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So we have been given a dispersion relation of the form: $$ E=6-2(\cos k_xa+\cos k_ya) $$ and asked to calculate the density of states.

The equation for the density of states is (eq 2.48 from here http://www.tcm.phy.cam.ac.uk/~pbl21/princ-condmat/v2.pdf):

$$ g(E)=\int \frac{ds}{2\pi^2}\frac{1}{|\nabla_k E(\mathbf{k})|} $$

where we integrate over a surface (or curve in 2d) of constant energy.

Let's work out the bits individually:

  • $|\nabla_k E(\mathbf{k})|=|\left(2a\sin (k_xa),2a\sin (k_ya)\right)|=2a\sqrt{\sin^2k_xa+\sin^2k_ya}$

  • $ds=\sqrt{dk_x^2+dk_y^2}$ where $dE=0=2a\sin (k_xa) dk_x+2a\sin (k_ya) dk_y$ and hence we get $ds=\sqrt{\sin^2k_xa+\sin^2k_ya} \frac{dk_x}{\sin(k_ya)}$

  • The bound of the integral are $0$ and $\cos^{-1}(3-E/2)$ where E goes from 2 to 10.

  • the total integral is then

$$g(E)=\frac{1}{a\pi^2}\int_0^{\cos^{-1}(3-E/2)/a}\frac{dx}{\sqrt{1-(3-\cos(xa)-E/2)^2}}$$

I then put this into mathematica and got either errors or complex answers

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