# I Dispersion relation in tight binding model

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1. Nov 28, 2016

### tobix10

Hamiltonian of tight binding model in second quantization is given as $$H = -t \sum_{<i,j>} a_i^{\dagger} a_j$$
After changing basis it is $$H = \sum_{\vec{k}} E_{\vec{k}} a_{\vec{k}}^{\dagger} a_{\vec{k}}$$
where $$E_{\vec{k}} = -t \sum_{\vec{b}} e^{i \vec{k} \cdot \vec{b}}$$
where $\vec{b}$ is a nearest neighbour shift vector.

I've used this relation to calculate $E_{\vec{k}}$ for a square lattice, where nearest neighbours are points $(a,0)$, $(-a,0)$, $(0,a)$, $(0,-a)$. The result is $$E_{\vec{k}} = -t[e^{i(k_x a + 0)} + e^{i(- k_x a + 0)} + e^{i(0 + k_y a)} + e^{i(0 - k_y a)}] = -2t (cos k_x a + cos k_y a)$$

For a traingular lattice I set point (0,0) in the middle so neighbours are $(a/2,a \sqrt{3}/2)$, $(-a/2,-a \sqrt{3}/2)$, $(a/2,-a \sqrt{3}/2)$, $(-a/2,a \sqrt{3}/2)$, $(a,0)$, $(-a,0)$. The result is $$-2t (cos k_x a + cos (k_x \frac{a}{2} +k_y \frac{\sqrt{3}}{2}a) + cos (k_x \frac{a}{2} -k_y \frac{\sqrt{3}}{2}a))$$

Are these calculations correct?

2. Dec 3, 2016