Carsten Timm: Theory of superconductivity

We now consider the force F = −eE and calulate the current density
∫
d 3 k k
j(r, t) = −e
(2π) 3 ρ(r, k, t). (4.19)
m
To that end, we have to solve the Boltzmann equation
( ∂
∂t + k m · ∂
∂r − eE ·
)
∂
∂k
ρ = − ρ − ρ 0
. (4.20)
τ
We are interested in the stationary solution (∂ρ/∂t = 0), which, for a uniform field, we assume to be spacially
uniform (∂ρ/∂r = 0). This gives
− eE ·
⇒
∂
∂k ρ(k) = ρ 0(k) − ρ(k)
τ
ρ(k) = ρ 0 (k) + eEτ
·
We iterate this equation by inserting it again into the final term:
ρ(k) = ρ 0 (k) + eEτ
(
· ∂ eEτ
∂k ρ 0(k) +
·
(4.21)
∂
ρ(k). (4.22)
∂k
)
∂ eEτ
∂k ·
∂
ρ(k). (4.23)
∂k
To make progress, we assume that the applied field E is small so that the response j is linear in E. Under this
assumption we can truncate the iteration after the linear term,
ρ(k) = ρ 0 (k) + eEτ
· ∂
∂k ρ 0(k). (4.24)
By comparing this to the Taylor expansion
(
ρ 0 k + eEτ )
= ρ 0 (k) + eEτ
· ∂
∂k ρ 0(k) + . . . (4.25)
we see that the solution is, to linear order in E,
(
ρ(k) = ρ 0 k + eEτ )
∝ n F (ϵ k+eEτ/ ). (4.26)
Thus the distribution function is simply shifted in k-space by −eEτ/. Since electrons carry negative charge, the
distribution is shifted in the direction opposite to the applied electric field.
E
eEτ/h
ρ
0
The current density now reads
∫
j = −e
d 3 k k
(2π) 3 m ρ 0
(
k + eEτ ) ∫
∼= −e
d 3 k k
(2π) 3 m ρ 0(k)
} {{ }
= 0
21
∫
− e
d 3 k k eEτ
(2π) 3 m
∂ρ 0
∂k . (4.27)