The Fermi Energy is defined as the highest energy occupied at $T=0$ by a system composed of Fermions.
When we study the Cooper problem as an introduction to BCS theory, we say: let's put a pair of electrons at the energy above the Fermi energy, thus with $k > k_F$.
We are more precisely looking for an eigenstate : $ | \Psi \rangle = \sum_{\boldsymbol{k}} \alpha_\boldsymbol{k} a^\dagger_{\boldsymbol{k}\uparrow}a^\dagger_{-\boldsymbol{k}\downarrow}|0\rangle$ with $\alpha_\boldsymbol{k}=0$ if $k<k_F$.
We do some calculations and we find using the Hamiltonian with the electron-phonon interaction term that the $| \Psi \rangle$ is an eigenstate of the hamiltonian with $E<2 E_F$.
And we say : so the Fermi sea is unstable as it is favorable to put two electrons above the Fermi energy.
What I don't understand :
What is precisely the $E_F$ energy we are talking about? Is it $\frac{h^2 k_F^2}{2m}$: the maximum energy of free electrons? But electrons are not free here as we have electron-phonon interaction. Why are we looking at the maximum energy of free electrons if we are studying an hamiltonian with interaction? This shouldn't be a relevant quantity…?
Best Answer
That's precisely what an instability is about... The Fermi surface of free particles is usually stable. But if you have attractive interaction between the free electrons it becomes instable. Cooper showed that the Fermi surface is unstable when some attractive electron-phonon interaction are present. So in fact the Fermi surface doesn't exist in superconductors.
The strategy is the following: you take a Hamiltonian whose ground state is known [the free electron gas for the Cooper problem, with known Fermi surface]. Then you switch on some interaction [the attractive electron-phonon interaction] and you ask whether the ground state of the two systems -- with or without the interaction -- is the same. In the case of the Cooper problem the two systems have not the same ground state.
It is nevertheless difficult to know how to write the ground state of the interacting system, so you're happy already knowing that the ground state of the interacting system has a lesser energy than the non-interacting one. It simply means that, whenever the attractive interaction is present, the system will favour a ground state which is not a Fermi surface [or a free electron gas in the Cooper problem].
In the Cooper problem, this new ground state is the Cooper fluid, or BCS ground state. Its approximate (= mean-field) form has been postulated by Bardeen, Cooper and Schrieffer a few months after Cooper understood that the Fermi surface is not the ground state of a superconductor.
Historically, the Cooper problem is important, since it clearly demonstrated that the Fermi surface is not always the ground state of interacting fermions. Before Cooper, it was commonly believed that the Fermi surface is a stable object, following some arguments by Luttinger, who showed that the Fermi surface can still be defined even with interaction. Of course, what Luttinger forget was the possibility for phase transition.