2
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.