I'm studying Solid State Physics. I know how to describe the Sommerfeld model, but I don't know how to apply it on Weyl semimetals.

The dispersion relation on Weyl semimetals is the following:

$$\varepsilon(\vec{k})=v_{F}\hbar k $$

I pretend do demonstrate from the semi classical equations for dynamics:

$$\vec{F}=\hbar \frac{d\vec{k}}{dt},\hspace{15pt} \vec{v}=\frac{1}{\hbar}\nabla_{\vec{k}}\varepsilon$$

that $v_F$ on the dispersion relation is the Fermi speed. How should I do this?


In a more general context than Weyl semimetals, in any theoretical condensed matter model we are always only interested in low-energy excitations. Thus assume you have an electron of momentum $\vec{k}$ that lies just above the Fermi surface. Most condensed matter models can be accurately described in the framework of Landau's Fermi-liquid theory at low energy, meaning that the electrons behave in very good approximation as free particles with interaction-renormalized parameters. Thus the dispersion relation of your electron is just

$\epsilon(\vec{k}) = \frac{k^2}{2 m^*}$,

where $m^*$ is the effective mass of the electron. We can decompose $\vec{k}$ in two colinear vectors $\vec{k} = \vec{k_F} + \delta \vec{k}$, where $\vec{k_F}$ lies on the Fermi surface and $\delta \vec{k}$ is small with respect to $k_F$. The dispersion relation then reads

$\epsilon(\vec{k}) = \frac{(\vec{k_F} + \delta \vec{k})^2}{2 m^*} = \frac{k_F^2}{2 m^*} + \frac{k_F \delta k}{m^*} + \frac{\delta k^2}{2 m^*}$.

You can then throw the $\delta k^2$ part because it is second order, $\frac{k_F^2}{2 m^*} = E_F$ is the Fermi energy which you take as a reference, and using the fact that $p_F = \hbar k_F$ you get the desired formula.

  • $\begingroup$ Thank you for your answer. This question was proposed by my teacher, and he said that to demonstrate that $v_F$ is the Fermi speed, we just need to see that using $\varepsilon=\hbar v_F k$ and $\vec{v}=\frac{1}{\hbar}\nabla_{\vec{k}}\varepsilon$, we get $v=v_F$. $\vec{v}$ is the velocity of a electron, and so, the Fermi speed. So we just get it proven. I didn't like this question at all. It seems too simple $\endgroup$ – Élio Pereira Mar 10 '16 at 9:33

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