DOS of Van Hove singularity in 2D square lattice tight binding model

Question

For the simplest example, 2D square lattice tight binding model gives the energy band as $$\varepsilon_k=-2t(\cos k_x+\cos k_y) \, .$$ We know that $\mathbf{k}=(0,\pi)$ and related momentum points are saddle points which give $|\nabla_k \varepsilon_k|=0$ and thus some kind of singularity in density of state (DOS) since $$\rho(\varepsilon)\propto \int_{\varepsilon=\text{const}} \frac{d S}{|\nabla_k \varepsilon_k|} \, .$$ How can I get the $\ln$ divergence for DOS near $\varepsilon=0$? Should I only care about those singularity points and omit the integral from normal parts and do Taylor expansion near those saddle points?

Moreover, why are points like $(\pi/2,-\pi/2)$ or $(2\pi/3,-\pi/3)$ not called Van Hove points when those points are also lie in the $\varepsilon=0$ line and give $|\nabla_k \varepsilon_k|=0$?

Answer 1

So I do not know if this is still relevant, but maybe someone else can profit from my answer?

To analyse the behaviour around the critical points $\mathbf k^*$, you can just Taylor expand $\varepsilon_\mathbf{k}$ around these points and calculate the DOS with your or this $$ \rho(\varepsilon) = \frac{1}{(2\pi)^d} \int_{BZ}d^dk\,\delta(\varepsilon - \varepsilon_\mathbf k) $$ formula (valid in $d$ dimensions). For a square 2D lattice, this will result in a logarithmic van-Hove singularity at $\varepsilon = 0$ (ie. $\mathbf k^* = (0,\pi),(\pi,0) $) and no divergence at $\varepsilon = \pm 4t$ (ie. $\mathbf k^* = (0,0),(\pi,\pi)$).

Doing so, you will see that the integrable zeroes of the gradient of the dispersion relation do not result in singularities in the DOS. For example, the integral of $ 1/x^2 $ over $[-\infty,\infty]$ is finite, even though the integrand diverges at 0.

However, the value of $\varepsilon$ does not matter, so the fact that $\varepsilon = 0$ does not tell us anything about a potential singularity. Also the last mentioned points do not result in a zero gradient! (See comments!)