763312A QUANTUM MECHANICS I - Solution set 6 - Autumn 2016
1. Imagine a bead of mass m that slides frictionlessly around a circular wire
ring of circumference L. Find the stationary states (with appropriate
normalization) and the corresponding allowed energies. Note that each
energy En is degenerate; corresponding to clockwise and counter-clockwise
circulation. How do you account for this degeneracy in terms of Problem
4 of Exercise 5?
Solution
The problem can be treated as a one dimensional, position of the bead
x is measured around the ring. Governing equation is the S.E. for a free
particle
−
~2 d2 ψ
= Eψ.
2m dx2
(1.1)
Rearranging gives
√
00
2
ψ = −k ψ;
k=
2mE
> 0.
~
(1.2)
General solution of this equation is
ψ(x) = Aeikx + Be−ikx ,
(1.3)
where A and B are constants. Since the bead is moving on a wire of
circumference L, the points x and x + L are the same for all x, that is
∀x.
ψ(x) = ψ(x + L)
(1.4)
Especially for x = 0 we have
A + B = AeikL + Be−ikL ,
(1.5)
and for x = π/(2k)
Aeiπ/2 + B −iπ/2
⇔ i(A − B)
⇔A−B
= Aei(π/2+kL) + B −i(π/2+kL)
= i(Ae
= Ae
ikL
ikL
+ Be
− Be
−ikL
−ikL
)
.
(1.6)
(1.7)
(1.8)
Now add (1.5) and (1.8) together to obtain
2A = 2AeikL .
(1.9)
That is, either A = 0 or eikL = 1. The latter condition imposes a restriction to k
eikL = 1 ⇔ kL = 2πn ⇔ k =
1
2πn
, n ∈ N.
L
(1.10)
If A = 0 we have
B = Be−ikL ,
(1.11)
and we get the same conditions as before
B=0
or k =
2πn
, n ∈ N.
L
(1.12)
This means that for every positive n we get two solutions (for n = 0 only
1 solution)
ψn+ (x) = Aei2πnx/L
and ψn− (x) = Be−i2πnx/L .
(1.13)
Normalization of the states (now the space of the particle is just the interval [0, L]):
L
Z
|ψn± (x)|2 dx = |N |2
1=
0
Z
L
dx = |N |2 L
(1.14)
0
that is
1
N=√ ,
L
(1.15)
where N = A, B, and we chose it to be real and positive.
The (degenerate) energy levels are
En =
~2 kn2
2~2 π 2 n2
=
, n ∈ N.
2m
mL2
(1.16)
Any other state with energy E can be formed as a linear combination of
ψn± (x).
The theorem of Ex.5.4 fails since now ψ(x) doesn’t go to zero at infinity, and we can’t determine the constant C in the proof.
2
2. Consider a quantum particle in the ”step” potential:
(
0,
x ≤ 0,
V (x) =
V0 , x > 0.
(2.1)
Assume that the particle has energy E > V0 .
a) Calculate the reflection coefficient.
b) Because the potential does not go to zero on the right-hand side
of the barrier, the transmitted wave has a different speed than the
incident wave. In such cases the transmission coefficient is not simply T 6= |F |2 /|A|2 , where A is the incident amplitude and F is the
transmitted amplitude. For instance, show that for potential step
the transmission coefficient is
r
E − V0 |F |2
T =
.
(2.2)
E |A|2
d) Show that R + T = 1.
Solution
a) First we need to solve the S.E. for the particle moving in this potential.
In the region x ≤ 0 S.E. is
√
00
2mE
ψ = −k 2 ψ;
k=
(2.3)
~
and the solution is
ψ(x) = Aeikx + Be−ikx ,
(2.4)
where A and B are constants.
When x > 0 S.E. reads
00
p
2
ψ = −l ψ;
l=
2m(E − V0 )
.
~
(2.5)
If the particle is originally coming from the negative x direction there is
no wave propagating to negative x direction in the region x > 0. Thus
the solution reads
ψ(x) = F eilx .
(2.6)
From the continuity of the wave function at x = 0 we get
A + B = F,
(2.7)
and from the continuity of its derivative
ik(A − B) = ilF.
3
(2.8)
Combine and simplify to obtain
k
k
= −B 1 +
.
A 1−
l
l
(2.9)
The reflection coefficient is defined as the ratio between the amplitude of
the reflected wave (part of the wave function proportional to e−ikx ) and
the incident wave (part of the wave function proportional to eikx ). Thus,
the reflection coefficient becomes
R=
(1 − k/l)2
(k − l)2
(k − l)4
|B|2
=
=
.
= 2
2
2
2
|A|
(1 + k/l)
(k + l) )
(k − l2 )2
Substituting the definitions of k and l gives
√
√
( E − E − V 0 )4
R=
.
V02
(2.10)
(2.11)
b) Transmission coefficient can be determined from the probability currents
in the regions x ≤ 0 and x > 0. It is defined as the ratio between probability current of the transmitted wave and incident wave.
Probability current is defined as
~
∗ dψ
,
S(x) = = ψ
m
dx
(2.12)
where = stands for the imaginary part. For the incident wave it is
~k 2
~
~ ∗ −ikx d
ikx
Si (x) = = A e
(Ae ) = = A∗ e−ikx ikAeikx =
|A|
m
dx
m
m
(2.13)
and for the transmitted wave (∝ eilx )
~l
d
~ ~
St (x) = = F ∗ e−ilx (F eilx ) = = F ∗ e−ilx ilF eilx = |F |2 .
m
dx
m
m
(2.14)
Thus the transmission coefficient is
l |F |2
St (x)
=
=
T =
Si (x)
k |A|2
r
E − V0 |F |2
.
E |A|2
(2.15)
c) From b) we have that
1 − k/l
F =A+B =A 1−
1 + k/l
4
=A
2k
,
k+l
(2.16)
thus,
T =
l 4k 2
4kl
=
.
2
k (k + l)
(k + l)2
(2.17)
So we have
R+T =
(k − l)2
4kl
k 2 + l2 − 2kl + 4kl
(k + l)2
+
=
=
= 1.
2
2
2
(k + l)
(k + l)
(k + l)
(k + l)2
(2.18)
5
3. A particle scatters from a potential step of height V0 . The energy of the
particle E < V0 .
a)
b)
c)
d)
Show that the probability current inside the potential step vanishes.
Show that total reflection happens.
Estimate the penetration depth by evaluating hxi inside the step.
Show that there is a phase shift between the incoming and the scattered wave.
Solution
a) Let us solve the S.E. for the particle moving in this potential.
In the region x ≤ 0 S.E. the solution is the same as in Problem 2
ψ(x) = Aeikx + Be−ikx ,
(3.1)
where A and B are constants.
In the region x > 0 S.E. reads
00
p
2m(V0 − E)
κ=
>0
~
2
ψ = κ ψ;
(3.2)
and the solution is
ψ(x) = Deκx + F e−κx ,
but D = 0 since e
κx
(3.3)
→ ∞ as x → ∞, so
ψ(x) = F e−κx .
Now we can evaluate the probability current inside the step:
~
~ d
S = = F ∗ e−κx (F e−κx ) = = |F |2 e−2κx = 0.
m
dx
m
(3.4)
(3.5)
b) As calculated above the probability current inside the potential step vanishes inside the potential step, i.e. there is no transmitted wave. This
of course means that there must be total reflection. Another way is to
calculate the reflection coefficient and show that it is equal to unity.
Take the solutions in the regions x ≤ 0 and x > 0 from a) and demand the
wave function ψ(x) and its derivative tot be continuous at x = 0. From
this we get two conditions
A+B =F
and ik(A − B) = −κF.
(3.6)
Combine the above to obtain
A+B
ik
⇔A 1+
κ
6
ik
(A − B)
κ
ik
= B
−1 .
κ
= −
(3.7)
(3.8)
Now we can calculate the reflection coefficient
ik/κ − 1 1 + (k/κ)2
|B|2
=
=
= 1,
R=
|A|2
ik/κ + 1 1 + (k/κ)2
(3.9)
as it should.
c) To estimate the penetration depth we assume that the particle is inside
the potential step and calculate the expectation value of x. From a) we
get the wave function in region x > 0 ψ(x) = F e−κx . First we need to
normalize ψ:
∞
Z ∞
Z ∞
|F |2
e−2κx 1=
|ψ(x)|2 dx = |F |
=
e−2κx dx = −|F |2
, (3.10)
2κ 0
2κ
0
0
thus,
F =
Now we can evaluate hxi:
Z ∞
Z
hxi =
x|ψ(x)|2 dx = 2κ
0
√
2κ.
∞
xe−2κx dx =
0
(3.11)
~
1
= p
.(3.12)
2κ
2 2m(V0 − E)
d) In a) we found the incoming wave to be
ψi (x) = Aeikx
(3.13)
ψs (x) = Be−ikx .
(3.14)
and the scattered wave
We also obtained the relation
B=
ik/κ − 1
κ + ik
k − iκ
A=
A=
A = Ae−2iθ ,
ik/κ + 1
ik − κ
k + iκ
(3.15)
where tan θ = κ/k. With this the scattered wave can be written like
ψs (x) = Ae−2iθ e−ikx = Ae−i(kx+2θ) ,
i.e. there is a phase difference of 2θ between ψi and ψs .
7
(3.16)
4. The theory of scattering can be generalized for localized potentials of form
(
V0 (x), −a ≤ x ≤ a,
V (x) =
(4.1)
0,
|x| ≥ a,
where 2a is the width of the potential. The solution of the general localized
potential differs from that of the delta-potential only in the region where
|x| ≤ a, where solution is of form
ψ(x) = Cf (x) + Dg(x).
(4.2)
Above, f (x) and g(x) are linearly independent particular solutions, whose
form depends on the exact form of the potential V0 (x). One ca proceed
similarly as with delta-potential and eliminate C and D, resulting in
B = S11 A + S12 G,
F = S21 A + S22 G,
(4.3)
or
B
S11
=
F
S21
S12
S22
A
.
G
(4.4)
This defines the so-called scattering matrix S. In the case of scattering
from left, G = 0 and
|F |2 |B|2 2
= |S11 | , Tl =
= |S21 |2 .
(4.5)
Rl =
|A|2 G=0
|A|2 G=0
In the case of scattering from right, A = 0 and
|F |2 |B|2 2
Rr =
=
|S
|
,
T
=
= |S12 |2 .
22
r
|G|2 A=0
|G|2 A=0
(4.6)
a) Construct the S-matrix for the scattering from delta-potential well.
b) Construct the S-matrix for the finite square well.
Solution
a) For the delta-potential well the scattering states are (lecture notes pages
20 and 21)
(
Aeikx + Be−ikx , x < 0
ψ(x) =
(4.7)
F eikx + Ge−ikx , x > 0.
From the continuity conditions one gets (lecture notes p. 20)
F +G =
A+B
(4.8)
F −G =
A(1 + 2iβ) − B(1 − 2iβ),
(4.9)
8
where β = mα/(~2 k). Subtract (4.9) from (4.8) to obtain
2G = −2iβA + B(2 − 2iβ) ⇔ B =
G + iβA
.
1 − iβ
(4.10)
Next we multiply (4.8) by 1 − 2iβ and add it to (4.9) and obtain
2A = 2(1 − iβ)F − 2iβG ⇔ F =
A + iβG
.
1 − iβ
(4.11)
Compare the above results to
B = S11 A + S12 G,
F = S21 A + S22 G.
One sees that the scattering matrix is
1
iβ
S=
1 − iβ 1
1
.
iβ
(4.12)
(4.13)
b) For an even potential scattering from the right is the same as scattering
from the left with substitutions
x ↔ −x,
A ↔ G,
B ↔ F.
(4.14)
F = S21 A + S22 G
(4.15)
B = S21 G + S22 A.
(4.16)
Thus, for right scattering we have
B = S11 A + S12 G,
and for left scattering
F = S11 G + S12 A,
Combine these to obtain
S11 = S22
and S12 = S21 .
(4.17)
Finite potential well is even, so we can apply the above for it. The case
of left scattering is solved in the lecture notes. So we have to just collect
the results. In the lecture notes at page 20 it is shown that
−1
l2 + k 2
F = e−2ika cos(2la) − i
sin(2la)
A.
(4.18)
2kl
From the eqs. (172) on page 22 of the lecture notes one can solve for B.
One obtains
sin(2la) 2
B=i
(l − k 2 )F.
(4.19)
2kl
Now by comparing these to (4.16), and remembering that for left scattering
G = 0, one sees that the scattering matrix reads
!
e−2ika
i sin(2la)
(l2 − k 2 )
1
2kl
S=
.
2 +k 2
2
2
1
i sin(2la)
cos(2la) − i l 2kl
sin(2la)
2kl (l − k )
(4.20)
9