Solutions to Problems in Merzbacher,
Quantum Mechanics, Third Edition
Homer Reid
June 24, 2000
Chapter 6
Problem 6.1
Obtain the transmission coefficient for a rectangular potential barrier of width 2a
if the energy exceeds the height V0 of the barrier. Plot the transmission coefficient
as a function E/V0 (up to E/V0 = 3), choosing (2ma2 V0 )1/2 = (3π/2)~.
In the text, Merzbacher treats this problem for the case where the particle’s
energy is less than the potential barrier. He obtains the result
i
M11 = cosh 2κa + sinh 2κa e2ika
(1)
2
where
κ=
and
r
2m(V0 − E)
~2
κ k
− .
(2)
k
κ
We can re-use the result (1) for the case where the energy is greater than
the potential barrier. To do this we note that κ becomes imaginary in this case,
and we write
r
2m(E − V0 )
κ = iβ = i
~2
so that (2) becomes
iβ
k
k
β
=
≡ iλ
−
=i
+
k
iβ
k
β
=
1
Homer Reid’s Solutions to Merzbacher Problems: Chapter 6
Plugging into (1) and noting that cosh ix = cos x, sinh ix = i sin x we have
iλ
sin 2βa e2ika .
M11 = cos 2βa −
2
and
1/2
λ2
2
|M11 | = cos 2βa +
sin βa
4
2
1/2
λ
= 1+
− 1 sin2 βa
4
1/2
4
1
β + k4
2
sin
βa
−
= 1+
4β 2 k 2
2
1/2
2
2 2
(β − k )
2
sin βa
= 1+
4β 2 k 2
1/2
V02
2
= 1+
sin βa
4E(E − V0 )
1/2
1
= 1+
sin2 βa
4γ(γ − 1)
1/2
4γ(γ − 1) + sin2 βa
=
4γ(γ − 1)
2
where γ = E/V0 .
We have
1/2
2m
a
(E
−
V
)
0
h2
1/2
2mV0 a2
=
(γ − 1)1/2
h2
3π
(γ − 1)1/2
=
2
βa =
so the transmission coefficient is
"
#
1
4γ(γ − 1)
T =
=
1/2
|M11 |2
4γ(γ − 1) + sin2 3π
2 (γ − 1)
This is plotted in Figure 1.
2
3
Homer Reid’s Solutions to Merzbacher Problems: Chapter 6
1
0.9
0.8
Transmission Coefficient
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.5
1
1.5
2
E/V0
2.5
3
3.5
Figure 1: Transmission coefficient versus E/V0 for Problem 6.1.
4
4
Homer Reid’s Solutions to Merzbacher Problems: Chapter 6
Problem 6.2
Consider a potential V = 0 for x > a, V = −V0 for a ≥ x ≥ 0, and V = +∞
for x < 0. Show that for x > a the positive energy solutions of the Schrödinger
equation have the form
ei(kx+2δ) − e−ikx
Calculate the scattering coefficient |1 − e2iδ |2 and show that it exhibits maxima
(resonances) at certain discrete energies if the potential is sufficiently deep and
broad.
We have
with

0,


Ψ(x) = Aeik1 x + Be−ik1 x ,

 ik2 x
Ce
+ De−ik2 x ,
x ≤0
0 ≤x ≤
a
a≤
x
r
r
2m(E + V0 )
2mE
k2 =
.
k1 =
~2
~2
Applying the requirement that Ψ be continuous at x = 0, we see we must take
A = −B, so Ψ(x) = γ sin k1 x for 0 ≤ x ≤ a. The other standard requirement,
that the derivative of Ψ also be continuous, does not hold at x = 0 because the
potential is infinite there. Hence γ is undetermined as yet. Eventually, we could
apply the normalization condition on Ψ to find γ if we wanted to.
Next applying continutity of Ψ and its derivative at x = a, we obtain
γ sin k1 a = Ceik2 a + De−ik
2
a
2
k1 γ cos k1 a = ik2 [Ceik2 a − De−ik a ]
Combining these yields
1 −ik2 a k1
γe
cos k1 a + i sin k1 a
2i
k2
1
k1
D = − γe+ik2 a
cos k1 a − i sin k1 a
2i
ik2
C=
(3)
(4)
It’s now convenient to write
k1
cos k1 a + i sin k1 a = αeiδ
k2
where
α2 =
k1
k2
2
cos2 k1 a + sin2 k1 a
(5)
(6)
Homer Reid’s Solutions to Merzbacher Problems: Chapter 6
and
δ = tan
−1
k2
tan k1 a
k1
5
so that α contains magnitude information, while δ represents phase information.
Then we can rewrite (3) and (4) as
γα −ik2 a iδ
e
e
2i
γα
D = − e+ik2 a e−iδ
2i
C=
Then the expression for the wavefunction to the left of x = a becomes
Ψ(x) = Ceik2 x + De−ik2 x
(x > a)
i
γα h ik2 (x−a) iδ
−ik2 (x−a) −iδ
e
e −e
e
=
2i
i
γα −iδ h ik2 (x−a) 2iδ
e
=
e
e − e−ik2 (x−a) .
2i
Using (5) and (6), the scattering coefficient is
2
2
2
k1
k1
2
cos
k
a
sin
k
a
−
sin
k
a
cos
k
a
+
2i
1
1
1
1
k2
k2
2iδ 2
|1 − e | = 1 −
2
2
k1
cos2 k1 a + sin k1 a
k2
2
h
i
2 sin k1 a sin k1 a − i kk1 cos k1 a 2
= 2
k1
cos2 k1 a + sin2 k1 a k2
We have
=
4 sin2 k1 a
k1
k2
2
2
(7)
cos2 k1 a + sin2 k1 a
E + V0
1
= 1+
E
λ
r
2ma2 (E + V0 )
k1 a =
= β(λ + 1)1/2
~2
k1
k2
=
where λ = E/V0 and β = (2ma2 V0 /~2 )1/2 in Merzbacher’s notation. Then the
scattering coefficient (7) is
scattering coefficient =
(1 +
1 2
λ)
4 sin2 [β(λ + 1)1/2 ]
cos2 [β(λ + 1)1/2 ] + sin2 [β(λ + 1)1/2 ]
In Figure 6.2 I have plotted this for β = 25.
6
Homer Reid’s Solutions to Merzbacher Problems: Chapter 6
4
3.5
Scattering coefficient
3
2.5
2
1.5
1
0.5
0
0
0.5
1
1.5
E/V0
2
2.5
Figure 2: Scattering coefficient versus E/V0 for Problem 6.2.
3
Homer Reid’s Solutions to Merzbacher Problems: Chapter 6
7
Problem 6.3
A particle of mass m moves in the one-dimensional double well potential
V (x) = −gδ(x − a) − gδ(x + a).
If g > 0, obtain transcendental equations for the bound-state energy eigenvalues of
the system. Compute and plot the energy levels in units of ~2 /ma2 as a function of
the dimensionless parameter mag/~2 . Explain the features of the plot. In the limit
of large separation, 2a, between the wells, obtain a simple formula for the splitting
∆E between the ground state (even parity) energy level, E+ , and the excited (odd
parity) energy level, E− .
In this problem, we can divide the x axis into three regions. In each region,
the wavefunction is just the solution to the free-particle Schrödinger equation,
but with energy E < 0 since we’re looking for bound states. Putting k =
p
2mE/~2 , we have

x
≤ −a
 Aekx + Be−kx ,
Cekx + De−kx ,
−a ≤ x ≤ a
Ψ(x) =

Eekx + F e−kx ,
a ≤ x.
Now, first of all, the wavefunction can’t blow up at infinity, so B = E = 0.
Also, since the potential in this problem has mirror-reversal symmetry, the
wavefunction will have definite parity. Considering first the even parity solution,

x
≤ −a
 Aekx ,
B cosh(kx), −a ≤ x ≤ a
Ψ(x) =
(8)

Ae−kx ,
a ≤ x.
Matching the value of the wavefunction at x = −a gives
Ae−ka = B cosh(−ka).
(9)
Since the potential becomes infinite at x = −a, the normal derivative-continuity
condition doesn’t hold there. Instead, we can write down the Schrödinger equation,
d2
2m
2m
Ψ(x) = 2 V (x)Ψ(x) − 2 EΨ(x),
2
dx
~
~
then integrate from −a − to −a + and take the limit as → 0. This gives
−a+
dΨ 2mg
= − 2 Ψ(−a).
(10)
dx −a−
~
Applying this condition to the wavefunction (8) yields
kB sinh(−ka) − kAe−ka = −
2mg
B cosh(−ka).
~2
Homer Reid’s Solutions to Merzbacher Problems: Chapter 6
8
Substituting from (9),
kB sinh(−ka) − kB cosh(−ka) = −
2mg
B cosh(−ka)
~2
or
tanh(ka) =
β
2mg
−1=
−1
~2 k
ka
with β = 2mag/~2 . This equation determines the energy eigenvalue of the evenparity state, which will be the ground state. On the other hand, the odd parity
state looks like

x
≤ −a
 Aekx ,
B sinh(kx), −a ≤ x ≤ a
(11)
Ψ(x) =

−Ae−kx ,
a ≤ x.
Matching values at x = −a gives
Ae−ka = B sinh(−ka)
and applying condition (10) gives
2mg
B sinh(−ka)
~2
2mg
kB cosh(ka) + kB sinh(ka) = 2 B sinh(ka)
~
β
−1
coth(ka) =
ka
kB cosh(−ka) − kAe−ka = −
so this is the condition that determines the energy of the odd parity state.
In Figure (3) I have plotted tanh(ka), coth(ka), and β/(ka) − 1 for the case
β = 3. As expected, the coth curve crosses the β/(ka) − 1 curve at a lower value
of ka than the tanh curve; that means that the energy eigenvalue for the odd
parity state is smaller in magnitude (less negative) than the even parity state.
Problem 6.4
Problem 3 provides a primitive model for a one-electron linear diatomic molecule
with interatomic distance 2a = |X|, if the potential energy of the “molecule” is
taken as E± (|X|), supplemented by a repulsive interaction λg/|X| between the wells
(“atoms”). Show that, for a sufficiently small value of λ, the system (“molecule”)
is stable if the particle (“electron”) is in the even parity state. Sketch the total
potential energy of the system as a function of |X|.
9
Homer Reid’s Solutions to Merzbacher Problems: Chapter 6
PSfrag replacements
2
tanh(ka)
coth(ka)
β/(ka) − 1
1.5
1
0.5
0
1
1.2
1.4
1.6
1.8
ka
Figure 3: Graphical determination of energy levels for Problem 6.3 with β = 3.
2
10
Homer Reid’s Solutions to Merzbacher Problems: Chapter 6
Problem 6.5
If the potential in Problem 3 has g < 0 (double barrier), calculate the transmission
coefficient and show that it exhibits resonances. (Note the analogy between the
system and the Fabry-Perot étalon in optics.)
Now we’re assuming that the energy E is positive, so
 ikx
−ikx
x ≤ −a

 Ae + Be
ikx
−ikx
Ψ(x) = Ce + De
−a≤x≤a

 ikx
−ikx
Ee + F e
a≤x
with k =
p
2mE/~2 . Matching values at x = −a, we have
Ae−ika + Beika = C −ika + Dika
(12)
(13)
Also, as before, we have the derivative condition
x=−a+
dΨ 2mg
= − 2 Ψ(−a)
dx x=a−
~
where g is now negative. Applying this to the wavefunction in (12), we have
ik[Ce−ika − Deika − Ae−ika + Beika ] = −
2mg
[Ae−ika + Beika ].
~2
Combining (13) and (14) yields
β 2ka
β
e B
A+
C = 1+
ika
ika
β −2ka
β
D=−
B
e
A+ 1−
ika
ika
(14)
(15)
(16)
with β = mag/~2 as before.
Now, applying the matching conditions to the wavefunction at x = +a will
give two equations exactly like (13) and (14), but with the substitutions A → C,
B → D, C → E, D → F , and a → −a. Making these substitutions in (15) and
(16) we obtain
β
β 2ka
e D
(17)
E = 1−
C−
ika
ika
β 2ka
β
F =+
D
(18)
e C + 1+
ika
ika
PSfrag replacements
11
Homer Reid’s Solutions to Merzbacher Problems: Chapter 6
1
0.9
0.8
0.7
T (k)
0.6
0.5
0.4
0.3
0.2
0.1
0
0
5
10
15
20
25
ka
Figure 4: Transmission coefficient in Problem 6.4 with β = 15.
Combining equations (15) through (18), we have
#
"
2
2β
β
β
−4ika
(e
− 1) A + +
E = 1+
1−
sin 2ka B
ika
ka
ika
#
"
2
2β
β
β
4ika
(e
− 1) B
F =
1+
sin 2ka A + 1 +
ka
ika
ika
This is the M matrix, and the transmission coefficient is given by T = 1/|M11 |2 ,
or
1
T =
2 2
β
β
+ 1 (1 − cos(4ka))
1 + 2 ka
ka
In Figure 4 I have plotted this for β = 15.
30
Homer Reid’s Solutions to Merzbacher Problems: Chapter 6
12
Problem 6.6
A particle moves in one dimension with energy E in the field of a potential defined
as the sum of a Heaviside step function and a delta function:
V (x) = V0 η(x) + gδ(x)
(with V0 and g > 0)
The particle is assumed to have energy E > V0 .
(a) Work out the matrix M , which relates the amplitudes of the incident and
reflected plane waves on the left of the origin (x < 0) to the amplitudes on
the right (x > 0).
(b) Derive the elements of the matrix S, which relates incoming and outgoing
amplitudes.
(c) Show that the S matrix is unitary and that the elements of the S matrix satisfy
the properties expected from the applicable symmetry considerations.
(d) Calculate the transmission coefficient for particles incident from the right and
for particles incident from the left, which have the same energy (buf different
velocities).
We have
Ψ(x) =
with
(
r
Aeik1 x + Be−ik1 x ,
Ce
ik2 x
2m
E
k1 =
~2
Matching values at x = 0 gives
+ De
k2 =
−ik2 x
r
,
x≤0
x≥0
2m
(E − V0 ).
~2
C +D =A+B
(19)
Also, the delta function at the origin gives rise to a discontinuity in the derivative
of the wavefunction as before:
0+
dΨ 2mg
= 2 Ψ(0)
dx 0−
~
so
ik2 (C − D) − ik1 (A − B) =
or
C −D =
2mg
(A + B)
~2
k1
2mg
(A − B) +
(A + B).
k2
ik2 ~2
(20)
Homer Reid’s Solutions to Merzbacher Problems: Chapter 6
13
Adding and subtracting (19) and (20), we can read off
1
2mg
2mg
1
k1
k1
A+
B
+
+
C=
1+
1−
2
k2
ik2 ~2
2
k2
ik2 ~2
2mg
2mg
1
1
k1
k1
−
−
D=
A 1+
B.
1−
2
k2
ik2 ~2
2
k2
ik2 ~2
We could also write this as
C
M11
=
∗
D
M12
M12
∗
M11
A
B
Or instead of the M matrix we could use the S matrix, which is defined by
A
S11 S12
B
=
S21 S22
D
C
Since we already know the M coefficients, we can calculate the elements of the
S matrix from the formula
S11
S21
M∗
= − M12
∗
11
1
= M∗
11
S12 = M1∗
11
M12
S22 = + M
∗
11
However, this is tedious and long and boring and I don’t want to do it.