Here is what I understood about Cooper pairs:
We have the "general" Hamiltonian describing electrons in a crystal :
$$ H =\sum_{k,\sigma} E_k \hat{a}^{\dagger}_{k, \sigma}\hat{a}_{k, \sigma} -|g_\mathrm{eff}|^2 \sum_{k_1, k_2, \sigma_1, \sigma_2} \hat{a}^{\dagger}_{k1+q, \sigma_1}\hat{a}^{\dagger}_{k2-q, \sigma_2} \hat{a}_{k2, \sigma_2}\hat{a}_{k1, \sigma_1} $$
Well it is not the most general as in fact the interaction term could have a dependence but it is the "almost general" Hamiltonian often used (by almost general I want to say that excepted the non dependency of $g_\mathrm{eff}$ we didn't do any other approximation).
When we look for eigenstates of this Hamiltonian, we can look for a general solution with an $N$-body wavefunction but it is too "hard" in this case.
So we can see what happens for a 2 body wavefunctions: $|\Psi\rangle=\sum_{k_1, k_2, \sigma_1, \sigma_2} \alpha_{k_1, k_2, \sigma_1, \sigma_2} \hat{a}^{\dagger}_{k_1, \sigma_1}\hat{a}^{\dagger}_{k_2, \sigma_2}|0\rangle$
We add three other assumptions: the spins of the electrons considered are opposed, they have opposed wavevectors and we assume that they are above the Fermi level.
The given Hamiltonian, and these assumptions on the wavefunction is the Cooper pair problem.
So the wavefunction we finally look for is :
$$|\Psi\rangle=\sum_{k,\sigma} \alpha_{k, \sigma} \hat{a}^{\dagger}_{k, \sigma}\hat{a}^{\dagger}_{-k, -\sigma}|0\rangle,$$
with the constraint that
$$||k||\geq ||k_F||.$$
And when we solve the eigenvalue problem we find that the energy associated to the wavefunction will be $E < 2E_F$, which suggests that it is energy-favorable to put two electrons above the Fermi Sea and to couple them via the electron-phonon interaction.
First question: do you agree with what I said?
Second question: How is it possible to have electrons "initially" at $E>E_F$, it is not clear for me...
