If I have a 3D dispersion relation
$E=E(k_x, k_y, k_z)$ I have an equation for the density of states, which is
$D(E)=\frac{1}{\nabla_k E}\int\frac{dS}{(2\pi)^3}$
1) I am confused about the integral. What am I integrating over, exactly? It must have units of the $k$ vector, so it's a k-space integral, but what is the surface that is relevant?
3) Where does that equation come from, how is it derived?
Best Answer
deriving the desired equation for the density of states we are reminded that the surface which we are integrating over is a surface of constant energy in the reciprocal space which is denoted by $S(e)$, where $e$ is the 3-dimensional dispersion relation
we know that the number of states between the surfaces $S(e)$ and $S(e + de)$ is given by the integral. $$D(e)\, \mathrm{d}e = \int_{S(e)} \frac{\mathrm{d}S}{(2\pi )^3} \mathrm{d}q_\perp(\boldsymbol q)$$ if we linearly expand the dispersion relation $e$ as $e+\mathrm{d}e = e + |\nabla_\boldsymbol q e(\boldsymbol q)|\mathrm{d}q_\perp(\boldsymbol q)$ where $\boldsymbol q$ is a vector, $\mathrm{d}q_\perp (\boldsymbol q)$ is the perpendicular distance between the surfaces and $\mathrm{d}q_\perp(\boldsymbol q)= \frac {\mathrm{d}e}{|\nabla_\boldsymbol q e(\boldsymbol q)|}$. this leads to $$D(e) = \frac{1}{(2\pi)^3} \int_{S(e)} \frac {\mathrm{d}S}{|\nabla_\boldsymbol q e(\boldsymbol q)|}$$ which is the desired equation for the density of states.