Example 5.1: Single s band

$$H = \sum_{i \neq 0}^{N-1} e^{i{\bf k}.{\bf R}_i} \gamma({\bf R}_i) + \epsilon$$ (199)

$$S = 1+\sum_{i\neq 0}^{N-1} e^{i{\bf k}.{\bf R}_i} \alpha({\bf R}_i)$$ (200)

$$\epsilon(k) = \frac{H}{S} = \frac{\epsilon + \sum_{i \neq 0}^{N-1} e^{i{\bf k}.... ...a({\bf R}_i)} {1+\sum_{i\neq 0}^{N-1} e^{i{\bf k}.{\bf R}_i} \alpha({\bf R}_i)}$$ (201)

Going back to the linear chain, we find that the conditions we need to impose to recover the previous results are:

$$\gamma = -t {\mathrm for} R_i = \pm a; \alpha = 0$$ (202)

An then we obtain

$$\epsilon(k) = \epsilon - 2t \cos{(ka)}.$$ (203)