Department of physics, NTNU TFY4340 Mesoscopic Physics Spring 2010

Solution to Exercise 6

As explained in the exercise text, the current density is

j = ∫ dj(k) = ∫ (−e) · v(k) · dn(k),

with

dn(k) = ϕ(k) · \frac{1}{4π^{3}} · d^{3}k ≃ [f(k) + \frac{eτ}{h̄} E · ∇kf(k)] · \frac{1}{4π^{3}} · d^{3}k,

and v(k) = h̄k/m^{∗}. Because of symmetry (f(k) is an even function of k),

∫ kf(k)d^{3}k = 0.

Furthermore, we have

∇kf(k) = \frac{∂f(k)}{∂E}∇kE(k)
= −\frac{1}{|∂E/∂k|} δ(k − k_{F}) · \frac{h̄k}{m^{∗}}
= −\frac{m^{∗}}{h̄k} δ(k − k_{F}) · \frac{h̄k}{m^{∗}}
= −δ(k − k_{F})k/k

which yields the expression

j = \frac{e^{2}τ}{4π^{3}m^{∗}} ∫ k (E · k) δ(k − k_{F}) k dk sin θ dθ dφ

for the current density, with d^{3}k in spherical coordinates. The conductivity then becomes

σ = \frac{j · E}{E^{2}}
= \frac{e^{2}τ}{4π^{3}m^{∗}} ∫ \begin{matrix}(E · k) & 2 \\ E & 2 \end{matrix} δ(k − k_{F}) k dk sin θ dθ dφ
= \frac{e^{2}τk_{F}^{3}}{4π^{3}m^{∗}} ∫_{0}^{π} dθ cos^{2} θ sin θ ∫_{0}^{2π} dφ
= \frac{e^{2}τ · 3π^{2}n}{4π^{3}m^{∗}} · |_{0}^{π} (−\frac{1}{3} cos^{3} θ) · 2π
= \frac{e^{2}τ · 3π^{2}n}{4π^{3}m^{∗}} · \frac{2}{3} · 2π
= \frac{e^{2}τn}{m^{∗}}