I am trying to build physical intuition about van Hove singularities. The density of states for a system with energy dispersion $E_\mathbf{k}$ is defined as

$$ D(E) = \int_{S(E)} \frac{dS}{4\pi^3} \frac{1}{|\nabla E_\mathbf{k}|},$$

where $S(E)$ is the constant energy surface in k-space.

For the simple case $E_\mathbf{k} = k^2$, we would expect a singularity at $k=0$, but this is not the case, because the surface area $dS$ goes to zero the same rate as the energy gradient at $k=0$, and the density of states is constant.

Then, we can consider the case $E_\mathbf{k} = k^4$. In this case, the gradient goes to zero faster than the area, and there is a divergence in the DOS (not a kink).

Clearly $\nabla E_\mathbf{k} = 0$ is a necessary condition for a van Hove singularity. But, what are the condition on the energy dispersion that will lead to 'picks' in the density of states?


1 Answer 1


Maybe a picture like this would help? It shows the density of states along with the band structure. You get a spike in the DOS most of the time when one of the dispersion curves goes flat. I don't think that focusing on the origin is that instructive since that's a special (albeit important) case.

Whenever the dispersion curve goes flat, you have "a lot" of states with more-or-less the same energy, so the states are density packed together, and there is a spike in the density of states. However, there are a finite number of states, so the singularities should be integrable.


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