Find the gap energy \(\Delta\) of the BCS superconductor


 * 1) Write the gap energy in terms of the BCS ground state: \(\Delta=g \sum_k \langle g.s. |\hat{c}_{-k\downarrow}\hat{c}_{k\uparrow}|g.s.\rangle\)
 * 2) Diagonalise with the Bogoliubov transformation: \(\Delta=g \sum_k \langle g.s. |\left(v\hat{\alpha}^{\dagger}_{k\uparrow}+u\hat{\alpha}_{-k\downarrow}\right)\left(u\hat{\alpha}_{k\uparrow}-v\hat{\alpha}^{\dagger}_{-k\downarrow}\right)|g.s.\rangle\)
 * 3) Expand out the brackets: \(\Delta=g \sum_k \langle g.s. |\left(uv\hat{\alpha}^{\dagger}_{k\uparrow}\hat{\alpha}_{k\uparrow} - v^2\hat{\alpha}^{\dagger}_{k\uparrow}\hat{\alpha}^{\dagger}_{-k\downarrow} + u^2 \hat{\alpha}_{-k\downarrow} \hat{\alpha}_{k\uparrow} - uv \hat{\alpha}_{-k\downarrow}\hat{\alpha}^{\dagger}_{-k\downarrow} \right)|g.s.\rangle\)
 * 4) Ignore all terms with a creation operator on the left or an annihilation operator on the right, since all of these evaluate to zero: \(\Delta=g \sum_k \langle g.s. | - uv \hat{\alpha}_{-k\downarrow}\hat{\alpha}^{\dagger}_{-k\downarrow}|g.s.\rangle\)
 * 5) Make the substitution \(u = \cos{\,\theta_k}\), \(v=\sin{\,\theta_k}\): \(\Delta=g \sum_k \langle g.s. | -\sin{\,\theta_k}\cos{\,\theta_k} \hat{\alpha}_{-k\downarrow}\hat{\alpha}^{\dagger}_{-k\downarrow}|g.s.\rangle\)
 * 6) Use the standard commutator to rearrange: \(u = \cos{\,\theta_k}\), \(v=\sin{\,\theta_k}\): \(\Delta=g \sum_k \langle g.s. | -\sin{\,\theta_k}\cos{\,\theta_k} \left(1 - \hat{\alpha}^{\dagger}_{-k\downarrow}\hat{\alpha}_{-k\downarrow}\right)|g.s.\rangle\)
 * 7) Since the annihilation operator now acts on the ground state, ignore: \(\Delta= -g \sum_k \sin{\,\theta_k}\cos{\,\theta_k} \langle g.s. |g.s.\rangle = -g \sum_k \sin{\,\theta_k}\cos{\,\theta_k} \)
 * 8) Use the trigonometric identity \(\sin{\,\theta}\cos{\,\theta} = \frac{1}{2}\sin{\,2\theta}\) to simplify: \(\Delta=-\frac{g}{2} \sum_k \sin{\,2\theta_k}\)
 * 9) Substitute in the standard identity for the Bogoliubov transformation, \(\sin{\,2\theta_k} = \frac{-\Delta}{\sqrt{\Delta^2 + \left(\epsilon_k - \mu^2\right)}}\): \(\Delta=\frac{g}{2} \sum_k \frac{\Delta}{\sqrt{\Delta^2 + \left(\epsilon_k - \mu^2\right)}}\)