Unstability of Fermi surface in superconductors & Cooper problem

Question

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...?

Answer 1

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.

Answer 2

Let me try to explain what I understand.

(1) Here $E_F$ is the energy of an electron sitting on the Fermi surface, and therefore it is the minimal (not the maximal!) energy if you add two free electrons on top of the Fermi surface.

According to FraSchele's comments, it should be emphasized that the system before introducing the electron pair is an interacting Fermi-liquid, rather a non-interacting Fermi gas. One should mind the difference. For the latter, one indeed has $E_f=\frac{h^2 k_F^2}{2m}$. However, although the derivation provided in the cited textbook uses the characteristic value of the density of Fermi-gas around (10.27b), the specific values of Fermi energy and/or density of state near Fermi are not essential in the proof addressed below in point (4).

(2) The electrons inside the Fermi surface are not considered as free, since they are mostly irrelevant to the conductivity. Their microscopic motions mostly cancel out and do not contribute to the measurable macroscopic current. In this context, the electron pair in question is considered free.

(3) For instance, in the textbook Solid State Physics by Itach and Luth, in section 10.3, it is argued that there might be some small effective attraction between electrons, and the interaction is only significant for free electron pairs with opposite momenta. Therefore, let us assume that the remaining electrons on the Fermi surface, while being subjected to other interactions, do not feel this type of attraction.

(4) Consequently, the above textbook provides a proof which shows that no matter how small the attraction between electrons is, the electron pair with opposite momentum may form a state whose energy is lower than that of Fermi energy.

(5) The above result is not trivial. If there is no interaction, and since the free two electrons are above the Fermi surface, their energy satisfies $E>2E_f$. Now, of course, attraction will always lower the energy, but there is a competition between the two causes, and it is not obvious why the attraction always wins and leads eventually $E<2E_f$ for interacting electron pair.

(6) The existence of the two-electron bound state, namely, Cooper pair, indicates that they are more stable in comparison to the other electrons on the Fermi surface.

(7) Regarding FraSchelle's answer, I think here the concept of instability is not exactly the same as that in classical mechanics. In the latter case, instability mostly refers to linear instability. It implies that any small perturbation will be amplified exponentially in time, as long as the context of small perturbation is still valid. Here for Cooper pair, to me, it only means the appearance of a bound state.

Edit

The points (1) and (3) above are modified according to FraSchelle's comments.