 ≡ ~k0. It then follows that

∆X =[〈

X2〉

− 〈X〉2]1/2

=1

k0, (13)

∆P =[〈

P 2〉

− 〈P 〉2]1/2

=~k0√2. (14)

Hence, ∆X∆P = 12

√2 ~, which is consistent with the Heisenberg uncertainty principle,

∆X∆P ≥ 12~ .

Finally, as time evolves, the wave packet in coordinate space spreads out. In principle,one can demonstrate this behavior by computing,

ψ(x, t) =1√2π~

∫ ∞

−∞φ(p, t)eipx/~ dp =

1√2π~p0

∫ ∞

−∞exp

{

i

~

(

px− p2t

2m

)}

e−|p|/p0 dp .

Unfortunately, the above integral cannot be evaluated in terms of elementary functionsalone [one would need to employ the error function erf(x)]. However, if one simply wants

4

to demonstrate the qualitative behavior of the spreading wave packet, one can make use ofthe uncertainty relation obtained in eq. (2.2.30) on p. 85 of Sakurai and Napolitano,

∆X(0)∆X(t) ≥ ~t2m

.

Using ∆X(0) = 1/k0 obtained in eq. (13), it follows that ∆X(t) ≥ ~k0t/(2m), whichindicates that the width of the wave packet increases (at least as fast as) linearly in thetime t.

2. (a) Consider a quantum mechanical ensemble characterized by a density matrix ρ.Suppose that the system is governed by a Hamiltonian H (which may be time dependent).Show that the time evolution of ρ (in the Schrödinger picture) is given by:

∂ρ

∂t= − i

~[H, ρ] .

The density operator for a quantum ensemble made up of N representatives is defined by

ρ =∑

i

pi |ψi〉 〈ψi| ,

where the state |ψi〉 appears ni times in the ensemble and pi ≡ ni/N . That is, pi is theprobability that an arbitrarily chosen element of the ensemble is in the state |ψi〉.

The time evolution of the state |ψi〉 (in the Schrödinger picture) is given by the Schrödingerequation,

i~d

dt|ψi〉 = H |ψi〉 . (15)

Under the assumption that the Hamiltonian is a self-adjoint operator, the adjoint of eq. (15)is given by,

−i~ ddt

〈ψi| = 〈ψi|H . (16)

Noting that the probabilities pi do not depend on the time t, it follows that

∂ρ

∂t=

∑

i

pi

[(

∂

∂t|ψi〉

)

〈ψi|+ |ψi〉(

∂

∂t〈ψi|

)]

=1

i~

∑

i

pi[

H |ψi〉 〈ψi| − |ψi〉 〈ψi|H]

= − i~

[

H , ρ]

.

It follows that∂ρ

∂t+i

~[H, ρ] = 0 ,

which is called the von Neumann equation.

5

(b) Let U(t, t0) be the time evolution operator. Find a general expression for ρ(t) interms of ρ(t0) and U .

Starting from

ρ(t) =∑

i

pi |ψi(t)〉 〈ψi(t)| , ρ(t0) =∑

i

pi |ψi(t0)〉 〈ψi(t0)| ,

with |ψi(t)〉 = U(t, t0) |ψ(t0)〉, it immediately follows that

ρ(t) = U(t, t0)ρ(t0)U†(t, t0) . (17)

REMARKS:

Note that U(t, t0) satisfies the differential equation

i~∂

∂tU(t, t0) = H(t)U(t, t0) , (18)

which is equivalent to the Schrödinger equation given by eq. (15). Taking the adjoint ofthis equation, under the assumption that the Hamiltonian is a self-adjoint operator,

−i~ ∂∂tU †(t, t0) = U

†(t, t0)H(t) . (19)

Taking the time derivative of eq. (17) and employing eqs. (18) and (19), one again derivesthe von Neumann equation obtained in part (a).

(c) Prove that Tr ρ2 is time-independent. Hence, show that a pure state cannot evolveinto a mixed state.

Using eq. (17),

Tr ρ2(t) = Tr{[

U(t, t0)ρ(t0)U†(t, t0)

][

U(t, t0)ρ(t0)U†(t, t0)

]}

= Tr{

U(t, t0)ρ2(t0)U

†(t, t0)}

= Tr ρ2(t0) .

In the penultimate step above, we cyclically permuted the arguments of the trace (whichdoes not change its value) and put U †U = I after noting that U is unitary. It follows thatTr ρ2 is time-independent.

We now recall that a pure state satisfies Tr ρ2 = 1, whereas a mixed state satisfiesTr ρ2 < 1. We can therefore conclude that pure states always evolve into pure states,whereas mixed states always evolve into mixed states.

6

3. In a one-dimensional problem, consider a particle of potential energy V (x) = −fx,where f > 0.

(a) Write Ehrenfest’s theorem for the mean values of the position 〈X〉 and the momen-tum 〈P 〉 of the particle. Integrate these equations and compare with the classical motion.

Ehrenfest’s theorem states that

d

dt〈X〉 =

〈

∂H

∂P

〉

,d

dt〈P 〉 = −

〈

∂H

∂X

〉

. (20)

For this problem, the Hamiltonian is given by

H =P 2

2m− fX ,

where f > 0. Using ∂H/∂P = P/m and ∂H/∂X = −fI (where I is the identity operator),eq. (20) yields,

d

dt〈X〉 = 1

m〈P 〉 , d

dt〈P 〉 = f . (21)

In obtaining the second result above, we used the fact that the expectation value of theidentity operator I with respect to normalized state is equal to 1. Integrating eq. (21) thenyields

〈P 〉 = 〈P 〉0+ ft , 〈X〉 = 〈X〉

0+ 〈P 〉

0

t

m+ft2

2m, (22)

where 〈P 〉0≡ 〈P 〉t=0 and 〈X〉0 ≡ 〈X〉t=0. As expected, eq. (22) is identical in form to the

corresponding classical mechanics results.

(b) Show that the root-mean-square deviation ∆P does not vary with time.

The root-mean square deviation satisfies the following relation,

(∆P )2 =〈

P 2〉

− 〈P 〉2 . (23)

To evaluate 〈P 2〉, we use the fact that d 〈H〉 /dt = 0, which is a consequence of the factthat H has no explicit time dependence.1 It then follows that

1

2m

d

dt

〈

P 2〉

+ fd

dt〈X〉 = 0 .

1Recall that the generalized Ehrenfest theorem is given by

d

dt〈Ω〉 = − i

~

〈

[Ω , H ]〉

+

〈

∂Ω

∂t

〉

,

for any operator Ω. In the case of Ω = H , we have d 〈H〉 /dt = 〈∂H/∂t〉 (since H commutes with itself).For a time-independent Hamiltonian, we have ∂H/∂t = 0. It then follows that d 〈H〉 /dt = 0.

7

Using eq. (21) for d〈X〉/dt, we immediately obtain,

d

dt

〈

P 2〉

= −2f 〈P 〉 . (24)

Next, we make use of eq. (21) for d〈P 〉/dt to conclude that

d

dt〈P 〉2 = 2 〈P 〉 d

dtP = 2f 〈P 〉 . (25)

Hence, the time derivative of eq. (23) yields

d

dt(∆P )2 = −2f 〈P 〉+ 2f 〈P 〉 = 0 ,

after using eqs. (24) and (25). Hence, ∆P does not vary in time.

(c) Write the Schrödinger equation in the p-representation and deduce a relation between

∂

∂t| 〈p|ψ(t)〉 |2 and ∂

∂p| 〈p|ψ(t)〉 |2

Solve the equation thus obtained and give a physical interpretation.

The Schrödinger equation [cf. eq. (15)] for this problem is,

i~d

dt|ψ(t)〉 =

(

P 2

2m− fX

)

|ψ(t)〉 . (26)

In the p-representation,

i~∂

∂t〈p|ψ(t)〉 = 〈p|

(

P 2

2m− fX

)

|ψ(t)〉 . (27)

Using the adjoint of P |p〉 = p |p〉 recalling that P is self-adjoint and hence its eigenvaluesare real), and recalling that2

〈p|X |ψ(t)〉 = i~ ∂∂p

〈p|ψ(t)〉 , (28)

it then follows from eq. (27) that

i~∂φ(p, t)

∂t=

(

p2

2m− i~f ∂

∂p

)

φ(p, t) , (29)

where φ(p, t) ≡ 〈p|ψ(t)〉 is the wave function in the p-representation.2Due to eq. (28), one says that the operator X is represented by the differential operator i~ ∂/∂p in the

p-representation.

8

Using eq. (29),

∂

∂t|φ(p, t)|2 = φ∗(p, t)∂φ(p, t)

∂t+ φ(p, t)

∂φ∗(p, t)

∂t

= −φ∗(p, t)(

ip2

2m~+ f

∂

∂p

)

φ(p, t) + φ(p, t)

(

ip2

2m~− f ∂

∂p

)

φ∗(p, t)

= −f(

φ∗(p, t)∂φ(p, t)

∂p+ φ(p, t)

∂φ∗(p, t)

∂p

)

= −f ∂∂p

|φ(p, t)|2 .

Denoting P(p, t) ≡ |φ(p, t)|2, the above equation reads,

∂P(p, t)∂t

= −f ∂P(p, t)∂p

. (30)

To solve eq. (30), it is convenient to change variables by introducing

r ≡ p− ft , s ≡ p+ ft .

Then∂P∂p

=∂P∂r

+∂P∂s

,∂P∂t

= −f ∂P∂r

+ f∂P∂s

.

Inserting these results into eq. (30) yields

∂P∂s

= 0 .

That is P(r, s) is a function of r alone. This means that

P(p, t) = f(p− ft) ,

where f is an arbitrary function. In particular, it follows that

P(p, t) = P(p− ft, 0) .

The interpretation of this result is that the probability distribution in p moves accordingto the classical equation of motion, p = p0 + ft.

(d) Write the Schrödinger equation in the x-representation. What are the energy eigen-functions?

The Schrödinger equation in the x-representation is obtained from eq. (26) by multiplyingfrom the left by 〈x|. The wave function in the x-representation is ψ(x, t) = 〈x|ψ(t)〉. Afterwriting ψ(x, t) = ψ(x)e−iEt/~, we obtain the time-independent Schrödinger equation in thex-representation,

(

− ~2

2m

d2

dx2− fx

)

ψ(x) = Eψ(x) ,

9

which we can rewrite asd2ψ

dx2+

(

2mf

~2x+

2mE

~2

)

ψ = 0 . (31)

It is convenient to introduce a new variable,

y ≡(

2mf

~2

)−2/3(2mf

~2x+

2mE

~2

)

. (32)

By changing variables from x to y, eq. (31) simplifies to

d2ψ

dy2+ yψ = 0 . (33)

Consulting, e.g., N.N. Lebedev, Special Functions and Their Applications (Dover Publica-tions, Inc., New York, 1972), pp. 136–139, we see that the solutions to eq. (33) are the Airyfunctions,

ψ(y) = c1Ai(−y) + c2Bi(−y) , (34)where c1 and c2 are arbitrary constants are determine by the boundary conditions relevantfor the problem.

(e) Is the energy spectrum continuous or discrete? What is the behavior of the energyeigenfunctions as |x| → ∞ ? If the potential were replaced by V (x) = f |x|, how wouldyour answer change?

To see whether the energy spectrum is discrete or continuous, let us examine the asymptoticbehavior of the Airy functions. On p. 138 of Lebedev (op. cit.), the following results aregiven:

Ai(x) ∼

1

2√πx−1/4 exp

(

−23x3/2

)

, for x→ ∞,

1√π(−x)−1/4 cos

(

2

3(−x)3/2 − 1

4π)

, for x→ −∞,

Bi(x) ∼

1√πx−1/4 exp

(

2

3x3/2

)

, for x→ ∞,

− 1√π(−x)−1/4 sin

(

2

3(−x)3/2 − 1

4π)

, for x→ −∞.

10

Noting that f > 0, the potential V (x) = −fX is exhibited below.

x

V (x)

E

Since V (x) > E as x → −∞, we demand that limx→−∞ ψ(x) = 0. Thus, we must choosec2 = 0 in the solution given in eq. (34). We conclude that ψ(y) = cAi(−y), where yis given in terms of x by eq. (32) and the constant c is determined by an appropriatenormalization condition. Note that as x → ∞, we expect oscillatory behavior of thewave function (in analogy to the wave function ψ(x) ∼ eikx of a free particle). This isindeed the case as is evident from the asymptotic form for Ai(−x) as x→ ∞ given above.Note that no condition on E needs to be imposed to satisfy the boundary conditions (i.e.,an exponentially vanishing wave function as x → −∞ and an oscillatory wave function asx→ ∞). Thus, the energy spectrum is continuous (and there are no bound state solutions).

In contrast, consider the potential V (x) = f |x|, with f > 0. The previous figure is nowmodified as follows.

x

V (x)

E

In this case, V (x) > E as x → ±∞. The spectrum must consist entirely of bound states,and the energy spectrum is discrete. To solve for the bound state energies, we must solve

d2ψ

dx2+

(

−2mf~2

|x|+ 2mE~2

)

ψ = 0 . (35)

11

separately for x > 0 and x < 0. Following our previous analysis, we obtain,

ψ(x) =

{

cAi(−y) , for x < 0,c′ Ai(y′) , for x > 0,

(36)

where y is given in terms of x by eq. (32) and

y′ ≡(

2mf

~2

)−2/3(2mf

~2x− 2mE

~2

)

.

Note that the wave function given by eq. (36) exponentially approaches zero as x→ ±∞, asrequired by the bound state solutions to this problem. To determine the relation betweenthe constants c and c′, we require that both ψ(x) and dψ/dx are continuous functions atx = 0. These two conditions can only be satisfied for a discrete set of energy eigenvalues Ewith a fixed value of c′/c. There remains one overall undetermined coefficient (whichdepends on the bound state energy E), which is determined by requiring that the norm ofthe bound state wave functions are unity.

4. Consider a particle in three dimensions whose Hamiltonian is given by:

H =~P

2

2m+ V ( ~X) . (37)

By calculating the commutator, [ ~X · ~P , H ], derive the quantum Virial Theorem,

d

dt

〈

~X · ~P〉

=

〈

~P2

m

〉

−〈

~X · ~∇V〉

. (38)

To identify eq. (38) with the quantum mechanical analog of the Virial Theorem, it isessential that the left-hand side of eq. (38) vanish. Under what condition does this happen?

We shall make use of the generalized Ehrenfest theorem,

d

dt〈Ω〉 = − i

~

〈

[Ω , H ]〉

+

〈

∂Ω

∂t

〉

, (39)

Using Ω = ~X · ~P , it follows that

d

dt

〈

~X · ~P

〉

= − i~

〈

[

~X · ~P , H]

〉

. (40)

For H given by eq. (37),

[

~X · ~P , H]

=[

~X · ~P ,~P

2

2m+ V ( ~X)

]

=[

~X · ~P ,~P

2

2m

]

+[

~X · ~P , V ( ~X)]

. (41)

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We evaluate separately the two commutators on the right hand side of eq. (41) First,

[

~X · ~P ,~P

2

2m

]

=1

2m

[

XiPi , PjPj]

=1

2m

[

Xi , PjPj]

Pi =1

2m

{

Pj[

Xi , Pj]

Pi+[

Xi , Pj ]PjPi

}

,

where there is an implicit sum over the pairs of repeated indices. Using[

Xi , Pj]

= i~δijand

[

Xi , Xj]

=[

Pi , Pj]

= 0, it follows that

[

~X · ~P ,~P

2

2m

]

=i~

m~P

2 . (42)

Second, we evaluate[

~X · ~P , V ( ~X)]

= Xi[

Pi , V ( ~X)]

. (43)

In light of problem 5(b) on Problem Set 1,

[

P , F (X)]

= −i~ ∂F∂X

.

The extension to three dimensions is straightforward,

[

Pi , F (X)]

= −i~ ∂F∂Xi

.

Hence, using eq. (43) we obtain

[

~X · ~P , V ( ~X)]

= Xi[

Pi , V ( ~X)]

= −i~Xi∂V

∂Xi= −i~ ~X · ~∇V . (44)

Combining eqs. (42) and (44), we end up with

[

~X · ~P , H]

=[

~X · ~P ,~P

2

2m+ V ( ~X)

]

= i~

[

~P2

m− ~X · ~∇V

]

.

Thus, eq. (40) yields

d

dt

〈

~X · ~P〉

=

〈

~P2

m

〉

−〈

~X · ~∇V〉

. (45)

To identify eq. (45) with the quantum mechanical analog of the Virial Theorem, it isessential that the left-hand side of eq. (45) vanish. This will happen if the expectationvalues are taken with respect to a stationary state. In this case,

〈

~X · ~P〉

t≡ 〈ψ(t)| ~X · ~P |ψ(t)〉 = 〈ψ(0)| eiHt/~ ~X · ~P e−iHt/~ |ψ(0)〉

= 〈ψ(0)| eiEt/~ ~X · ~P e−iEt/~ |ψ(0)〉 = 〈ψ(0)| ~X · ~P |ψ(0)〉 =〈

~X · ~P〉

0

.

13

That is,〈

~X · ~P〉

is time-independent when taken with respect to a stationary state. In

this case,d

dt

〈

~X · ~P〉

= 0 , (46)

and we arrive at the quantum mechanical virial theorem,

〈

~P2

m

〉

=〈

~X · ~∇V〉

. (47)

REMARKS:

A technical remark is in order. The classical mechanical virial theorem is derivedunder the assumption that all the particle orbits are bounded. In the quantum mechanicalcontext, this condition translates into the requirement that the stationary states used incomputing the expectation values above are bound states. That is, the states used toevaluate expectation values must be discrete energy levels rather than continuous energylevels. To see where this condition arises in the above analysis, we note that if Ω is time-independent, then eq. (39) yields,

d

dt〈Ω〉 = − i

~

〈

[Ω , H ]〉

.

With respect to an energy eigenstate,

〈

[Ω , H ]〉

= 〈E|ΩH −HΩ |E〉 = E 〈E|Ω |E〉 −E 〈E|Ω |E〉 = 0 , (48)

after using H |E〉 = E |E〉 and the corresponding adjoint relation. But this last computa-tion is correct as long as 〈E|Ω |E〉 is finite. Note that for discrete energy levels, we have〈E|E ′〉 = δEE′. Thus, we expect that 〈E|Ω |E〉 is also finite, in which case eq. (48) is valid.

In contrast, for energy eigenvalues in the continuum corresponding to scattering states,we have 〈E|E ′〉 = δ(E − E ′). That is, for continuum energy levels, 〈E|E〉 = ∞. Likewise〈E|Ω |E〉 is also infinite, so we cannot conclude as in eq. (48) that

〈

[Ω , H ]〉

= 0. Hence,eq. (46) is not expected to be valid for expectation values with respect to stationary scat-tering states (in the continuum). That is, the quantum mechanical virial theorem given byeq. (47) applies only for the vacuum expectation values with respect to stationary boundstates.

5. In this problem, you are asked to derive the Feynman-Hellmann Theorem.

(a) If the Hamiltonian H(λ) depends on a real parameter λ, i.e., H(λ) |ψ〉 = E(λ) |ψ〉,then show that:

∂E

∂λ=

〈

ψ

∣

∣

∣

∣

∂H

∂λ

∣

∣

∣

∣

ψ

〉

.

14

Consider E(λ) ≡ 〈ψ|H(λ) |ψ〉. For a normalized state, 〈ψ|ψ〉 = 1. Then,dE

dλ=

d

dλ〈ψ|H(λ) |ψ〉 =

〈

dψ

dλ

∣

∣

∣

∣

H(λ)

∣

∣

∣

∣

ψ

〉

+

〈

ψ

∣

∣

∣

∣

H(λ)

∣

∣

∣

∣

dψ

dλ

〉

+

〈

ψ

∣

∣

∣

∣

∂H

∂λ

∣

∣

∣

∣

ψ

〉

= E(λ)

{〈

dψ

dλ

∣

∣

∣

∣

ψ

〉

+

〈

ψ

∣

∣

∣

∣

dψ

dλ

〉}

+

〈

ψ

∣

∣

∣

∣

∂H

∂λ

∣

∣

∣

∣

ψ

〉

. (49)

Note that if one differentiates the normalization condition, 〈ψ|ψ〉 = 1 with respect to λ, weimmediately obtain,

0 =d

dλ〈ψ|ψ〉 =

〈

dψ

dλ

∣

∣

∣

∣

ψ

〉

+

〈

ψ

∣

∣

∣

∣

dψ

dλ

〉

.

Using this last result in eq. (49), we end up with

∂E

∂λ=

〈

ψ

∣

∣

∣

∣

∂H

∂λ

∣

∣

∣

∣

ψ

〉

, (50)

which is the Feynman-Hellman theorem.

(b) Consider the one-dimensional problem with the Hamiltonian:

H = − ~2

2m

d2

dx2+ V (x) ,

where V is independent of the parameter m. Suppose one finds that this Hamiltonianpossesses a particular energy eigenstate with energy eigenvalue E. Describe the behaviorof E as m decreases.

If we rewrite the Hamiltonian in terms of the momentum operator P and the positionoperator X , then

H =P 2

2m+ V (X) .

Since V (x) is independent of m, it follows that

∂H

∂m= − P

2

2m2.

Hence, using eq. (50),

dE

dλ=

〈

ψ

∣

∣

∣

∣

∂H

∂m

∣

∣

∣

∣

ψ

〉

= − 12m2

〈ψ|P 2 |ψ〉 < 0 ,

since the operator P 2 is non-negative.3 More explicitly,〈

ψ

∣

∣

∣

∣

∂H

∂m

∣

∣

∣

∣

ψ

〉

= − ~2

2m

∫ ∞

−∞

∣

∣

∣

∣

dψ

dx

∣

∣

∣

∣

2

< 0 .

We conclude that ∂E/∂m < 0. That is, as m decreases, the bound state energies increase.3If we define |φ〉 ≡ P |ψ〉 where P is self-adjoint (P = P †), then 〈ψ|P 2 |ψ〉 = 〈ψ|P †P |ψ〉 = 〈φ|φ〉 > 0

since the norm of any non-zero state vector is positive.

15

6. Consider the operator:

a =mωX + iP√

2m~ω, (51)

where X and P are the position and momentum operators, respectively.

(a) Let |z〉 be an eigenvector of a with eigenvalue z. This state is called a coherent state.Compute 〈x|z〉. Using this result, show that 〈z|z′〉 6= 0. Why does the lack of orthogonalityof these states not violate any of our quantum mechanics postulates?

The eigenvalue equation for the operator a is

a |z〉 =(

mωX + iP√2m~ω

)

|z〉 = z |z〉 .

Consider the eigenvalue equation with respect to the x-basis. Multiplying on the left by〈x| then yields,

〈

x

∣

∣

∣

∣

(

mωX + iP√2m~ω

)∣

∣

∣

∣

z

〉

= z |z〉 .

To evaluate this equation, we make use of

〈x|X |z〉 = x 〈x|z〉 , 〈x|P |z〉 = −i~ ddx

〈x|z〉 .

It then follows that(mω

2~

)1/2(

x+~

mω

d

dx

)

〈x|z〉 = z 〈x|z〉 .

This is a differential equation fro 〈x|z〉. Denoting ψz(x) ≡ 〈x|z〉, we proceed to solve[

d

dx+mωx

~−

(

2mω

~

)1/2

z

]

ψz(x) = 0 .

The solution to this equation is

〈x|z〉 ≡ ψz(x) = Cz exp[

−mω2~

x2 +

(

2mω

~

)1/2

zx

]

, (52)

where Cz is a normalization constant (which will depend on the eigenvalue z).We can check to see whether states of different values of z are orthogonal by computing

〈z|z′〉. After inserting a compete set of position eigenstates,,

〈z|z′〉 = C∗zCz′∫ ∞

−∞exp

[

−mω2~

x2 +

(

2mω

~

)1/2

zx

]

exp

[

−mω2~

x2 +

(

2mω

~

)1/2

z′x

]

= C∗zCz′

∫ ∞

−∞exp

[

−mω~x2 +

(

2mω

~

)1/2

(z′ + z∗)x

]

= C∗zCz′

(

π~

mω

)1/2

exp[

1

2(z′ + z∗)2

]

. (53)

16

For z 6= z′, it is clear that 〈z|z′〉 6= 0. Hence, the eigenstates of the operator a are notmutually orthogonal. This does not violate any of the quantum mechanics postulates sincea is not a self-adjoint (nor is it an hermitian) operator. Thus, its eigenvalues need not bereal and its eigenstates need not be mutually orthogonal.

(b) Consider the operator a in the context of the one-dimensional harmonic oscillator.Compute 〈n|z〉, where |n〉 is the nth energy eigenstate. (Assume that |z〉 is normalized tounity.) Given a coherent state |z〉, find the most probable value of n (and correspondingenergy E).

Repeated application of a† |n〉 =√n+ 1 |n+ 1〉 yields

|n〉 = (a†)n√n!

|0〉 . (54)

Taking the adjoint of eq. (54), it then follows that

〈n|z〉 =〈

0

∣

∣

∣

∣

an√n!

∣

∣

∣

∣

z

〉

=zn√n!

〈0|z〉 . (55)

Since {|n〉} are complete set of states, we can use the completeness relation to obtain

|z〉 =∞∑

n=0

|n〉 〈n|z〉 = 〈0|z〉∞∑

n=0

zn√n!

|n〉 . (56)

Using eq. (56), we can compute the norm of |z〉,

〈z|z〉 = |〈0|z〉|2∞∑

n=0

|z|2nn!

= |〈0|z〉|2 e|z|2 .

We shall normalize |z〉 to unity, i.e. we set 〈z|z〉 = 1. It then follows that

〈0|z〉 = e−|z|2/2 , (57)

after (conventionally) setting an overall phase factor to 1. Inserting this last result backinto eq. (55) then yields,

〈n|z〉 = zn

√n!e−|z|

2/2 .

Given the state |z〉, the probability of measuring a value n is given by

|〈n|z〉|2 = |z|2

n!e−|z|

2

.

The most probable value of n depends on z. For |z| ≤ 1, we see that n = 0 is the mostprobable value. As |z| becomes larger, so does the most probable value of n. For |z| ≫ 1we can make use of Stirling’s approximation,

n! ≃√2πnnn e−n .

17

Thus, for |z| ≫ 1, we approximately have,

|〈n|z〉|2 ≃ 1√2πn

( |z|2n

)n

en−|z|2

.

We would like to maximize the function,

f(n) = ln

{

1√n

( |z|2n

)n

en−|z|2

}

= n− |z|2 + n ln |z|2 − (n + 12) lnn .

Taking a derivative and setting f ′(n) = 0 yields ln(|z|2/n) = 1/(2n). For |z| ≫ 1, it is clearthat n ≃ |z|2 is an approximate solution to f ′(n) = 0. Moreover, f ′′(n) ≃ −1/n < 0 at theextremum, which indicates that n ≃ |z|2 is a maximum.

We conclude that the maximum value of |〈n|z〉|2 occurs for n ≃ |z|2, That is, forlarge |z|, the most probable energy is given by E ≃ ~ω|z|2.

(c) Prove that the normalized coherent state can be written as, |z〉 = e−|z|2/2eza† |0〉.

Combining eqs. (56) and (57), it follows that

|z〉 = e−|z|2/2∞∑

n=0

zn√n!

|n〉 = e−|z|2/2∞∑

n=0

zn

n!(a†)n |0〉 , (58)

after making use of eq. (54) to obtain the final result above. The sum over n can now beperformed. The end result is

|z〉 = e−|z|2/2 eza† |0〉 .

(d) Prove that the coherent state is a state of minimum uncertainty, i.e., ∆X∆P = ~/2.

We first compute the expectation values of X , X2 P and P 2 with respect to the coherentstate |z〉. From eq. (51) and its adjoint, we can solve for the operators X and P ,

X =

(

~

2m0ω

)1/2[

a+ a†]

, P = i

(

~

2m0ω

)1/2[

a† − a]

, (59)

where we have used the fact that X and P are self-adjoint operators. It then follows that

X2 =~

2m0ω(a + a†)2 =

~

2m0ω

[

a2 + (a†)2 + 2aa† + 1]

, (60)

P 2 = − ~2m0ω

(a† − a)2 = − ~2m0ω

[

a2 + (a†)2 − 2aa† − 1]

. (61)

Employing the above results and making use of,

a |z〉 = z |z〉 , 〈z| a† = 〈z| z∗ , (62)

18

it follows that

〈X〉 ≡ 〈z|X |z〉 =(

~

2mω

)1/2(

z + z∗)

,〈

X2〉

≡ 〈z|X2 |z〉 = ~2mω

[

(z + z∗)2 + 1]

,

〈P 〉 ≡ 〈z|P |z〉 = i(

h~ω

2

)1/2(

z∗ − z)

,〈

P 2〉

≡ 〈z|P 2 |z〉 = −12m~ω

[

(z∗ − z)2 − 1]

.

Using the above results, we obtain

(∆X)2 =〈

X2〉

−[

〈X〉]2

=~

2mω, (∆P )2 =

〈

P 2〉

−[

〈P 〉]2

= 12m~ω .

Hence, (∆X)2(∆P )2 = 14~2. Taking the positive square root yields

∆X∆P = 12~ ,

corresponding to a state of minimum uncertainty.

(e) Consider coherent states in which the parameter z is a real positive number muchlarger than 1. Evaluate the expectation value of the quantum Hamiltonian with respectto a coherent state with |z| ≫ 1. In what way is this state a good approximation to theclassical limit of the harmonic oscillator?

Recall that the quantum Hamiltonian of the harmonic oscillator can be written in terms ofthe raising and lowering operators as

H = ~ω(

a†a + 12

)

,

where the identity operator multiplying the factor of 12has been suppressed in the notation

above. Taking the expectation value of H with respect to the coherent state |z〉 and makinguse of eq. (62),

〈H〉 ≡ 〈z|H |z〉 = ~ω(

〈z| a†a |z〉+ 12

)

= ~ω[

|z|2 + 12

]

. (63)

In the case of |z| ≫ 1, we have〈H〉 ≃ ~ω|z|2 , for |z| ≫ 1. (64)

REMARK: One can also compute 〈H〉 starting from

H =P 2

2m+ 1

2mω2X2 ,

Using the results of part (d), it follows that

〈H〉 ≡ 〈z|H |z〉 = 12m

〈

P 2〉

+ 12m2ω2

〈

X2〉

= −14~ω

[

((z∗ − z)2 − 1]

+ 14~ω

[

(z + z∗)2 + 1]

= ~ω[

|z|2 + 12

]

,

thereby reproducing the result of eq. (63).

19

In order to appreciate why |z| ≫ 1 corresponds to the classical limit, we need toexamine the time dependence of the coherent states. Suppose that we examine the state |z〉[cf. eq. (58)] at time t = 0.

|z〉 ≡ |z, t = 0〉 = e−|z|2/2∞∑

n=0

zn√n!

|n〉 .

Then at time t,

|z, t〉 = e−|z|2/2∞∑

n=0

zn√n!e−iEnt/~ |n〉 ,

where En =1

2~ω(n + 1

2) are the eigenvalues of the harmonic oscillator Hamiltonian. That

is,

|z, t〉 = e−|z|2/2e−iωt/2∞∑

n=0

(ze−iωt)n√n!

|n〉 = e−iωt/2∣

∣ze−iωt〉

. (65)

The wave function corresponding to the coherent state in the x-representation is

ψz(x, t) ≡ 〈x|z, t〉 = e−iωt/2〈

x|ze−iωt〉

. (66)

The wave function 〈x|z〉 was previously obtained in eq. (52). We can now evaluate Cz byrequiring that 〈z|z〉 = 1. Using eq. (53), it follows that

Cz =(mω

π~

)1/4

exp[

−14(z + z∗)2

]

,

where we have (conventionally) set the arbitrary phase of Cz to unity. Inserting this backinto eq. (52) yields

〈x|z〉 =(mω

π~

)1/4

exp

[

−mω2~

x2 +

(

2mω

~

)1/2

zx− 14(z + z∗)2

]

. (67)

Replacing z with ze−iωt in eq. (67) and using the resulting expression in eq. (66) yields

ψz(x, t) = e−iωt/2

(mω

π~

)1/4

exp

[

−mω2~

x2 +

(

2mω

~

)1/2

xze−iωt − 14(ze−iωt + z∗eiωt)2

]

.

We now evaluate the squared magnitude of ψ(z, t) which yields the probability densityof the coherent state wave packet. Remarkably, it takes on a very simple form,

|ψz(x, t)|2 =(mω

π~

)1/2

exp[

−mω~

(x− xc)2]

, (68)

where

xc ≡(

~

2mω

)1/2(

ze−iωt + z∗eiωt)

. (69)

20

We see that the center of the wave packet, x = xc, oscillates with the frequency of theharmonic oscillator. But the shape of the wave packet is time-independent! In particular,this wave packet does not spread. Indeed, we can easily repeat the calculation of part (d)by taking expectation values with respect to |z, t〉. The end result is the same, namely∆X∆P = 1

2~, independently of the time t. Thus, the coherent state wave packet is a

minimum uncertainty wave packet for all times t.Using the results of part (d), we can easily work out the expectation values of X and P

with respect to |z, t〉,

〈X〉 ≡ 〈z, t|X |z, t〉 =(

2~

mω

)1/2

Re(ze−iωt) , (70)

〈P 〉 ≡ 〈z, t|P |z, t〉 =(

2m~ω)1/2

Im(ze−iωt) . (71)

It is convenient to write z = |z|eiδ. The above equations then take the form,

〈X〉 = A cos(ωt− δ) , 〈P 〉 = −mωA sin(ωt− δ) , (72)

where

A ≡(

2~

mω

)1/2

|z| . (73)

As expected from Ehrenfest’s theorem, the mean position and momentum of the coherentstate wave packet follows precisely the position and momentum of the classical harmonicoscillator as a function of time.4 Moreover, for |z| ≫ 1, we have 〈H〉 = ~ω|z|2 [cf. eq. (64)],which can be rewritten as

〈H〉 ≃ 12mω2A2 , for |z| ≫ 1,

which is the expected behavior of the classical harmonic oscillator (where quantum fluctu-ations are negligible).5

Hence, the coherent state wave packet follows the classical motion of the harmonicoscillator without changing its form. Moreover as suggested above, the relative size of thequantum fluctuations vanish in the limit of |z| → ∞. For example, using eq. (65) and theresults of part (d), we easily compute

〈

X2〉

= A2 cos2(ωt− δ) + ~2mω

,〈

P 2〉

= m2ω2A2 sin2(ωt− δ) + 12~mω .

It then follows that6

(∆X)2 =〈

X2〉

−[

〈X〉]2

=~

2mω, (∆P )2 =

〈

P 2〉

−[

〈P 〉]2

= 12~mω .

4This should be compared with the expectation values of X and P with respect to the energy eigen-states |n〉. Indeed, in contrast to eqs. (70) and (71), 〈n|X |n〉 = 〈n|P |n〉 = 0 since X and P are parity-oddoperators and the |n〉 are states of definite parity. Moreover, these expectation values are time-independentsince the |n〉 are stationary states.

5For the classical variables, H = P 2/(2m)+ 12mω2X2 = 1

2mω2A2

[

cos2(ωt−δ)+sin2(ωt−δ)]

= 12mω2A2.

6As asserted below eq. (69), one immediately obtains ∆X∆P = 12~, independently of the time t.

21

Comparing ∆X to the amplitude A of harmonic motion [given by eq. (73)],

∆X

A=

1

2|z| ≪ 1 ,

in the limit of |z| ≫ 1. Likewise,

∆P

mωA=

1

2|z| ≪ 1 .

Hence, the quantum fluctuations are suppressed in the limit of |z| ≫ 1.A similar conclusion can be obtained by computing the root-mean square of the Hamil-

tonian.

〈

H2〉

= 〈z|H2 |z〉 = 〈z|(

a†a+ 12

)2 |z〉 = ~2ω2 〈z|(

a†a†aa + 2a†a+ 14

)

|z〉

= ~2ω2[

|z|4 + 2|z|2 + 14

]

=[

〈H〉]2

+ ~2ω2|z|2 ,

after making use of [a , a†] = 1 and employing eq. (63). Hence, it follows that

∆H2 =〈

H2〉

−[

〈H〉]2

= ~2ω2|z|2 ,

and∆H

〈H〉 ≃1

|z| ≪ 1 ,

in the limit of |z| ≫ 1.In conclusion, a coherent state with |z| ≫ 1 is a very good approximation to the classical

limit of the harmonic oscillator.

22