# How do photons decay in superconductors?

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?

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Does this relate to monopole confinement in superconductors? Could the photons decay into more exotic stuff, like neutrinos? These questions make no sense to me. What is this ? Is a neutrino a more exotic stuff ? –  FraSchelle Jan 27 '14 at 10:18
I deleted those questions. "This" referred to potential multi-photon vertices. Neutrinos in this context would be quite exotic I'd imagine i.e. unexpected. –  Sebby Jan 27 '14 at 12:28
Can you first answer the same question for a conventional conductor, which also has evanescent waves within the skin depth? Is the answer for a superconductor completely different> –  akrasia Sep 3 '14 at 22:23

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 ...).

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Fair enough, but what are the decay modes of the massive photon? –  Sebby Jan 27 '14 at 16:18
@Sebby I think this is an interesting question indeed :-) I believe what you call a mode is just a $\left(k,\omega\right)$ description for the photon. This mode comes from the possibility to decompose in Fourier modes the solutions of the wave-equation $\left(\partial_{ct}^{2}-\nabla^{2}\right)A=0$. Here is the complication: this equation is valid in vacuum. So I could answer you back with a question: how do you quantise the electromagnetic-modes in materials ? This could be the object of an other interesting question. (cont.) –  FraSchelle Jan 28 '14 at 16:49
(cont.) For superconductors the equation is the London's one $\left(\partial_{ct}^{2}-\nabla^{2}+\mu^{2}\right)A=0$ (I restored the time component, which is not relevant for frequencies below a few GHz, in order for you to understand the transition between the photon equation in vacuum and in superconductors). I would say a mode is $e^{-x/\mu}$. There is only one and it is frequency independent as long as $\mu$ is frequency independent. But I would not call this a photon mode, for I didn't quantised the field yet. By the way, I have no idea how to quantise the field inside materials... –  FraSchelle Jan 28 '14 at 16:55
Sorry, maybe the sentence This mode comes from the possibility to decompose in Fourier modes was not really explicit. A mode in the Fourier sense is the integer which multiplies the measure function. For instance $n$ in $e^{\mathbb{i} n\omega t}$ –  FraSchelle Jan 28 '14 at 16:58