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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 $k$).

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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.

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  • $\begingroup$ Is it even true if there is no time-reversal symmetry? $\endgroup$
    – fqq
    Commented Aug 22, 2019 at 14:55
  • $\begingroup$ @fqq I am not sure, what is the flaw of above argument if there is only inversion symmetry? $\endgroup$
    – an offer can't refuse
    Commented Aug 23, 2019 at 0:05
  • $\begingroup$ The question seems very old and I am not sure why it popped back up in my list. At any rate, I don't think that answers the question. The answer actually made no assumption about equilibrium. Instead, the implicit assumption that was made is that all k states are occupied so that they cancel out by symmetry. What this shows is that a fully occcupied band cannot carry current (equilibrium or not). This is a valid point but pretty different from what was asked I think $\endgroup$
    – Barbaud Julien
    Commented Jul 13, 2021 at 5:00
  • $\begingroup$ A 2D/1D strip with in plane time-reversal-breaking magnetic field and hall voltage, can have hall current. Not totally sure if this is a counter example that shows time reversal or inversion symmetry is necessary. The hall voltage might have some weird boundary condition. $\endgroup$
    – Bohan Xu
    Commented Apr 7, 2023 at 17:46
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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.

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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.

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