Physics 463
Winter 2012
HW # 06 Solutions
1. Show
V
(
r
) =
∑
G
V
G
e
i
G
·
R
where
G
are the reciprocal lattice vectors.
V
(
r
) =
V
(
r
+
T
) =
∑
G
V
G
e
i
G
·
R
e
i
G
·
T
Therefore
e
G
·
T
= 1 for this to be satisfied.
If
G
is a reciprocal lattice vector, then
e
i
G
·
T
=
e
i
Q
v
j
n
i
a
i
·
b
j
=
e
2
πi
(
Integer
)
= 1. If
G
is a vector such
e
i
G
·
T
= 1 then
G
is a reciprocal lattice vector by definition of reciprocal
lattice vectors.
2. a. In general, the bands will have this kind of structure (Figure 9.4)
This is degenerate perturbation theory where the unperturbed Hamiltonian is the free electron (only kinetic
energy). Therefore, each Fourier coefficient must be small compared to the kinetic energy in order for this
approximation to be reasonable.
b. We can write:
V
(
x
) =
V
0
cos(2
πx/a
) =
V
0
2
e
i
2
πx/a
+
e

i
2
πx/a
=
U
(
e
igx
+
e

igx
)
Where
g
= 2
π/a
. To find the energy as a function of
k
, we can use the
central equation
(Equation 7.27),
which is reminiscent of manyfold degenerate perturbation theory.
Since our potential only contains two
Fourier components nearby the diagonal, expanding the central equation yields Equation 7.32
λ
k

2
g

E
U
0
0
0
U
λ
k

g

E
U
0
0
0
U
λ
k

E
U
0
0
0
U
λ
k
+
g

E
U
0
0
0
U
λ
k
+2
g

E
C
(
k

2
g
)
C
(
k

g
)
C
(
k
)
C
(
k
+
g
)
C
(
k
+ 2
g
)
=
0
0
0
0
0
Where
λ
k
=
~
2
k
2
2
m
(Equation 7.28) and the
C
’s are from the wavefunction
ψ
=
∑
k
C
(
k
)
e
ikx
(Equation 7.25).
Why do we use the central equation (degenerate perturbation theory)?
Recall the energy correction in
second order
non
degenerate perturbation theory:
1
Δ
E
(2)
=
X
k
0
6
=
k
h
ψ
k
0

V

ψ
k
i
2
E
k

E
k
0
This condition is no longer valid where
E
k
0
=
E
k
, with the added condition from the central equation that
k
0
+
k
+
G
= 0 (comes from the expansion of V(x) in the reciprocal lattice). Now, in the case for 1D lattices,
there are at most two degenerate points, which create the bands; however, for 2D or 3D lattices, generally
there can be more degenerate energy points, requiring diagonalization of larger matrices.
For the gap between the first and second band, there is mixing between the two states at the edge of the first
Brillouin zone; which is separated by one reciprocal lattice vector. Therefore, use the submatrix (the blue
shaded region or
M
[2
,
3]
×
[2
,
3]
), corresponding to the
C
(
k
) and
C
(
k

g
) components. For the first Brillouin
zone, the Reduced zone scheme and Extended zone scheme are the same. For the periodic zone scheme, one
Want to read all 5 pages?
Want to read all 5 pages?
You've reached the end of your free preview.
Want to read all 5 pages?