Use the mean field approximation to simplify 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) Apply the mean field approximation.
 * 3) Assume that \(\Delta = g\sum_k \langle g.s.|\hat{c}^{\,}_{k\downarrow}\hat{c}^{\,}_{-k\uparrow}|g.s.\rangle \neq 0\), using the approximation \(g \sum_k \hat{c}^{\,}_{k\prime\downarrow}\hat{c}^{\,}_{-k\prime\uparrow} = \Delta + g\sum_k \hat{c}^{\,}_{k\downarrow}\hat{c}^{\,}_{-k\uparrow} - \Delta\)
 * 4) Substitute this in to the interaction part of the Hamiltonian: \(g\sum \limits_{k,k\prime} \hat{c}^{\dagger}_{k\uparrow}\hat{c}^{\dagger}_{-k\downarrow}\hat{c}^{\,}_{k\prime\downarrow}\hat{c}^{\,}_{-k\prime\uparrow} = \frac{1}{g}\left(\Delta^{*} + g\sum_k \hat{c}^{\dagger}_{k\uparrow}\hat{c}^{\dagger}_{-k\downarrow} - \Delta^{*}\right)\left(\Delta + g\sum_k \hat{c}^{\,}_{-k\downarrow}\hat{c}^{\,}_{k\uparrow} - \Delta\right)\)
 * 5) Expand, using the fact that \(\Delta \gg \sum_k \hat{c}^{\,}_{k\uparrow}\hat{c}^{\,}_{-k\downarrow} - \Delta\) to simplify: \(g\sum \limits_{k,k\prime} \hat{c}^{\dagger}_{k\uparrow}\hat{c}^{\dagger}_{-k\downarrow}\hat{c}^{\,}_{k\prime\downarrow}\hat{c}^{\,}_{-k\prime\uparrow} = \frac{1}{g}\left(\Delta^{*}\Delta + \Delta\left(g\sum_k \hat{c}^{\dagger}_{k\uparrow}\hat{c}^{\dagger}_{-k\downarrow} - \Delta^{*}\right) + \Delta^{*}\left(g\sum_k \hat{c}^{\,}_{-k\downarrow}\hat{c}^{\,}_{k\uparrow} - \Delta\right)\right)\)
 * 6) Subtract out \(\Delta^{*}\Delta\): \(g\sum \limits_{k,k\prime} \hat{c}^{\dagger}_{k\uparrow}\hat{c}^{\dagger}_{-k\downarrow}\hat{c}^{\,}_{k\prime\downarrow}\hat{c}^{\,}_{-k\prime\uparrow} = \frac{1}{g}\left(-|\Delta|^2 + \Delta g\sum_k \hat{c}^{\dagger}_{k\uparrow}\hat{c}^{\dagger}_{-k\downarrow} + \Delta^{*}g\sum_k \hat{c}^{\,}_{-k\downarrow}\hat{c}^{\,}_{k\uparrow}\right)\)
 * 7) Simplify: \(g\sum \limits_{k,k\prime} \hat{c}^{\dagger}_{k\uparrow}\hat{c}^{\dagger}_{-k\downarrow}\hat{c}^{\,}_{k\prime\downarrow}\hat{c}^{\,}_{-k\prime\uparrow} = \frac{-|\Delta|^2}{g} + \Delta \sum_k \hat{c}^{\dagger}_{k\uparrow}\hat{c}^{\dagger}_{-k\downarrow} + \Delta^{*}\sum_k \hat{c}^{\,}_{-k\downarrow}\hat{c}^{\,}_{k\uparrow}\)
 * 8) Substitute into the Hamiltonian: \(\mathcal{H} = \sum \limits_{k, \sigma} \epsilon_k \hat{c}^{\dagger}_{k\sigma} \hat{c}^{\,}_{k\sigma} - \frac{-|\Delta|^2}{g} + \Delta \sum_k \hat{c}^{\dagger}_{k\uparrow}\hat{c}^{\dagger}_{-k\downarrow} + \Delta^{*}\sum_k \hat{c}^{\,}_{-k\downarrow}\hat{c}^{\,}_{k\uparrow}\)
 * 9) Sum over \(\rho = \uparrow, \downarrow\): \(\mathcal{H} - \mu\hat{N} = \sum_k \left(\epsilon_k - \mu\right) \left(\hat{c}^{\dagger}_{k\uparrow} \hat{c}^{\,}_{k\uparrow} + \hat{c}^{\dagger}_{k\downarrow} \hat{c}^{\,}_{k\downarrow} \right) + \Delta \hat{c}^{\dagger}_{k\uparrow}\hat{c}^{\dagger}_{-k\downarrow} - \Delta^{*} \hat{c}^{\,}_{-k\downarrow}\hat{c}^{\,}_{k\uparrow} + \frac{|\Delta|^2}{g}\)
 * 10) Rearrange using (fermionic) commutation relations: \(\mathcal{H} - \mu\hat{N} = \sum_k \left(\epsilon_k - \mu\right) \left(\hat{c}^{\dagger}_{k\uparrow} \hat{c}^{\,}_{k\uparrow} - \hat{c}^{\,}_{k\downarrow} \hat{c}^{\dagger}_{k\downarrow} \right) + \Delta \hat{c}^{\dagger}_{k\uparrow}\hat{c}^{\dagger}_{-k\downarrow} - \Delta^{*} \hat{c}^{\,}_{-k\downarrow}\hat{c}^{\,}_{k\uparrow} + \sum_k \left(\epsilon_k - \mu\right) + \frac{|\Delta|^2}{g}\)
 * 11) Write this in matrix form: \(\mathcal{H} - \mu\hat{N} = \sum_k \begin{pmatrix} \hat{c}^{\dagger}_{k\uparrow} & \hat{c}^{\,}_{-k\downarrow} \end{pmatrix} \begin{pmatrix} \epsilon_k - \mu & -\Delta \\ -\Delta^{*} & -\epsilon_k + \mu \end{pmatrix} \begin{pmatrix} \hat{c}^{\,}_{k\uparrow} \\ \hat{c}^{\dagger}_{-k\downarrow} \end{pmatrix} + \sum_k \left(\epsilon_k - \mu\right) + \frac{|\Delta|^2}{g}\)