# Why is the Fermi surface stable?

As a condensed matter physicist, I take it for granted that a Fermi surface is stable.

But it is stable with respect to what?

For instance, Cooper pairing is known as an instability of the Fermi surface.

I'm simply wondering what makes the Fermi surface stable?

Possible way of thinking: Is it a topological property of the Fermi gas (only of the free one ?, only robust against disorder?)? What is the modern, mathematical definition of the Fermi surface (shame on me, I don't even know this, and all my old textbooks are really sloppy about that, I feel)? What can destroy the Fermi surface, and what does destroy mean?

Any idea / reference / suggestion to improve the question is welcome.

Addenda / Other possible way to discuss the problem: After writing this question, I noted this answer by wsc, where (s)he presents a paper by M. Oshikawa (2000), Topological Approach to Luttinger’s Theorem and the Fermi Surface of a Kondo Lattice PRL 84, 3370–3373 (2000) (available freely on arXiv), and a paper by J. Luttinger & J. Ward Ground-State Energy of a Many-Fermion System. II. Phys. Rev. 118, 1417–1427 (1960). An other interesting reference to start with is a paper by J. Luttinger, Fermi Surface and Some Simple Equilibrium Properties of a System of Interacting Fermions, Phys. Rev. 119, 1153–1163 (1960), where he shows (eq.33) that the volume of the Fermi surface is conserved under interaction, using analytic properties of the Green function including the self-energy as long as the total number of particles is conserved. I'm not sure if it's sufficient to proof the stability of the Fermi surface (but what does stability means exactly, I'm now confused :-p ) Is there absolutely no modern (topological ?) version of this proof ?

• Joe Polchinski's lecture notes, arxiv.org/abs/hep-th/9210046 ("Effective Field Theory and the Fermi Surface"), are good for explaining why Cooper pairing is the only instability you usually have to worry about. Everything else is an irrelevant operator. – Matt Reece Jun 27 '13 at 14:23
• @MattReece Thanks a lot for the reference. I just finnish to read it. I particularly loved the maginality of the Cooper pairing. But still, Polchinski does not really explain the stability for other interaction than phonons. I nevertheless think your comment deserves to become an answer. – FraSchelle Jun 27 '13 at 22:28
• Suppose there is a periodic potential. The Fermi surface would change, right? – poisson Jun 20 '16 at 4:35

There are answers in the note by Polchinski linked by Matt, and an article by Shankar in Review of Modern Physics: Renormalization-group approach to interacting fermions. Just to flesh out was it meant by "stability" and "Fermi surface". The Fermi-liquid can be thought of as a phase characterized by several properties: arbitrarily long-lived, gapless electron-like excitations, preservation of various symmetries, the presence of the discontinuity that characterizes the Fermi surface, and in the end by a certain analytic structure of the correlators as elucidated by Landau.

We know that the free electron gas is in this phase, in a trivial way. If we start adding interaction what happens? In the usual sense, we want to know if this phase is stable - that is: if we add an arbitrarily small interaction of some kind will we change the phase at zero temperature? Note that because of the Fermi surface there are an infinite number of different interactions. As the articles show the "normal" interactions do not change the phase. However the "pairing" interactions change the phase at zero temperature, even when they are arbitrarily small. This you know from BCS theory already - the superconductor is the ground state for all attractive interaction, regardless how weak (although the transition temperature goes to zero rapidly with interaction strength).

A couple more points: the Fermi surface can be unstable to large values of interactions such as the Pomeranchuk instability (unless I'm getting the names confused), or because of particular geometric structures like nesting Fermi surfaces. This is somewhat different from the question of: "is the Fermi liquid generally stable?"

You ask about disorder: This is a technical topic which I'm not expert in, but my understanding is that the appropriately defined disordered Fermi-liquid is stable in 3-dimensions (i.e. it takes a finite amount of disorder to turn it to an insulator). See for example this paper by Basko, Aleiner and Altshuler.

• Thanks for your answer, and thanks for the edit of the question, too :-) I've read the Shankar review long ago, thanks for pointing toward this paper, I forget what's in. I'll have a look later. When I asked the question, I was thinking that the Fermi surface should mathematically be defined as a discontinuity of $n_{k}(T=0)$ (mode occupation). Then, immediately, I thought: "hey, what about interaction, disorder, superconductors, temperature... ? Does it make sense to define something as a discontinuity ?" etc... Thanks again for your answer :-) – FraSchelle Jun 27 '13 at 21:32
• (cont.) It's really interesting to compare a Fermi surface with a phase. But I'm not sure if it really helps me. I should then say: "what is the phase transition mechanism ?" I'll try to check on Shankar paper. Thanks again. – FraSchelle Jun 27 '13 at 21:37
• @Oaoa: But to be clear, it is a phase, as it is characterized properties by which cannot be changed continuously. A system either has long-lived, gapless, electronic quasi-particles or it does not. I mean we usually think of the mechanisms that take us out of the FL phase, but those should all work in reverse. – BebopButUnsteady Jun 27 '13 at 21:57
• Also, I think fundamentally you should characterize the FL by the analytic structure of the correlators, rather than the discontinuity. That would tell you that there could be a disorded FL even though there is no momentum conservation and so no real FS. And the usual definition as a discontinuity of $n_k(T=0)$ comes for free. But temperature is a good question. Can you go continuously from a $T=0$ Fermi-liquid to a $T=0$ non Fermi-liquid by going through finite temperature? – BebopButUnsteady Jun 27 '13 at 21:57
• Oh, ok, I get your point ! You discuss the Fermi surface as the maximal $k$-state occupied in the Fermi liquid, am I correct ? I need to refresh my memory about the details of the FL then (is there really a phase transition towards the FL ? is it really a generic behaviour of any fermionic liquid ? and some other stupid questions like these ones ... I remember it's valid at low density... well, anyway, tomorrow is an other day :-). Ok, ok, I progress... Thanks a lot :-) – FraSchelle Jun 27 '13 at 23:00

2) One can indeed classify Fermi surfaces using topological arguments, see http://arxiv.org/abs/hep-th/0503006, and also G. Volovik The universe in a Helium droplet'', available for free on his homepage at Aalto university).