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The common way to discuss BCS superconductivity, as I have seen in several books, is to assume that we have a Fermi liquid, on top of which we add an interaction of the form $$H_{int}=\sum_{p,p^\prime,q} V(q)\psi^\dagger_\uparrow(-p) \psi^\dagger_\downarrow(-p^\prime) \psi_\downarrow(p^\prime+q)\psi_\uparrow(p-q)$$ The assumption is that as long as $V$ is negative we get a Cooper instability and a condensate will form at low enough temperatures. This doesn't make sense to me as if $V\propto 1/q^2$ (in 3D) we just get a correction to the Coulomb interaction and remain with a Fermi liquid.

In that case, what do we actually need to demand on $V(q)$ to get superconductivity? Do we explicitly need a constant term?

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  • $\begingroup$ You mean $V$ is negative, since the interaction is attractive. $\endgroup$ Dec 18 '20 at 14:12
  • $\begingroup$ Sure. I corrected it, thank you. $\endgroup$ Dec 18 '20 at 14:13
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The origin of the interaction term the phonon exchange between the electrons, so its shape is quite different from the Coulomb interaction. In fact, this attraction dominates at long distances, wheres at short distances the Coulomb repulsion remains dominant, which is why the electrons constituting a cooper pair are usually separated by at least a few lattice distances.

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  • $\begingroup$ Thank you, but I am not sure I understand how it explains the answer to my question. In principle I can have all sorts of particles mediating the interaction, right? So I could cook up all sorts of q-dependences. For the second part, I think I understand what you are saying phenomenologically, but how should it work mathemtically? $\endgroup$ Dec 18 '20 at 14:25
  • $\begingroup$ Indeed, one could imagine all kinds of particles, like in QED (incidentally, the math of phonon exchange is nearly the same as that of photon exchange, but with the speed of sound.) However, here we are dealing with a crystal lattice and rather low energies, so phonons is what one really have. $\endgroup$ Dec 18 '20 at 14:29

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