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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 $g(E)$?

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    $\begingroup$ 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
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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

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

Hence

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

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.

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