Use the Bogoliubov transformation to diagonalise the BCS superconductivity Hamiltonian


 * 1) Identify the BCS Hamiltonian: \(\mathcal{H} = \sum \limits_{k, \sigma} \epsilon_k \hat{c}^{\dagger}_{k\sigma} \hat{c}^{\,}_{k\sigma} - g\sum \limits_{k,k\prime} \hat{c}^{\dagger}_{k\uparrow}\hat{c}^{\dagger}_{-k\downarrow}\hat{c}^{\,}_{k\prime\downarrow}\hat{c}^{\,}_{-k\prime\uparrow}\)
 * 2) Use the mean field approximation to approximate the Hamiltonian in matrix form, as described here: \(\mathcal{H} - \mu\hat{N} = \sum_k \begin{pmatrix} \hat{c}^{\dagger}_{k\uparrow} & \hat{c}^{\,}_{-k\downarrow} \end{pmatrix} \begin{pmatrix} \epsilon_k - \mu & -\Delta \\ -\Delta^{*} & -\epsilon_k + \mu \end{pmatrix} \begin{pmatrix} \hat{c}^{\,}_{k\uparrow} \\ \hat{c}^{\dagger}_{-k\downarrow} \end{pmatrix} + \sum_k \left(\epsilon_k - \mu\right) + \frac{|\Delta|^2}{g}\)
 * 3) Apply the Bogoliubov transformation \(\begin{pmatrix} \hat{\alpha}^{\dagger}_{k\uparrow} & \hat{\alpha}^{\,}_{-k\downarrow} \end{pmatrix} = \begin{pmatrix}\cos{\,\theta_k} & \sin{\,\theta_k} \\ -\sin{\,\theta_k} & \cos{\,\theta_k}\end{pmatrix} \begin{pmatrix} \hat{c}^{\dagger}_{k\uparrow} \\ \hat{c}^{\,}_{-k\downarrow} \end{pmatrix}\): \(\mathcal{H} - \mu\hat{N} = \sum_k \begin{pmatrix} \hat{\alpha}^{\dagger}_{k\uparrow} & \hat{\alpha}^{\,}_{-k\downarrow} \end{pmatrix} \begin{pmatrix}\cos{\,\theta_k} & \sin{\,\theta_k} \\ -\sin{\,\theta_k} & \cos{\,\theta_k}\end{pmatrix}\begin{pmatrix} \epsilon_k - \mu & -\Delta \\ -\Delta^{*} & -\epsilon_k + \mu  \end{pmatrix}\begin{pmatrix}\cos{\,\theta_k} & -\sin{\,\theta_k} \\ \sin{\,\theta_k} & \cos{\,\theta_k}\end{pmatrix} \begin{pmatrix} \hat{\alpha}^{\,}_{k\uparrow} \\ \hat{\alpha}^{\dagger}_{-k\downarrow} \end{pmatrix} + \sum_k \left(\epsilon_k - \mu\right) + \frac{|\Delta|^2}{g}\)
 * 4) Multiply out the matrices, using the identities \(\cos^2{\,x} - \sin^2{\,x} = \cos{\,2x}\) and \(2\cos{\,x} \,\sin{\,x} = \sin{\,2x}\): \(\mathcal{H} - \mu\hat{N} = \sum_k \begin{pmatrix} \hat{\alpha}^{\dagger}_{k\uparrow} & \hat{\alpha}^{\,}_{-k\downarrow} \end{pmatrix} \begin{pmatrix} \left(\epsilon_k - \mu\right)\cos\left(2\theta_k\right) - \frac{1}{2}\sin\left(2\theta_k\right)\left(\Delta + \Delta^*\right) & 0 \\ 0 & \left(\epsilon_k - \mu\right)\cos\left(2\theta_k\right) + \frac{1}{2}\sin\left(2\theta_k\right)\left(\Delta + \Delta^*\right)  \end{pmatrix} \begin{pmatrix} \hat{\alpha}^{\,}_{k\uparrow} \\ \hat{\alpha}^{\dagger}_{-k\downarrow} \end{pmatrix} + \sum_k \left(\epsilon_k - \mu\right) + \frac{|\Delta|^2}{g}\)
 * 5) Use the identities \(\cos{\,2\theta_k} = \frac{\epsilon_k - \mu}{\sqrt{\Delta^2 + \left(\epsilon_k - \mu\right)^2}}\) and \(\sin{\,2\theta_k} = \frac{\Delta}{\sqrt{\Delta^2 + \left(\epsilon_k - \mu\right)^2}}\) to simplify: \(\mathcal{H} - \mu\hat{N} = \sum_k \frac{1}{\sqrt{\Delta^2 + \left(\epsilon_k - \mu\right)^2}} \begin{pmatrix} \hat{\alpha}^{\dagger}_{k\uparrow} & \hat{\alpha}^{\,}_{-k\downarrow} \end{pmatrix} \begin{pmatrix} \left(\epsilon_k - \mu\right)^2 - \frac{\Delta}{2}\left(\Delta + \Delta^*\right) & 0 \\ 0 & \left(\epsilon_k - \mu\right)^2 + \frac{\Delta}{2}\left(\Delta + \Delta^*\right)  \end{pmatrix} \begin{pmatrix} \hat{\alpha}^{\,}_{k\uparrow} \\ \hat{\alpha}^{\dagger}_{-k\downarrow} \end{pmatrix} + \sum_k \left(\epsilon_k - \mu\right) + \frac{|\Delta|^2}{g}\)
 * 6) Multiply out the matrix: \(\mathcal{H} - \mu\hat{N} = \sum_k \sqrt{\Delta^2 + \left(\epsilon_k - \mu\right)^2}\left(\hat{\alpha}^{\dagger}_{k\uparrow}\hat{\alpha}_{k\uparrow} - \hat{\alpha}^{\,}_{-k\downarrow}\hat{\alpha}^{\dagger}_{k\downarrow}\right) + \sum_k\frac{|\Delta|^2 - \Delta^2}{2}\left(\hat{\alpha}^{\dagger}_{k\uparrow}\hat{\alpha}_{k\uparrow} - \hat{\alpha}^{\,}_{-k\downarrow}\hat{\alpha}^{\dagger}_{k\downarrow}\right) + \sum_k \left(\epsilon_k - \mu\right) + \frac{|\Delta|^2}{g}\)
 * 7) Simplify with the assumption that \(\frac{|\Delta|^2 - \Delta^2}{2}\hat{\alpha}^{\dagger}_{k\uparrow}\hat{\alpha}_{k\uparrow} + \frac{|\Delta|^2 - \Delta^2}{2}\hat{\alpha}^{\,}_{-k\downarrow}\hat{\alpha}^{\dagger}_{k\downarrow} \approx \sqrt{\Delta^2 + \left(\epsilon_k - \mu\right)^2}\): \(\mathcal{H} - \mu\hat{N} = \sum_k \sqrt{\Delta^2 + \left(\epsilon_k - \mu\right)^2}\left(\hat{\alpha}^{\dagger}_{k\uparrow}\hat{\alpha}_{k\uparrow} - \hat{\alpha}^{\,}_{-k\downarrow}\hat{\alpha}^{\dagger}_{k\downarrow}\right) + \sum_k \left(\epsilon_k - \mu - \sqrt{\Delta^2 + \left(\epsilon_k - \mu\right)^2}\right) + \frac{|\Delta|^2}{g}\)
 * 8) Tidy up by summing over the spins: \(\mathcal{H} - \mu\hat{N} = \sum \limits_{k,\sigma} \sqrt{\Delta^2 + \left(\epsilon_k - \mu\right)^2}\left(\hat{\alpha}^{\dagger}_{k\sigma}\hat{\alpha}^{\,}_{k\sigma}\right) + \sum_k \left(\epsilon_k - \mu - \sqrt{\Delta^2 + \left(\epsilon_k - \mu\right)^2}\right) + \frac{|\Delta|^2}{g}\)