Find the ground state of the BCS superconductor


 * 1) Identify the state which is annihilated by all of the \(\hat{\alpha}_{k\sigma}\) operators: \(|g.s.\rangle = \prod_k \hat{\alpha}_{k\uparrow}\hat{\alpha}_{-k\downarrow}|0\rangle\)
 * 2) Transform this state with the Bogoliubov transformation: \(|g.s.\rangle = \prod_k \left(u\hat{c}^{\,}_{k\uparrow} + v\hat{c}^{\dagger}_{-k\downarrow}\right)\left(-v\hat{c}^{\dagger}_{k\uparrow} + u\hat{c}^{\,}_{-k\downarrow}\right)|0\rangle\)
 * 3) Multiply out the brackets: \(|g.s.\rangle = \prod_k \left(-uv\hat{c}^{\,}_{k\uparrow}\hat{c}^{\dagger}_{k\uparrow} - v^2\hat{c}^{\dagger}_{-k\downarrow}\hat{c}^{\dagger}_{k\uparrow} + u^2\hat{c}^{\,}_{k\uparrow}\hat{c}^{\,}_{-k\downarrow} + uv\hat{c}^{\dagger}_{-k\downarrow}\hat{c}^{\,}_{-k\downarrow}\right)|0\rangle\)
 * 4) Rearrange using the commutation relation: \(|g.s.\rangle = \prod_k \left(-uv\left(1 - \hat{c}^{\dagger}_{k\uparrow}\hat{c}^{\,}_{k\uparrow}\right) - v^2\hat{c}^{\dagger}_{-k\downarrow}\hat{c}^{\dagger}_{k\uparrow} + u^2\hat{c}^{\,}_{k\uparrow}\hat{c}^{\,}_{-k\downarrow} + uv\hat{c}^{\dagger}_{-k\downarrow}\hat{c}^{\,}_{-k\downarrow}\right)|0\rangle\)
 * 5) Ignore all terms with an annihilation operator on the left: \(|g.s.\rangle = \prod_k \left(- uv - v^2\hat{c}^{\dagger}_{-k\downarrow}\hat{c}^{\dagger}_{k\uparrow}\right)|0\rangle\)