Solutions to Problems in Merzbacher,
Quantum Mechanics, Third Edition
Homer Reid
November 20, 1999
Chapter 2
Problem 2.1
A one-dimensional initial wave packet with a mean wave number
kx and a Gaussian amplitude is given by
x2
+
ik
x
.
Ψ(x, 0) = C exp −
x
4(∆x)2
Calculate the corresponding kx distribution and Ψ(x, t), assuming
free particle motion. Plot |Ψ(x, t)|2 as a function of x for several
values of t, choosing ∆x small enough to show that the wave packet
spreads in time, while it advances according to the classical laws.
Apply the results to calculate the effect of spreading in some typical
microscopic and macroscopic experiments.
The first step is to compute the Fourier transform of Ψ(x, 0) to find the
distribution of the wave packet in momentum space:
Φ(k) = (2π)−1/2
Z
∞
Ψ(x, 0)e−ikx dx
x2
+
i(k
−
k)x
dx
= (2π)−1/2 C
exp −
0
4(∆x)2
−∞
−∞
Z ∞
1
(1)
(I have dropped the x subscripts, and I write k0 instead of k).
To proceed we need to complete the square in the exponent:
−
x2
+ i(k0 − k)x
4(∆x)2
x2
2
2
2
2
−
i(k
−
k)x
−
(k
−
k)
(∆x)
+
(k
−
k)
(∆x)
0
0
0
4(∆x)2
2
x
− i(k0 − k)∆x − (k0 − k)2 (∆x)2
= −
2(∆x)
2
1
x − 2i(k0 − k)(∆x)2 − (k0 − k)2 (∆x)2
(2)
= −
2
4(∆x)
= −
Now we plug (??) into (??) to find:
Φ(k) = (2π)−1/2 C exp[−(k0 −k)2 (∆x)2 ]
Z
∞
−∞
exp −
1
2 2
]
[x
−
2i(k
−
k)(∆x)
dx
0
4(∆x)2
In the integral weR can make the shift x → x − 2i(k0 − k)(∆x)2 and use the
∞
standard formula −∞ exp(ax2 )dx = (π/a)1/2 . The result is
Φ(k) =
√
2C∆x exp −(k0 − k)2 (∆x)2
To put this into direct correspondence with the form of the wave packet in
configuration space, we can write
√
(k0 − k)2
Φ(k) = 2C∆x exp −
4(∆k)2
where ∆k = 1/(2∆x). This is the minimum possible k width attainable for
a wave packet with x width ∆x, which is why the Gaussian wave packet is
sometimes referred to as a minimum uncertainty wave packet.
The next step is to compute Ψ(x, t) for t > 0. Since we are talking about
a free particle, we know that the momentum eigenfunctions are also energy
eigenfunctions, which makes their time evolution particularly simple to write
down. In the above work we have expressed the initial wave packet Ψ(x, 0)
as a linear combination of momentum eigenfunctions, i.e. as a sum of terms
exp(ikx), with the kth term weighted in the sum by the factor Φ(k). The wave
packet at a later time t > 0 will be given by the same linear combination, but
now with the kth term multiplied by a phase factor exp[−iω(k)t] describing its
time evolution. In symbols we have
Z ∞
Ψ(x, t) =
Φ(k)ei[kx−ω(k)t] dk.
−∞
For a free particle the frequency and wave number are connected through
ω(k) =
2
h̄k 2
.
2m
Using our earlier expression for Φ(k), we find
Ψ(x, t) =
Z
√
2C∆x
h̄k 2
t dk.
exp −(k0 − k)2 (∆x)2 + ikx − i
2m
−∞
∞
(3)
Again we complete the square in the exponent:
h̄k 2
ih̄
−(k0 − k)2 (∆x)2 + ikx − i
t = − [(∆x)2 +
t]k 2 − [2k0 (∆x)2 + ix]k + k02 (∆x)2
2m
2m
2 2
= − α k − βk + γ
β
β2
β2
= −α2 k 2 − 2 k + 4 − 4 − γ
α
4α
4α
2
β
β
= −α2 [k − 2 ]2 + 2 − γ
2α
4α
where we have defined some shorthand:
α2 = (∆x)2 +
ih̄
t
2m
β = 2k0 (∆x)2 + ix
γ = k02 (∆x)2 .
Using this in (??) we find:
√
Ψ(x, t) = 2C∆x exp
β2
−γ
4α2
Z
β 2
exp −α (k − 2 ) dk.
2α
−∞
∞
2
The integral evaluates to π 1/2 /α. We have
Ψ(x, t) =
=
√
2
1
β
2πC∆x
exp
−γ
α
4α2
"
#1/2
#
"
√
(∆x)2
(ix + 2k0 (∆x)2 )2
−k02 (∆x)2
2πC
e
exp
ih̄
ih̄
(∆x)2 + 2m
t
4[(∆x)2 + 2m
t]
This is pretty ugly, but it does display the relevant features. The important
point is that term ih̄t/2m adds to the initial uncertainty (∆x)2 , so that the
wave packet spreads out with time.
In the figure, I’ve plotted this function for a few values of t, with the following parameters: m=940 Mev (corresponding to a proton or neutron), ∆x=3 Å,
k0 =0.8 Å−1 . This value of k0 corresponds, for a neutron, to a velocity of about
5 · 104 m/s; and note that, sure enough, the center of the wave packet travels
about 5 nm in 100 fs. The time scale of the spread of this wave packet is ≈ 100
fs.
3
Gaussian wave packet example
350
t=0 s
t=30 fs
t=70 fs
t=100 fs
4
Wavefunction (arbitrary units)
300
250
200
150
100
50
0
-1e-08
-5e-09
0
Distance (meters)
5e-09
1e-08
Problem 2.2
Express the spreading Gaussian wave function Ψ(x, t) obtained
in Problem 1 in the form Ψ(x, t) = exp[iS(x, t)/h̄]. Identify the
function S(x, t) and show that it satisfies the quantum mechanical
Hamilton-Jacobi equation.
In the last problem we found
Ψ(x, t)
√
2
∆x
β
=
2πC
exp
−γ
α
4α2
√
β2
∆x
+ 2 −γ
2πC
= exp ln
α
4α
(1)
where we’ve again used the shorthand we defined earlier:
α2 = (∆x)2 +
ih̄
t
2m
β = 2k0 (∆x)2 + ix
γ = k02 (∆x)2 .
From (??) we can identify (neglecting an unimportant additive constant):
β2
h̄
− ln α + 2 .
S(x, t) =
i
4α
Things are actually easier if we define δ = α2 . Then
ih̄
1
h̄
β2
2
δ = (∆x) +
− ln δ +
t
S(x, t) =
2m
i
2
4δ
Now computing partial derivatives:
∂S
∂t
=
=
∂S
∂x
=
=
∂2S
∂x2
=
h̄
1
β 2 ∂δ
− − 2
i
2δ 4δ ∂t
β2
h̄2 1
+ 2
−
2m 2δ 4δ
h̄ β ∂β
i 2δ ∂x
h̄β
2δ
ih̄
2δ
The quantum-mechanical Hamilton-Jacobi equation for a free particle is
2
1 ∂S
ih̄ ∂ 2 S
∂S
+
−
=0
∂t
2m ∂x
2m ∂x2
5
(2)
(3)
(4)
Inserting (??), (??), and (??) into this equation, we find
h̄2
h̄2 β 2
h̄2 β 2
h̄2
−
+
+
=0
4mδ 8mδ 2
8mδ 2
4mδ
so, sure enough, the equation is satisfied.
−
Problem 2.3
Consider a wave function that initially is the superposition of two
well-separated narrow wave packets:
Ψ1 (x, 0) + Ψ2 (x, 0)
chosen so that the absolute value of the overlap integral
Z +∞
γ(0) =
Ψ∗1 (x)Ψ2 (x)dx
−∞
is very small. As time evolves, the wave packets move and spread.
Will |γ(t)| increase in time, as the wave packets overlap? Justify
your answer.
It seems to me that the answer to this problem depends entirely on the
specifics of the particular problem.
One could well imagine a situation in which the overlap integral would not
increase with time. Consider, for example, the neutron wave packets plotted
in the figure from problem 2.1 If one of those wave packets were centered in
Chile and another in China, the overlap integral would be tiny, since the wave
packets only have appreciable value within a few angstroms of their centers.
Furthermore, even if the neutrons are initially moving toward each other, their
wave packets spread out on a time scale of ≈ 100 fs, long before their centers
ever come close to each other.
On the other hand, if the two neutron wave packets were each centered, say,
20 angstroms apart, then they would certainly overlap a little before collapsing
entirely.
Problem 2.4
A high resolution neutron interferometer narrows the energy spread
of thermal neutrons of 20 meV kinetic energy to a wavelength dispersion level of ∆λ/λ= 10−9 . Estimate the length of the wave
packets in the direction of motion. Over what length of time will
the wave packets spread appreciably?
6
First let’s compute the average momentum of the neutrons.
p0
= [ 2mE ]1/2
≈ [ 2 · (940Mev · c−2 ) · (20mev) ]1/2
= 6.1 kev / c
We’re given the fractional wavelength dispersion level; what does this tell us
about the momentum dispersion level?
p =
dp =
so
dp =
p
so the momentum uncertainty is
2πh̄
λ
−2πh̄
dλ
λ2
dλ = 10−9
λ
∆p = p0 · 10−9 = 6.1 · 10−6 ev.
This implies a position uncertainty of
∆x
≈
=
h̄
∆p
6.6 · 10−16 ev · s
6.1 · 10−6 ev / c
≈ c · 10−10 s
≈ 30 cm.
This is HUGE! So the point is, if we know with this precision how quickly
our thermal neutrons are moving, we have only the most rough indication of
where in the room they might be.
To estimate the time scale of spreading of the wave packets, we can imagine
that they are Gaussian packets. In this case we start to get appreciable spreading
when
or
h̄
t ≈ (∆x)2
2m
t ≈ 2m(∆x)2 /h̄
2 · 9.4 · 108 ev · (0.3 m)2
≈
c2 · 6.6 · 10−16 ev · s
≈ 32 ks ≈ 10 hours.
7