[Physics] Connection between fermi velocity and mean (square) velocity of diffusion current

electronic-band-theorysolid-state-physics

Is there any connection between $v_{F}$ and the $<v^{2}_{diff}>$, lets say for electrons in metals? I have come to a conclusion, that they are the same order of magnitude and their equivalence (well up to the square root) makes sense on the verbal level, yet I still lack the needed theoretical expertise to fully analyse this concept. It might be just simply wrong, too. A full answer is not necessary, a reference to the relevant literature is sufficient, since I seem to fail to find any source discussing this topic.

Best Answer

The Fermi velocity is related to the Fermi energy $\epsilon_F=\frac12m_{\text e}v_F^2$. Actually, the Fermi energy $\epsilon_F$ is the chemical potential of the electron gas (that is the minimum energy required to add an extra electron to the gas).

The energies of the electrons are distributed, according to the Fermi-Dirac distribution, as $$ f(\epsilon)=\frac{1}{\mathrm e^{\beta(\epsilon-\epsilon_F)}+1}.$$ At low temperature (large $\beta=1/k_BT$), this distribution is approximately a step $$f(\epsilon)\approx\left\{\begin{array}{ll}1&\text{if $\epsilon<\epsilon_F$}\\ 0&\text{otherwise}\end{array}\right.\qquad(T\ll\epsilon_F/k_B).$$ So at low temperature, the energy distribution is fully controlled by the Fermi energy, and so does the velocity. The mean square velocity depends on the density of states $D(\epsilon)$. If $D(\epsilon)\propto \epsilon^\alpha$ (in 2 dimensions, $\alpha=0$ and in 3 dimensions, $\alpha=1/2$), then $\left\langle v^2\right\rangle=\frac{\alpha+1}{\alpha+2}v_F^2$.

At large temperature, $\beta$ is small, such that energies $\epsilon\gg\epsilon_F$ are allowed and the high energy tail of $f(\epsilon)$ is approximately a Maxwell-Boltzmann distribution $$f(\epsilon)\approx\mathrm e^{-\beta\epsilon}\qquad(T\gg\epsilon_F/k_B).$$

In the second situation, the Fermi energy does not control the distribution, the mean square velocity is $\left\langle v^2\right\rangle=\frac{\Gamma(\alpha+2)}{\Gamma(\alpha+1)}\frac{2k_BT}m$.

So the answer is that $\left\langle v^2\right\rangle$ and $v_F^2$ are related only at low temperature and the factor between them depends on the density of states.