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

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 $|\nabla_k \varepsilon_k|=0$?

• ε=0 at (π/2,−π/2) (2π/3,−π/3), while $|\nabla_kε_k|$ is not equal 0. Note that $\nabla_k$ is an vector. Feb 2, 2018 at 8:29

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!)