Ashcroft and Mermin: Derivation from (13.24) to (13.27)

Question

In the A&M, the conductivity tensor $\sigma$ can be expressed as $$ \sigma = \sum_n e² \int \frac{dk}{4 \pi³ \hbar} \tau_n(\epsilon_n(k)) v_n(k) v_n(k) \left( -\frac{\partial f} {\partial \epsilon} \right) \tag{13.25} $$ where $\tau_n(\epsilon_n(k))$ is the relaxation time for the band $n$, $v_n(k)$ is the mean velocity of an electron in the definite Bloch level $n$ and $f$ is the Fermi-Dirac distribution. An approximation has been previously made that $\tau$ depends on $k$ only through the energy $\epsilon_n(k)$. To an accuracy of order $(k_BT/\epsilon_F)²$ at $T=0$ can reduce to $$ \sigma = e² \tau(\epsilon_F) \int \frac{dk}{4 \pi³} \left(\frac{\partial } {\partial k} v(k) \right)f(\epsilon (k). \tag{13.27} $$ I'm pretty sure there's an errata, and the term $f(\epsilon(k))$ should not be there.

But, even if I correct it, I can either integrate by part the Fermi-Dirac distribution to get a delta dirac $\delta(\epsilon - \epsilon_F)$ since we evaluate at $T=0$. $$ \left(\frac{\partial f}{\partial \epsilon} \right) = \delta(\epsilon - \epsilon_F) $$ which transform $\tau_n(\epsilon_n(k)) \rightarrow \tau(\epsilon_F)$. Or I can use the identity $$ v(k) \left( -\frac{\partial f} {\partial \epsilon} \right) = -\frac{1}{\hbar} \frac{\partial} {\partial k} f(\epsilon(k)) \tag{13.26} $$ to get the $\hbar^{-1}$. In the derivation, it looks like they are first changing $\tau_n(\epsilon_n(k)) \rightarrow \tau(\epsilon_F)$ which should make disappear the $f$. Then, another $f$ magically appear so they can use the second identity.

Is there really an error. If so, how do you do the derivation correctly?

Answer 1

The subtility lies in the fact that $\tau(\epsilon)$ and the integral is on k. That means you can transform $$ \left(\frac{\partial f}{\partial \epsilon} \right) = \delta(\epsilon - \epsilon_F) $$ With it, you know you'll only evaluate $\tau(\epsilon)=\tau(\epsilon_F)$ so you change it. But since you integrate on k instead of $\epsilon$, the integral doesn't disappear. You can retransform the $\delta(\epsilon - \epsilon_F)$ into the Fermi function using the same identity. With the Fermi function, you use the identity (13.26) (stated in the question) to solve the problem. There's a subtility right here $$ \sigma = e² \tau(\epsilon_F) \int \frac{dk}{4 \pi³} \left(-\frac{\partial } {\partial k} f(\epsilon (k)) \right)v(k) $$ where you need to use $$ \int_C dr (u\nabla v) = -\int_C dr (v\nabla u) \tag{I.1} $$ which is true if u(r) and v(r) have the periodicity of the Bravais lattice and C is taken over a primitive cell C.