0
$\begingroup$

The energy dispersion relation of graphene with the tight-binding approximation and interactions up to nearest neighbors is $$E(k_1,k_2)=\pm |t|\sqrt{3+2\cos(ak_1) + 4\cos\left(\frac{a}{2}k_1\right)\cos\left(\frac{\sqrt{3}a}{2}k_2\right)}.$$ I am trying to show that around $K:=\frac{2\pi}{3a}(1,\sqrt{3})$, the energy relation is linear, that is, up to first order $E(K+v)\propto \|v\|.$ The problem I'm having is that I can't do a Taylor expansion around $K$ since the function is not differentiable at $K$. Any help would be appreciated!

$\endgroup$
0
$\begingroup$

You expect $E(K+v)$ to be proportional to $\lVert v\rVert$ which is not differentiable at $0$. Then why should $E$ be differentiable at $K$? In any case, you can Taylor expand everything under the root and get the result you need.

$\endgroup$

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.