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Physics

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  • $\begingroup$ But in the usual (textbook) treatment of BCS, or weakly interacting superconductors, the Coulomb interaction is simply negleccted, as in the BCS Hamiltonian that BCS wrote down, and one only considers Cooper pair condensation as the signature of forming a superconductor. $\endgroup$
    – cx1114
    Commented Jun 28 at 15:15
  • $\begingroup$ Sure, Coulomb interaction is not needed for the formation of Cooper pair condensate. But that's not related to the presence/absence of the Goldstone mode mass. I would say, Cooper pair condensation is not the whole story of superconductivity: superconductor is essentially a Cooper pair condensate of charged particles. And the physical properties of superconductor are largely determined by the interaction with EM field. Take Meissner effect as an example: it is present just because the EM field acquires mass in a superconductor. $\endgroup$
    – E. Anikin
    Commented Jun 28 at 15:40