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It is known, for example Equation (14) in the graphene review of Castro Neto (arXiv), that the full expression for the density of states (DOS) of graphene is in terms of an elliptic integral.

Close to the Dirac point, the well known DOS which goes linearly with energy is found.

How can one recover Eq. (14)? Or, more precisely, how to show the following integral for DOS leads to an elliptic integral?

$\rho (E) = \int \frac{d^2 k }{(2 \pi)^2} \delta(E-E_{\pm}(\boldsymbol{k}))$

where $E_{\pm}(\boldsymbol{k}) = \pm t \sqrt{3+f(\boldsymbol{k})}$ and $f(\boldsymbol{k}) = 2 \cos (\sqrt{3}k_y a) + 4 \cos (\tfrac{\sqrt{3}}{2}k_y a) \cos (\tfrac{3}{2} k_x a)$

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I've also stucked with such problem. Since this is old question, but I didn't find the full answer on it, I'll write down my attemption. It doesn't represent the full solution; however, I think that it almost gives the answer.

The density of states $\rho (E)$ is the imaginary part of the self-energy $\Sigma (\mathbf r , \mathbf r, E+i\epsilon)$, where $\epsilon \to 0^{+}$: $$ \rho (E) = -\frac{1}{\pi}\text{Im}\left(\lim_{\epsilon \to 0^{+}}\Sigma (\mathbf r , \mathbf r , E + i\epsilon) \right) $$ How can we determine $\Sigma (\mathbf r , \mathbf r , E + i\epsilon)$?

By the definition, the Green operator is $$ \hat{G} (T) = \frac{1}{T\hat{I} - \hat{H}} \equiv \sum_{\mathbf k}\frac{|\mathbf k \rangle \langle \mathbf k|}{T - E(\mathbf k)}, \quad T \equiv E + i\epsilon $$ Next, the Green function which connects the site $l$ of the lattice with itself (which is exactly the self energy) is $$ \tag 1 \Sigma (T, l, l) = \sum_{\mathbf k}\frac{\langle \mathbf{l}|\mathbf k \rangle \langle \mathbf k| \mathbf{l}\rangle}{T - E(\mathbf k)} $$ Lets talk about graphene in the nearest neigbours approximation. Its lattice is hexagonal (honeycomb), which can be represented by two interpenetrating triangular lattices with the strength of interaction given by $t$. Only the nearest cites (say, $A$ and $B$) of these lattices interact, so $\hat{H}$ lives in the space which it the direct product of spaces of two triangular lattices. Now this results in the fact that the hamiltonian can be given in the form of sum of two-dimensional matrix. Thus, for given cite the denominator of $(1)$ is $$ \tag 2 \begin{pmatrix} T & -\mu t \\ -\mu^{*}t & T\end{pmatrix} $$ Here $\mu$ defines the character of lattice, being $$ \mu = e^{ik_{x}a} + e^{i\left(\frac{\sqrt{3}k_{y}a}{2} - \frac{k_{x}a}{2} \right)} + e^{-i\left(\frac{\sqrt{3}k_{y}a}{2} + \frac{k_{x}a}{2} \right)} $$ Substituting $(2)$ into $(1)$, you can convert $(1)$ to the form $$ \tag 3 \rho (E) = -\frac{1}{\pi}\text{Im}\left[ \lim_{\epsilon \to 0^{+}}\int \limits_{\text{1st Br. zone}}\frac{d^{2}\mathbf k}{(2 \pi)^{2}}\frac{T}{T^{2} - t^{2}|\mu|^{2}}\right] = -\frac{E}{8t^{2}\pi}\text{Im} \left[\lim_{\epsilon \to 0^{+}}\tilde{G}\left(\frac{T}{t}\right)\right], $$ where $$ \tilde{G}\left(\frac{T}{t}\right) \equiv \frac{1}{\pi^2}\int \limits_{-\pi}^{\pi}\frac{dxdy}{ \frac{\frac{T^{2}}{t^{2}} - 3}{2} - \cos(2y) - 2\cos (y)cos(3x)} $$ Such quantity can be computed (there is no derivation of this result) for $\frac{\frac{T^{2}}{t^{2}} - 3}{2} > 3$, and $$ \tag 4 \tilde{G}\left(\frac{T}{t}\right) = \frac{T}{t\pi}\frac{1}{\sqrt{\left(\frac{T}{t} - 1\right)^3\left(\frac{T}{t} + 3\right)}}K\left( \frac{4\sqrt{\frac{T}{t}}}{\sqrt{\left(\frac{T}{t} - 1\right)^3\left(\frac{T}{t} + 3\right)}}\right), $$ where $K(x)$ is the elliptic integral of the first kind: $$ K(x) = \int \limits_{0}^{\frac{\pi}{2}}\frac{dy}{\sqrt{1 - x^{2}\sin^{2}(y)}} $$ The only thing which you have to do is to compute analytic continuation of $(4)$ and then to compute its imaginary part multiplied by four (which corresponds to the degeneracy of spins and two sites).

An edit

Here is the full derivation of the density of states in graphene.

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  • 1
    $\begingroup$ @Nigel1 : if You are interested in the full derivation, I may add it, since I haven't found it anywhere... $\endgroup$
    – Name YYY
    Mar 26, 2016 at 20:10
  • $\begingroup$ I also searched quite widely and could not find full derivation so I am sure it would be of interest $\endgroup$
    – Nigel1
    Mar 27, 2016 at 12:00
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    $\begingroup$ @Nigel1 : I've added the full derivation, as promised. [:)]. $\endgroup$
    – Name YYY
    Jun 9, 2016 at 11:22
  • $\begingroup$ Very nice - spasibo! $\endgroup$
    – Nigel1
    Jun 10, 2016 at 16:38
  • $\begingroup$ @NameYYY Hi, I know it's being a long time but I an studying the properties of graphene and am stuck here as well. The link you provided for the full derivation isn't working. Can you please look into it. Also as both of you have experienced studying it, can you suggest some materials for understanding it more. Thank You!! $\endgroup$
    – Gandalf73
    Aug 29, 2021 at 6:48

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