Phys 268r Fall 2013
Problem Set 5 Solutions  Corrected v2
B. Halperin
Solutions to Problem Set 5
Problem 1.
a)
We have the Hamiltonian
H
=
v~
τ
·
~
p
+
V
(
~
r
), and we can write out the momentum term explicitly:
~
τ
·
~
p
=
0
p
x

ip
y
p
x
+
ip
y
0
!
.
We write the wave function as a spinor:
Ψ =
ψ
A
ψ
B
!
,
and letting
k
≡
E/v
, we find a system of equations:
kψ
A
= (
p
x

ip
y
)
ψ
B
kψ
B
= (
p
x
+
ip
y
)
ψ
A
.
(1.1)
We can combine these equations this to find
k
2
Ψ = (
p
2
x
+
p
2
y
)Ψ, which has clearly has solutions
which are plane waves, with arbitrary direction of
~
k
and arbitrary form of the spinor. However,
the original pair of equations (1.1) is more restrictive. For
E >
0, the direction of
~
k
determines
the spinor up to an overall phase factor, so we have, for a given
~
k
=
k
(cos
θ,
sin
θ
)
Ψ
~
k
=
A
~
k
e
i
~
k
·
~
r
φ
A
φ
B
!
,
where the spinor is an eigenstate of
~
k
·
~
τ
with eigenvalue 1.
To be specific,we choose
φ
A
=
1
/
√
2
, φ
B
= (
e
iθ
/
√
2)
For
x >
0, we have
vk
0
=
V

E >
0, which again allows plane wave solutions at wave vectors
~
k
0
. Now, however, the solution with energy
E
has a velocity in the direction opposite to
~
k
0
, so
it is convenient to write
~
k
0
=

k
0
(cos
θ
0
,
sin
θ
0
), where
θ
0
is the direction of propagation. The
plane wave solutions with energy E may then be written
A
~
k
0
e
i
~
k
0
·
~
r
Φ
0
, where Φ = (1
, e
iθ
0
)
T
/
√
2.
This spinor is an eigenstate of
~
k
0
·
~
τ
with eigenvalue 1.
Now, suppose we have an incident wave with incident angle
θ
. The momentum
k
y
is conserved
by the Hamiltonian, so the transmitted and reflected waves must have the same
k
y
as the
incident wave.
This means that

k
0
sin
θ
0
=
k
sin
θ
.
For the transmitted wave, we require

π/
2
< θ
0
< π/
2, which means that the sign of
θ
0
is
opposite
to the sign of
θ
.
Thus if the
1
Phys 268r Fall 2013
Problem Set 5 Solutions  Corrected v2
B. Halperin
incident wave is moving in a direction between
x
and
y
, the transmitted wave will be moving
in a direction between
x
and

y
. This means that the planar boundary acts as a focusing lens
when
V
is greater than
E
.
Now we can use conservation of energy to see that
vk
0
=
V
0

vk
. There will be an acceptable
solution

sin
θ
0

<
1
0
if and only if (
k/k
0
) sin
θ <
1 . If
V
0
>
2
E
, we have
k
0
> k
, so this condition
is fulfilled for any angle of incidence. However if
E < V
0
<
2
E
, there will be no transmitted
wave (total reflection), unless the incident angle is smaller than a critical value, given by
θ
c
= sin

1
[(
V
0

E
)
/E
]
.
(1.2)
b)
Since the Hamiltonian is a firstorder differential operator, the boundary condition at
x
= 0 is
that the wave functions should match at the boundary. If the amplitudes of the incident, reflected
and transmitted waves are denoted respectively by
A, B, C
, then the matching component for
the upper component of the spinor is
A
+
B
=
C
, while the lower component gives
Ae
iθ
+
Be
iθ
r
=
Ce
iθ
0
(1.3)
where
θ
r
=
π

θ
is the direction of propagation of the reflected wave.
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 Fall '13
 BertrandI.Halperin
 Physics, Momentum, Schrodinger Equation, Fundamental physics concepts, B. Halperin, Corrected v2