Given that graphene has linear energy dispersion near the Fermi level and the dispersion is given by $E=\hbar \nu_F|\vec{K}|$, I would like to determine the density of states. I think it is equal to $$g(E)=\frac{E}{2\pi\hbar^2\nu_F^2},$$ but how could I show that?

In fact, all I am asking is an explanation why the following relation holds: $$N=\text{total number of states}=\frac{A}{2\pi}\int_0^{k(E)}d\mathbf{k}$$ in the vicinity of the dirac points. Also, what are the units of $g(E)$?

  • 1
    $\begingroup$ Could you post your two questions separately? You can edit this post to just contain one of the questions, and then put the other up as a new post. $\endgroup$ – David Z Jan 7 '14 at 5:47

Let we have $L \times L$ graphene sheet. Then we have $4(L/2\pi)^2$ density of states in the reciprocal space ($4$ comes from $2$ spin states and $2$ valleys in graphene). Therefore the total number of states is

$$ N =4\frac{L^2}{(2\pi)^2}\int_0^{k(E)}dk_x dk_y. $$

In polar coordinates it is

$$ N =4\frac{L^2}{(2\pi)^2}\int_0^{k(E)}2\pi k\, dk = \frac{2L^2}{\pi}\int_0^{k(E)}k\, dk. $$

The density of states per unit energy and unit area is

$$ g(E)dE = 4\frac{2\pi k\, dk}{(2\pi)^2} = \frac{2|E|dE}{\pi \hbar^2 {v\vphantom{\hbar}}_f^2}. $$


$$ g(E) = \frac{2|E|}{\pi \hbar^2 {v\vphantom{\hbar}}_f^2}. $$

It seem that you forget the 4-fold electron degeneracy. This result agrees with known formula. See, for example, A. H. Castro Neto et al. The electronic properties of graphene. Rev. Mod. Phys. 81, 109 (2009), arXiv:0709.1163.


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.