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:
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?
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.
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.
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?