Diagonalise the cubic lattice tight-binding Hamiltonian


 * 1) Identify the cubic lattice tight-binding Hamiltonian, where \(\mathbf{e}_i\) denotes nearest neighbours: \(\hat{\mathbf{H}}^{(0)}=-\sum \limits_{\langle \mathbf{x}_i, \mathbf{e}_i \rangle} t \hat{c}^{\dagger}_{\mathbf{x}_i + \mathbf{e}_i}\hat{c}_{\mathbf{x}_i}\)
 * 2) Take the Fourier transform of this expression: \(\hat{\mathbf{H}}^{(0)}=-t\sum \limits_{\langle \mathbf{x}_i, \mathbf{e}_i \rangle}\sum \limits_{\langle \mathbf{k}, \mathbf{q} \rangle} \frac{1}{V}e^{i\mathbf{k}\left(\mathbf{x}_i + \mathbf{e}_i\right)}\hat{c}^{\dagger}_{\mathbf{k}}e^{-i\mathbf{k}\mathbf{x}_i}\hat{c}_{\mathbf{k}}\)
 * 3) Simplify: \(\hat{\mathbf{H}}^{(0)}=-t\sum \limits_{\langle \mathbf{k}, \mathbf{q} \rangle} \hat{c}^{\dagger}_{\mathbf{k}} \hat{c}_{\mathbf{k}} \sum \limits_{\langle \mathbf{x}_i, \mathbf{e}_i \rangle}\frac{1}{V}e^{i\mathbf{x}_i\left(\mathbf{k} - \mathbf{q}\right)}e^{i\mathbf{k}\mathbf{e}_i}\)
 * 4) Substitute in the identity \(\frac{1}{V}e^{i\mathbf{x}_i\left(\mathbf{k} - \mathbf{q}\right)} = \delta\left(\mathbf{k} - \mathbf{q}\right)\): \(\hat{\mathbf{H}}^{(0)}=-t\sum \limits_{\langle\mathbf{k}, \mathbf{q} \rangle} \hat{c}^{\dagger}_{\mathbf{k}} \hat{c}_{\mathbf{k}} \sum \limits_{\langle \mathbf{x}_i, \mathbf{e}_i \rangle}\delta\left(\mathbf{k} - \mathbf{q}\right)e^{i\mathbf{k}\mathbf{e}_i}\)
 * 5) Use the delta function to equate \(\mathbf{k}\) and \(\mathbf{q}\): \(\hat{\mathbf{H}}^{(0)}=-t\sum \limits_{\mathbf{k}} \hat{c}^{\dagger}_{\mathbf{k}} \hat{c}_{\mathbf{k}} \sum \limits_{\mathbf{e}_i}e^{i\mathbf{k}\mathbf{e}_i}\)
 * 6) Create a new operator \(\hat\epsilon_{\mathbf{k}} = \sum \limits_{\mathbf{e}_i}e^{i\mathbf{k}\mathbf{e}_i}\): \(\hat{\mathbf{H}}^{(0)}=-t\sum \limits_{\mathbf{k}} \hat{c}^{\dagger}_{\mathbf{k}} \hat{c}_{\mathbf{k}} \hat\epsilon_{\mathbf{k}}\)