# What's the true ground state of electrons in metals? Cooper Pairing or Fermi surface?

When I learn quantum many-body, Landau Fermi liquid theory told me that even though there are rather strong repulsive interactions between electrons, the true ground state of electrons in metals is still the Fermi surface. And the gapless excitation is the quasiparticle above the Fermi sea.

However the electron-phonon interaction says that there is always an attractive effective interaction between electrons. And Cooper's argument says any small attractive interaction between electrons will result in the instability of Fermi surface. And the true ground state should be Cooper pairing.

So my questions:

1. Given any Hamiltonian $\hat H$, there should only one class of ground states with same lowest energy and the ground state of $\hat H$ is independent of temperature. So what's the true ground state of electrons in metal? Cooper Pairing or Fermi surface?

2. From Cooper's argument, does it mean that any metal (at least elemental metals) can be superconductor in enough low temperature? Because in any metal there is always electron-phonon effective attraction.

3. I don't know whether any metal can be superconductor. But at least many elemental metals ( see this) can become superconductor below some $T_c$. So in principle their true ground state should be Cooper Pairing. But how can Fermi liquid theory still successfully explain them above $T_c$ because the basic assupmtion that ground state is Fermi surface is wrong for them. How to explain this? I'm very puzzled about this question. Every textbook about BCS theory only talk that for BCS Hamitonian there is phase transition from order to disorder across $T_c$. But they do not explain why above $T_c$ it will reappear the Fermi liquid and Fermi surface for metals.

4. Or does it mean that metals can be explained by Fermi Liquid thoery cannot be superconductor? And the the above listed metals which can be superconductor in fact are not Fermi Liquid above $T_c$. However if it's true, how to explain the phase transition of He-$3$ from normal Fermi liquid to cooper pairing?

• I think your essential misunderstanding comes from your first question. At finite energy you need to find the state that minimizes the free energy, not Hamiltonian. In other words, you would have similar trouble understanding the paramagnetic to ferromagnetic transition if you kept thinking along these lines, because strictly speaking the Hamiltonian has a ground state that is ferromsgnetic at all temperatures. – KF Gauss Jul 20 '17 at 4:56
• @user157879 So you mean that Fermi surface is a state that minimize the free energy above $T_c$？ It seems reasonable. How to prove it? – user153663 Jul 20 '17 at 4:59
• (related) physics.stackexchange.com/q/33366 – vik Jul 20 '17 at 5:42
• @vik For BCS Hamitonian, I know above $T_c$ it should be disordered phase. But the crutial problem is how to prove this disordered phase can be discribed by Fermi Liquid Theory which assume the ground state is Fermi surface. Only from BCS Hamitonian which neglects the Coulomb potential, the disorder phase obviously cannot be describled by Fermi Liquid. – user153663 Jul 20 '17 at 8:17
• 1) You should read what vik mentioned, since clearly you misunderstood the statistical aspect of the question. 2) Cooper instability is about 2 electrons on top of the Fermi surface, not about the complete many-body problem which is called the Eliashberg problem. 3) Almost all metallic pure (not alloy) materials are superconducting at sufficiently low temperatures. I do not remember where is the periodic classification showing all superconducting elements, so I said almost all metallic elements. You can check e.g. en.wikipedia.org/wiki/List_of_superconductors to start. – FraSchelle Jul 21 '17 at 5:24