In BCS theory they break particle number conservation and show the existence of a gap, which would explain why groundstate properties stay relatively the same even for higher temperatures (until beta is of the order of the gap).

However, as long as you don't couple to the electromagnetic field, there is no Higgs mechanism and therefore there are also massless excitations along with the Cooper pairs (these massless excitations are fluctuations in the order parameter https://arxiv.org/abs/cond-mat/0607493)

But without such a gap (so without the anderson mechanism), why are groundstate properties still kinda invariant at nonzero temperature? Or maybe better, can you analytically show that after applying a simple potential difference you will have infinite current even with the goldstone-mode corrections?

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    $\begingroup$ "as long as you don't couple to the electromagnetic field" ... superconductors are coupled to the EM field by definition $\endgroup$ – d_b Mar 29 '19 at 18:42
  • $\begingroup$ I can just write down an effective theory for a superconductor with only electron-electron interactons (due to em) and electron-phonen interactions. This is precisely the starting point in the original bcs paper and no higgs mechanism was even mentioned. Starting from this hamiltonian you apparantly already have superconductivity, and I want to understand why. $\endgroup$ – camel Mar 30 '19 at 13:28
  • $\begingroup$ I think you might be conflating the superconducting gap to excited states of single particles and excited pairs with the gap in the photon spectrum that develops because of the Higgs mechanism. The spectrum of the BCS hamiltonian has a gap whether you couple to external EM field or not, whereas you only observe the photon gap by applying a field. $\endgroup$ – d_b Mar 30 '19 at 22:20
  • $\begingroup$ "The spectrum of the BCS hamiltonian has a gap whether you couple to external EM field or not" I don't think this is true. You have to break U(1) and without EM field there will always be a goldstone mode. It also contradicts the answers on this question : physics.stackexchange.com/questions/294618/… "The BCS excitation spectrum is completely wrong" $\endgroup$ – camel Mar 31 '19 at 14:27

The Goldstone modes of a neutral superfluid are sound waves, and the Goldstone particles themselves are phonons. At any non-zero temperature the neutral fluid will consist of a gas of phonons moving though a background superfluid condensate. This is the origin of the "two fluid" model of a superfluid.

  • $\begingroup$ I don't understand how you addressed my question. The typical argument for superconductivity is "there is a gap so you need an energy tresshold to change anything (break up pairs, scatter of impurities, ...)" but with these goldstone modes this is simply no longer true. You can create low energy goldstone modes and these will interact with the couper pairs, modifying the current. What is then still the argument for zero resistivity? $\endgroup$ – camel Mar 30 '19 at 13:32
  • $\begingroup$ The gap argument is false, or at least greatly oversimplified. There are plenty of gapless superconductors and superfluids. The "super" criterion depends on whether or a not a superflow can decay. This usually involves discussion of available phase space. A simple example that is relevent to the original question is the Landau argument for the critical velocity in superfluid $^4$He. This asks whether the phonon spectrum allows an object moving though the fluid can experience wave drag. If it can't, the object experiences no friction. $\endgroup$ – mike stone Mar 30 '19 at 19:14
  • $\begingroup$ But one way to show that superflow won't decay is by writing down the hamiltonian with such a flow as the groundstate, and then showing it has a gap. Then this superflow will in first order persist at low temperatures. But indeed, without this gap, how is it possible to show that this superflow won't decay? There's a small paragraph in the bcs paper on this, that I don't understand. I would've assumed that small interactions with gapless modes (and therefore no energy tresshold, so immediatly relevant at low temperatures) would provide resistance scaling with T. $\endgroup$ – camel Mar 31 '19 at 14:32

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