Solid State Theory
Solution Sheet 11
SS 11
Prof. M. Sigrist
Exercise 11.1
Relaxation time approximation
a) The transition rates are computed by use of Fermi’s golden rule, relating the proba
bility for a process to occur to the quantum mechanical matrix elements of scattering
off a single impurity a weighting with the impurity density (cf. section 6.3 of the
lecture notes)
W
(
k
,
k
0
) =
2
π n
imp
~
h
k
0

V
imp

k
i
2
δ
(
ε
k

ε
k
0
)
(1)
The matrix elements
1
h
k
0

V

k
i
=
V
(
k

k
0
) are given by the Fourier transform of
the potential
V
(
r
)
∼
δ
(
r
)
V
(
k

k
0
) =
1
2
π
Z
d
2
r e
i
(
k

k
0
)
·
r
V
0
δ
(
r
)
=
1
2
π
V
0
.
(2)
Thus, the transition rates are given as
W
(
k
,
k
0
) =
n
imp
2
π
~
V
2
0

{z
}
W
0
δ
(
ε
k

ε
k
0
)
,
(3)
and given the isotropy of the energy
ε
k
∼
k
2
they do not depend on
k
or
k
0
as long
as the energy is conserved.
b) In the static limit, the left hand side of the Boltzmann equation
∂f
(
r
,
k
, t
)
∂t
+
∇
r
f
(
r
,
k
, t
)
·
d
r
d
t
+
∇
k
f
(
r
,
k
, t
)
·
(
e
E
)
(4)
reduces with
f
(
r
,
k
, t
) =
f
(
k
) to the third term proportional to the driving external
field.
We restrict our considerations to linear order in
E
since we are interested
in linear response to the driving force. In consequence, we must replace
∇
k
f
with
∇
k
f
0
because
δf
is already of order
O
(
E
).
This term represents the socalled drift term. The right hand side of the Boltzmann
equation is the collision integral, which for impurity scattering acquires the form
(cf. Eq. (6.23) in the lecture notes)
∂f
(
k
)
∂t
coll
=

Z
d
2
k
(2
π
)
2
W
(
k
,
k
0
)(
f
(
k
)

f
(
k
0
))
=

Z
d
2
k
(2
π
)
2
W
0
δ
(
ε
k

ε
k
0
) (
f
0
(
k
) +
δf
(
k
)

f
0
(
k
0
)

δf
(
k
0
))
=

Z
d
θ
d
k k
(2
π
)
2
W
0
m
~
2
k
δ
(
k

k
0
) (
f
0
(
k
) +
δf
(
k
)

f
0
(
k
0
)

δf
(
k
0
))
=

m
~
2
Z
d
θ
(2
π
)
2
W
0
[
δf
(
θ
)

δf
(
θ
0
)]
.
(5)
1
Note that the potential only depends on the difference between
k
and
k
0
due to the homogeneity of
the system.
1
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Taking (4) and (5) together we arrive at
∇
k
f
0
·
(
e
E
) =

Z
d
θ
0
˜
W
0
[
δf
(
θ
)

δf
(
θ
0
)]
,
(6)
where
˜
W
0
=
mW
0
/
(2
π
~
)
2
.
c) We start with the Boltzmann equation,
X
l
d
l
e
ilθ
=

Z
d
θ
0
˜
W
0
X
l
f
l
e
ilθ

X
l
f
l
e
ilθ
0
!
,
(7)
which we project onto the
m
th Fourier mode with multiplication of both sides with
exp(

imθ
) and integration over
θ
Z
d
θe

imθ
X
l
d
l
e
ilθ
=

Z
d
θe

imθ
Z
d
θ
0
˜
W
0
X
l
f
l
e
ilθ

X
l
f
l
e
ilθ
0
!
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 Spring '11
 Sigrist
 Fourier Series, Joseph Fourier

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