5
$\begingroup$

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 $|\nabla_k \varepsilon_k|=0$?

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$
    – Zili Von
    Feb 2, 2018 at 8:29

1 Answer 1

1
$\begingroup$

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!)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.