Then, the above expression reduces to,
R
xx
(
ω
) =

1
Z
0
X
μ,ν
h
μ

j
x
P

ν
i
2
f
(
ε
ν
)

f
(
ε
μ
)
~
ω
+
ε
μ

ε
ν
+
i
0
+
,
(32)
where the expression inside the [
...
] brackets is known as the Lindhardfunction and
f
(
...
) is the
Fermifunction. (The factor
f
(
ε
ν
)

f
(
ε
μ
) arises from
f
(
ε
ν
)[1

f
(
ε
μ
)]

f
(
ε
μ
)[1

f
(
ε
ν
)].)
5
Problem 3:
We already saw in the previous problem that the conductivity is given by,
σ
xx
(
ω
) =
1
iω
1
Ω
X
μ,ν
h
μ

p
x

ν
i
2
ε
μ

ε
ν
+
ω
+
i
0
+
[
f
(
ε
ν
)

f
(
ε
μ
)]

1
iω
Nq
2
mc
2
.
(33)
Let us now look at the real part of
σ
xx
(
ω
),
Re
σ
xx
(
ω
) =
π
ω
X
μ,ν
h
μ

p
x

ν
i
2
δ
(
ε
μ

ε
ν
+
ω
)[
f
(
ε
ν
)

f
(
ε
μ
)]
.
(34)
We now use a trick to express the matrix elements of
p
x
in terms of matrix elements of
x
as follows:
h
μ

~
p

ν
i
=
m
i
~
h
μ

[
~
r, H
0
]

ν
i
=
m
i
~
(
ε
ν

ε
μ
)
h
μ

~
r

ν
i
.
(35)
Therefore, the expression for Re
σ
xx
, in the limit of small
ω
(so that we can taylor expand
f
(
ε
ν
)
about
ε
μ
), simplifies to,
Re
σ
xx
(
ω
) =
πm
2
~
2
ω
2
X
μ
f
0
(
ε
μ
)
h
μ

x
2

μ
i  h
μ

x

μ
i
2
.
(36)
However, we have been told in the problem that the last quantity in the brackets above remains
finite in the limit of
L
→ ∞
. Therefore, in the limit
ω
→
0, Re
σ
xx
(
ω
)
→
0.
Now note that Re
σ
xx
(

ω
) =Re
σ
xx
(
ω
).
If we compute the imaginary part, Im
σ
xx
(
ω
), via a
KramersKronig transformation, we obtain,
Im
σ
xx
(
ω
) =
1
π
Z
∞
∞
P
Re
σ
xx
(
ω
0
)
ω
0

ω
dω
0
=
1
π
2
ω
Z
∞
0
P
Re
σ
xx
(
ω
0
)
ω
0
2

ω
2
dω
0
.
(37)
Therefore, Im
σ
xx
(
ω
→
0) = 0.
Problem 4:
We would like to compute the scattering amplitude, which is going to be a matrix in spinspace,
using firstorder Born approximation for the following Hamiltonian:
H
=
~
p
2
2
m
+
U
(
~
r
) +
λ
∇
U
(
~
r
)
·
(
~
p
×
~
S
)
.
(38)
The scattered wave solution at large
r
is given by,
ψ
ˆ
n,τ
(
~
r, σ
)
∼
e
e
ik
ˆ
n.~
r
δ
στ
+
e
ikr
r
f
στ
(ˆ
r,
ˆ
n
)
,
(39)
where ˆ
n
and
τ
=
±
1 denote the incident direction and spin orientation of the incoming particle
and
f
is a 2
×
2 matrix which depends on ˆ
r
, the direction in which the amplitude is measured. To
lowest nonvanishing order,
f
is given by (see Sakurai, chap.7),
f
στ
(ˆ
r,
ˆ
n
) =

1
4
π
2
m
~
2
Z
d
3
r
0
e

ik
ˆ
r.~
r
0
[
U
(
~
r
0
)
δ
στ
+
λ
∇
r
0
U
(
~
r
0
)
·
(
~
p
×
~
S
)
στ
]
e
ik
ˆ
n.~
r
0
.
(40)
6
If we now integrate the second term by parts and use the fact that it is a shortranged potential
(vanishes fast enough as
~
r
0
→ ∞
), then
f
στ
(ˆ
r,
ˆ
n
) =

1
4
π
2
m
~
2
Z
d
3
r
0
e
ik
(ˆ
n

ˆ
r
)
.~
r
0
U
(
~
r
0
)[
δ
στ

iλk
2
(ˆ
n

ˆ
r
)
·
(ˆ
n
×
~
S
)
στ
]
.
(41)
It is now useful to write the above amplitude as
f
=
f
0
u
, where,
f
0
=

1
4
π
2
m
~
2
Z
d
3
r
0
e
ik
(ˆ
n

ˆ
r
)
.~
r
0
U
(
~
r
0
)
,
(42)
u
=
e
iλk
2
(ˆ
r
×
ˆ
n
)
.
~
S
.
(43)
We have exponentiated the matrix in the [
...
] brackets for small
λ
.
The matrix
u
is clearly an
SU(2) matrix since
~
S
represent the spin1/2 operators.
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 Summer '19
 Physics, Trigraph, Jx, matrix elements, py − Ay