Demonstration that electric current at equilibrium is zero in crystals

Question

As it is well known, electrons at equilibrium (no external field) do not conduct electric current, i.e.

$\int_{BZ} dk\,v_{k}\,f(\epsilon_k)=0$

where $f(\epsilon_k)$ is the Fermi-Dirac distribution

$f(\epsilon)=\frac{1}{e^{\beta(\epsilon-\mu)}+1}$

$v_k$ is the velocity

$v_k=\frac{\partial \epsilon_k}{\partial k}$

and $BZ$ is the Brillouin zone.

Is it possible to demonstrate that just using the explicit expression of the Fermi-Dirac distribution and the fact that we are considering functions periodic on the reciprocal lattice? I mean, is it possible to say that the integral is zero just from the general properties of the quantities involved? (for example, in the free-electron case it comes from the fact that it is the integral of an odd function in $k$).

Answer 1

As it is well known, electrons at equilibrium (no external field) do not conduct electric current

This phrase may be justified by the subsequent context, but, taken separately, it sounds strange, as there are superconductors.

Answer 2

Good question, I am trying to answer this question(partially) from symmetry considerations.

If the system has the time reversal or inversion symmetry, then $\varepsilon_{-k}=\varepsilon_{k}$. From this $v(-k)=-v(k)$, and $f(\varepsilon_{-k})=f(\varepsilon_k)$. Since for any point $k$ in the BZ, we can always find its counterpart $-k$, therefore the integration over the entire BZ is zero.

I don't know how to argue if the the system breaks both inversion and time-reversal symmetry.

Answer 3

This follows from the periodicity of the band energy $\epsilon_{\mathbf{k}}=\epsilon_{\mathbf{k}+\mathbf{G}}$. Define $$ F(\epsilon) = \frac{1}{\beta}\log\left(\frac{1+\mathrm{e}^{-\beta(\epsilon-\mu)}}{\beta}\right) $$ to be the antiderivative of $f(\epsilon)$, i.e. $F'(\epsilon)=f(\epsilon)$. Then in one dimension $$ J = \int_{\text{BZ}} \frac{\mathrm{d}k}{2\pi}\,v(k)f(\epsilon_{\mathbf{k}}) = \int \frac{\mathrm{d}k}{2\pi}\,\frac{\partial\epsilon(k)}{\partial k_i}F'(\epsilon_{\mathbf{k}}) = \int\frac{\mathrm{d}k}{2\pi}\,\frac{\partial F(\epsilon_{\mathbf{k}})}{\partial k_i}=F(\epsilon_{\pi})-F(\epsilon_{-\pi})=0 $$ as $\epsilon_{k=\pi}=\epsilon_{k=-\pi}$. This argument extends to higher dimensions too.