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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...?

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  • $\begingroup$ Even with electron phonon interaction the Fermi surface is usually stable, this is called the Luttinger theorem. And Landau liquid theory tells you that the electron gas with phonon interaction, electromagnetic interaction (screening and co), behaves as a free electron gas. Only Cooper instability realises the instability of the Fermi surface. All relevant reference are in the post physics.stackexchange.com/q/69358/16689 about the stability of the Fermi surface. In that case, one needs to compare the energy of the electron-phonon interacting system (behaving like a free gas) (...) $\endgroup$ – FraSchelle Oct 14 '17 at 3:51
  • $\begingroup$ (...) and the other possible one, that is, the superconducting state in case of a Cooper instability. $\endgroup$ – FraSchelle Oct 14 '17 at 3:52
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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.

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  • $\begingroup$ I think I don't understand. If I have an interaction term, my electrons are no longer free so the Fermi surface will not be relevant at all. It would be like to try to use Fermi Dirac distribution for bosons (to give an image of my misunderstanding). " In the case of the Cooper problem the two systems have not the same ground state." In fact without doing any calculation I don't see how they could be the same as the Hamiltonian is different. $\endgroup$ – StarBucK Apr 18 '17 at 10:07
  • $\begingroup$ Fermi surface is not an easy concept, since it is topologically stable, see physics.stackexchange.com/q/69358/16689 In any case instability are not interesting if you consider too restricted properties like the energy ground state $E_{0}$. Surely $E_{0}$ will change anytime you will add any small perturbation to the system. In general one talks about instability when a global property of the system is no more realised, as e.g. the surface of a classical fluid, or the Fermi surface of a degenerate fermionic gas. $\endgroup$ – FraSchelle Apr 18 '17 at 15:27
  • $\begingroup$ Perhaps you should first read about instability in classical systems, then about Fermi systems, before trying to understand the instability of the Fermi surface. $\endgroup$ – FraSchelle Apr 18 '17 at 15:27
  • $\begingroup$ @FraSchelle As I understand, in classical mechanics, if a configuration (for instance, a steady solution) is unstable, any small perturbation will be amplified exponentially in time (of course, when the context of small perturbation is still valid) therefore the configuration is usually dramatically affected. I guess it is not the same concept here when one say the Fermi surface is unstabilized(?) I think StarBuck was arguing if there is an attractive interaction, it is natural the calculated eigenstate has smaller energy then that without the interaction. Why the textbook calls that instable? $\endgroup$ – gamebm Oct 12 '17 at 23:12
  • $\begingroup$ @gamebm Because it is. It is not stable. Stable always means that the system is in the minimal energy state. Now, both in classical and quantum physics, you can change the minimal energy state under interaction. The interaction is what you call the perturbation in classical mechanics. To get the amplification you're talking about, you need at least a non-linear term in your description. For the Cooper problem, this non-linearity is the electron-phonon interaction. When you linearise this non-linearity, you get the exponential amplification towards the Cooper condensate. $\endgroup$ – FraSchelle Oct 14 '17 at 2:25
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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.

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  • $\begingroup$ (1) The initial gas is not the free electron gas, it is the Landau liquid. It has all properties of the free electron gas, especially the Fermi surface still exists, whatever interaction you may add to your description ... except a Cooper pairing. (7) The appearance of a bound state on top of the Fermi liquid due to attractive interaction mediated by phonon goes along with a lower energy : the two concepts are strictly equivalent in this context. Note the Cooper problem is about two electrons on top of a Fermi gas, not about a genuine Fermi gas / Landau liquid. $\endgroup$ – FraSchelle Oct 14 '17 at 2:36
  • $\begingroup$ For the Cooper problem, you can realise the two electrons lower their energy due to the interaction, resulting in a bound state of these two electrons. In the superconducting problem (namely a fermionic gas with attractive interaction in the mean field approximation), you no more have bound state, you have correlations between the electronic modes. These correlations lower the energy, and grow up as an order parameter. Hence the notion of phase transition. $\endgroup$ – FraSchelle Oct 14 '17 at 2:39
  • $\begingroup$ @FraSchelle Thanks for the comments! I modified the text accordingly. $\endgroup$ – gamebm Oct 15 '17 at 1:56

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