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Usually, the absence of Goldstone modes in a superconductor is seen as an example of the Anderson-Higgs mechanism, related to the fact that there is gauge invariance due to the electromagnetic gauge field coupled to the charged electrons.

However, this is puzzling in light of the fact in (3d) neutral atomic gas the fermionic degrees of freedom also undergo BCS transition, and there will be Goldstone modes since Goldstone theorem applies.

Mathematically the same model with an attractive $ g\bar{\psi}\bar{\psi} {\psi} {\psi}$ term is used for both superconductor and atomic gas BCS transtion. Suppose no external electromagnetic field is applied (so the Anderson-Higgs mechanism does not apply), then how come in one case there are Goldstones, while in a superconductor there are only gapped plasmons?

I guess another way to put the question is, when no external EM field applied, are there Goldstone modes in a superconductor, and why?

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    $\begingroup$ you seem to be confusing SSB for a gauge symmetry vs SSB for a flavor symmetry. these are radically different phenomena (even if conflated in the literature, perhaps more often in the cond-mat world). hopefully this will help a bit: physics.stackexchange.com/q/800131/84967 $\endgroup$
    – AccidentalFourierTransform
    Commented 2 days ago
  • $\begingroup$ I understand this distinction, and in fact the exact puzzle for me here is that when no gauge field and thus no gauge invariance is present, such as a BCS superconductor left alone, why there are no Goldstones. Or rather why the condclusion of the absence of gapless modes on their spectrum drawn from using Anderson-Higgs mechanism still applies. $\endgroup$
    – cx1114
    Commented 2 mins ago

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The Goldstone modes of a superconductor (considered as an electron gas with attractive potential) are physically the oscillations of electron density. As electrons are intrinsically charged, this will inevitably create the EM field. So we get plasma oscillations and nonzero mass of the Goldstone mode.

If we somehow turn off the Coulomb interaction between electrons, I guess that the Goldstone modes (density waves) will indeed become massless. But that's physically meaningless for electrons. So, answering your question, I would say that the external EM field does not play role here. Indeed, Goldstone modes acquire mass due to dynamic coupling with EM field.

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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 2 days ago
  • $\begingroup$ 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 2 days ago
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A neutral fermionic superluid has gapless soundwaves that can be regarded as a Goldstone mode due to the spontaneous breakdown of translational symmetry, but the fluctuating quantity in the sound wave is not the order parameter $\langle \psi\psi\rangle$ but rather the combination
$$ \Phi(x,t)= \sqrt{\rho} e^{i\theta} $$ where $\rho$ is the fluid density and $\theta$ phase of $\langle \psi\psi\rangle$.

Further the non-relativistic system does not have separate radial and tangential modes. Instead the radial and tangential part are coupled with the $\rho-\rho_0$ being the density fluctuations and ${\bf v}\propto \nabla \theta$ being the back-and-forward motion of the fluid.

There is some experimental evidence for fluctuations in the magnitude of $\langle \psi\psi\rangle$ but this is usually strongly damped.

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