How do photons decay in superconductors?

Question

If $A$ is the vector potential, the London equations imply that:

$$(\nabla^{2}-\mu^{2})A=0$$

if there is no external current. This can be interpreted as an effective photon mass and so, light cannot propagate indefinitely within a superconductor. Let $\lambda = 1/\mu$ be the London penetration depth. As a thought experiment, assume we had a thin (< $\lambda$) slab of superconducting material and shined high frequency light on it. Some of the energy would be lost as heat (photons hitting atoms, etc.). The rest would ``decay" exponentially but manage to get to other side of the slab. My thoughts and chain of questions:

How would the light emerge? Would the exiting light have lower frequency but proportionally higher intensity? Would it have the same frequency and just be the surviving fraction of photons? Then, what did the photons which didn't survive decay into? Or, does the decay just mean absorption of the photons e.g. heat generation?

The $U(1)$ gauge symmetry of quantum electrodynamics is broken/hidden. How do the Feynman diagrams look inside a superconductor?

Answer 1

You're asking too much questions, not necessarily connected among themselves. Below I give quick answers. Please ask further separate questions if you need more details.

How would the light emerge? Would the exiting light have lower frequency but proportionally higher intensity? Would it have the same frequency and just be the surviving fraction of photons?

Where did you see a coupling to the frequency in the London's equation ? So there is no change in frequency.

Then, what did the photons which didn't survive decay into? Or, does the decay just mean absorption of the photons e.g. heat generation?

The amplitude of the wave-function (say $\phi$) of the photon will decay (since $A\propto \phi + \phi^{\dagger}$), almost the same way the electronic wave-function decay in a tunnel barrier. You might be perturbed because the London's equation is a classical equation, but the phenomenon behind is purely quantum: a Higgs-Brout-Englert-(Anderson)-mechanism if you wish, first describe by the London brothers when they imposed a $j\propto A$ super-current (a purely diamagnetic current in modern language).

How do the Feynman diagrams look inside a superconductor?

Feynman diagrams exist for superconductor, you just have to invent new rules for the two-electrons correlations functions $\left\langle c^{\dagger}c^{\dagger}\right\rangle $. There is no formal QED theory for superconductor though, since the condensed matter theory is only valid at low velocity of the bulk. The exact form of the Feynman diagrams depend on the interaction ($\varphi^{4}$ theory, electron-phonon coupling, electron-electron coupling, to cite a few of them ...).