Solution set 3

763312A QUANTUM MECHANICS I - solution set 3 - autumn 2012
1. Time Derivative of Momentum Expectation Value
A particle moves in a potential V = V (x). Show that the quantum mechanical
expectation values satisfy
d
dt
px = −
∂V
∂x
where px = −i ∂
∂x is the momentum operator in the x-direction.
Solution
In the lecture notes it was shown that d
dt r = p
m . Here we will proceed in an
analogous manner. First we note that
d
dt
px =
d
dt
ψ∗
pxψ dV
=
∂ψ∗
∂t
pxψ + ψ∗
px
∂ψ
∂t
dV
= −i
∂ψ∗
∂t
∂ψ
∂x
− ψ∗ ∂2
ψ
∂x∂t
dV, (1.1)
where the integral is taken over all space. The time evolution of a wave function
is governed by the Schr¨odinger equation. That is say,
−
2
2m
2
ψ + V ψ = i
∂ψ
∂t
. (1.2)
Therefore
−
2
2m
2
ψ∗
+ V ψ∗
= −i
∂ψ∗
∂t
(1.3)
and
−
2
2m
2 ∂ψ
∂x
+
∂V
∂x
ψ + V
∂ψ
∂x
= i
∂2
ψ
∂x∂t
. (1.4)
As we substitute Eqs. (1.3) and (1.4) into Eq. (1.1), we obtain
d
dt
px = −
2
2m
2
ψ
∗
+ V ψ
∗ ∂ψ
∂x
+ ψ
∗
2
2m
2 ∂ψ
∂x
−
∂V
∂x
ψ − V
∂ψ
∂x
dV
=
2
2m
ψ
∗ 2 ∂ψ
∂x
−
∂ψ
∂x
2
ψ
∗
dV − ψ
∗ ∂V
∂x
ψ dV
=
2
2m
ψ
∗ 2 ∂ψ
∂x
−
∂ψ
∂x
2
ψ
∗
dV −
∂V
∂x
. (1.5)
1

Now we have to show that the ﬁrst term at the last line vanishes. For that
purpose, we state the identities
· ψ∗ ∂ψ
∂x
= ψ∗
·
∂ψ
∂x
+ ψ∗ 2 ∂ψ
∂x
, (1.6)
·
∂ψ
∂x
ψ∗
=
∂ψ
∂x
· ψ∗
+
∂ψ
∂x
2
ψ∗
. (1.7)
By employing these identities and the divergence theorem, we obtain
ψ∗ 2 ∂ψ
∂x
−
∂ψ
∂x
2
ψ∗
dV = · ψ∗ ∂ψ
∂x
−
∂ψ
∂x
ψ∗
dV
= ψ∗ ∂ψ
∂x
−
∂ψ
∂x
ψ∗
· d ¯A. (1.8)
If this surface integral vanishes, our result is proved.
2

2. Infinite Potential Well
Let us consider a particle in an infinite potential well. Calculate the expectation
value of its
a) kinetic energy,
b) potential energy.
Solution
In the lecture notes, we solved the time-independent Schrödinger equation for
a particle in an infinite potential well. We found out that the energy eigenfunc-
tions are
un(x) =
2
a sin nπ
a x 0 < x < a,
0 otherwise,
(2.1)
where n ∈ {1, 2, 3, · · · }. Let us now calculate the expectation value of kinetic
energy for the nth energy eigenfunction. We obtain
un|T|un =
∞
−∞
u∗
n(x)Tun(x) dx (2.2)
=
a
0
u∗
n(x)Tun(x) dx,
where the last equality follows from the fact that un(x) vanishes if x /∈ (0, a).
On the other hand, if x ∈ (0, a),
Tun(x) = −
2
2m
∂2
∂x2
2
a
sin
nπ
a
x
=
2
2m
n2
π2
a2
2
a
sin
nπ
a
x
=
2
2m
n2
π2
a2
un(x). (2.3)
Thus
un|T|un =
2
2m
n2
π2
a2
a
0
u∗
n(x)un(x) dx. (2.4)
We recall that we have normalized the functions un(x) (cf. lecture notes). Thus
the integral at the right-hand side is equal to unity. It immediately follows that
un|T|un =
n2
π2 2
2ma2
. (2.5)
3

The expectation value of potential energy is
un|V |un =
∞
−∞
u∗
n(x)V (x)un(x) dx
=
0
−∞
u∗
n(x)V (x)un(x) dx +
a
0
u∗
n(x)V (x)un(x) dx
+
∞
0
u∗
n(x)V (x)un(x) dx. (2.6)
The second integral vanishes because V (x) = 0 when x ∈ (0, a). However, both
the first and the last integral are undefined because the integrands are undefined
(the integrands are of the form ∞·0·∞). However, when we consider the infinite
potential well as a limiting case of the finite potential well, we obtain the result
that there is no contribution to the expectation value of potential energy from
the outside of the well 1
. Therefore we set the first and the last integral equal
to zero. We obtain the result that the expectation value of potential energy
vanishes. That is,
un|V |un = 0. (2.7)
1You may want to return to this point after the finite potential well has been discussed in
the lectures.
4

3. An Eigenvalue Problem
Let us consider the eigenvalue equation
Âu = λu, Â = −
d2
dx2
,
under the boundary condition u(−a) = u(a) = 0 (a is a positive real number).
Determine those eigenfunctions u that correspond to a positive eigenvalue λ .
Solution
Our eigenvalue equation
−
d2
dx2
u = λu (3.1)
is a second order homogeneous equation with constant coefficients. The associ-
ated auxiliary equation reads
r2
= −λ. (3.2)
The roots of the auxiliary equation are
r1 = i
√
λ,
r2 = −i
√
λ. (3.3)
The general solution
u(x) = C1er1x
+ C2er2x
= C1ei
√
λx
+ C2e−i
√
λx
= C1 cos
√
λx + i sin
√
λx + C2 cos
√
λx − i sin
√
λx
= A cos
√
λx + B sin
√
λx. (3.4)
Let us next consider the solutions we obtain for certain values of A and B.
A = 0 and B = 0
We obtain the zero function, which is not considered to be an eigenfunction.
A ∈ C arbitrary and B = 0
For the resulting eigenfunctions u(x) = A cos
√
λx the boundary conditions read
cos
√
λa = 0. (3.5)
This implies that
√
λa =
π
2
+ nπ ⇐⇒ λ =
π2
4a2
(1 + 2n)
2
. (3.6)
5

A = 0 and B ∈ C arbitrary
For the resulting eigenfunctions u(x) = B sin
√
λx the boundary conditions read
sin
√
λa = 0. (3.7)
This implies that
√
λa = nπ ⇐⇒ λ =
π2
a2
n2
. (3.8)
A = 0 and B = 0
It is only a minor task to show that the eigenvalues must now be of the form
(3.6) and of the form (3.8). Therefore we have to ﬁnd out whether there are
n1, n2 ∈ {1, 2, 3, · · · } satisfying
π2
4a2
(1 + 2n1)2
=
π2
a2
n2
2 ⇐⇒ 1 + 2n1 = 2n2. (3.9)
We see that the left-hand side is odd for all n1, whereas the right-hand side is
even for all n2. Therefore there are no eigenvalues that are of the form (3.6)
and of the form (3.8).
Summary
The eigenvalues λn and the corresponding eigenfunctions un:
λn = π2
4a2 (1 + 2n)
2
: un(x) = A cos
√
λnx,
λn = π2
a2 n2
: un(x) = B sin
√
λnx.
6

4. Scattering From a Potential Step
Particles scatter from a potential step. Show that from the conservation of the
probability current it follows that the reflection and transmission coefficients R
and T satisfy the condition
R + T = 1.
Solution
In the lecture notes the probability currents outside and inside the potential
step are denoted by SI and SII, respectively. Furthermore, SI is divided into
two parts, namely into currents towards the positive and negative x-axis. Math-
ematically,
SI = S+
I + S−
I . (4.1)
Similarly, we divide SII into currents towards the positive and negative x-axis:
SII = S+
II + S−
II. (4.2)
When the energy of the particle is greater than the step height, there is no
probability flow towards the negative x-axis inside the step. On the other hand,
when the energy of the particle is smaller than the step height, there is no prob-
ability flow to either direction deep inside the step (near the surface the flows
to opposite directions cancel out resulting in a zero net flow). To summarize,
in both cases deep inside the step
SII = S+
II. (4.3)
Let us denote the incoming, reflected and transmitted intensities by I+
I , I−
I
and I+
II, respectively. In terms of these intensities, the reflection and transmis-
sion coefficients are
R =
I−
I
I+
I
and T =
I+
II
I+
I
, (4.4)
respectively. On the other hand,
I+
I = N|S+
I |, I−
I = N|S−
I |, I+
II = N|S+
II|, (4.5)
where N is the density of the incoming particles and the probability current
densities are measured far away from the step (at the so called asymptotical
regions mentioned in the lecture notes).
The conservation of probability reads
SI = SII. (4.6)
7

Now
R + T =
I−
I
I+
I
+
I+
II
I+
I
=
N|S−
I |
N|S+
I |
+
N|S+
II|
N|S+
I |
=
−S−
I + SII
S+
I
, (4.7)
where the last equality holds because we consider S+
II to be measured deep inside
the step [cf. Eq. (4.3)]. By using the conservation of probability [Eq. (4.6)], we
obtain
R + T =
−S−
I + S+
I + S−
I
S+
I
= 1. (4.8)
8

5. Scattering From a Potential Step Revisited
A particle scatters from a potential step of height V0. The energy of the particle
E < V0.
a) Show that the probability current inside the potential step vanishes.
b) Show that total reflection happens.
c) Estimate the penetration depth by evaluating x inside the step.
d) Show that there is a phase shift between the incoming and scattered wave.
e) Study the behavior of the probability density outside the potential step.
Solution
a) If the total energy E of the particle is lower than the step height V0, the
wave function is exponentially decaying inside the potential step. To be more
precise,
uII(x) = Ce−hx
, (5.1)
where
h = −iκ
=
2m(V0 − E)
2
. (5.2)
Inside the potential step the probability density current
S(x) =
m
Im uII(x)∗ duII
dx
(x)
=
m
Im C∗
e−hx
−hCe−hx
=
m
Im −|C|2
he−2hx
real
= 0. (5.3)
b) We have total reflection if the reflection coefficient R = 1. On the other
hand, it was pointed out in the lecture notes that
R =
B
A
2
, (5.4)
where B and A are the amplitudes of the scattered and incoming waves, re-
spectively. Moreover, we chose A=1 and obtained B = e−2iθ
. As we substitute
these into the previous equation, we obtain
R = 1. (5.5)
That is to say, total reflection happens.
9

c) Here we indeed assume that the particle is inside the potential step. Thus
the wave function associated with the particle is uII(x) = 2 cos θe−hx
e−iθ
. As
always, it has to be normalized before calculating expectation values. The
normalization factor
N =
∞
0
|uII(x)|2
dx
= 4 cos2
θ
∞
0
e−2hx
dx
=
2 cos2
θ
h
. (5.6)
Now
x =
1
N
∞
0
uII(x)∗
xuII(x) dx
= 2h
∞
0
xe−2hx
dx
=
1
2h
=
1
2
2
2m(V0 − E)
. (5.7)
d)
Figure 1: Schematic picture of k + ih in the complex plane.
Let us present the wave function in two parts (just like before). The parts
outside and inside the step are
uI(x) = eikx
+ Be−ikx
, (5.8)
uII(x) = Ce−hx
, (5.9)
respectively. The wave function and its derivative must be continuous at x = 0.
Therefore
1 + B = C
ik(1 − B) = −hC
⇐⇒
B = k−ih
k+ih
C = 2k
k+ih
. (5.10)
10

Let us write k + ih = reiθ
. We obtain
B =
k − ih
k + ih
= e−2iθ
. (5.11)
As we substitute this into Eq. (5.8), we obtain
uI(x) = eikx
+ e−ikx−2iθ
. (5.12)
Now we see that the phase diﬀerence between the incoming and the scattered
wave is 2θ.
e) First we note that the wave function can not be normalized (due to its
form outside the step). Outside the step the probability density
ρ(x) = |uI(x)|2
= eikx
+ e−ikx−2iθ ∗
eikx
+ e−ikx−2iθ
= 2 (1 + cos(2kx + 2θ))
= 4 cos2
(kx + θ) . (5.13)
We see that it is an oscillating (i.e. periodic) function of x.
Figure 2: Plot of ρ(x) when k = 2 and θ = π/5.
11

Solution set 3

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