Physics Asked by Gippo on December 28, 2020
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 $k$).
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.
Answered by an offer can't refuse on December 28, 2020
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