In the crystal, infinitesimal translational symmetry breaking makes the phonon, In ferromagnet, time-reversal symmetry breaking makes magnon. I know that in superconductor there is a spontaneous symmetry breaking of $U(1)$. In the consequence, which kind of quasi-particle is in superconductor? At first, I thought it is Bogoliubov quasi-particles, but it has mass. What's this?
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2$\begingroup$ The key difference is that the U(1) symmetry in a superconductor is a gauge symmetry, unlike the other symmetries you mention. When you work this out, that means that the goldstone boson is 'eaten' by the gauge field which becomes massive. See physics.stackexchange.com/q/213401 or many other questions on the Higgs mechanism. $\endgroup$– RococoCommented Aug 5, 2016 at 21:18
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2 Answers
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This question is answered in Wikipedia, so let be quick about it.
In a superfluid, the condensate is made of atoms. These particles are non charged, and therefore the breaking of the U(1) symmetry generates some Goldstone modes, which are some kinds of phonons, i.e. deformations of the condensate. Since superfluids are fluids, the Galilean invariance is also broken, and the Goldstone modes resulting from this symmetry breaking are also some phonons.
In a superconductors, the condensate is built on electrons, which are intrinsically charged, and thus couple to the electromagnetic field via a minimal coupling (or Weyl, or covariant). The breaking of the U(1) redundancy (the gauge redundancy is not a symmetry in that case, sometimes it is called a local symmetry, to avoid the confusion with the case of superfluidity which breaks the global U(1) symmetry) then leads to the Higgs mechanism. Instead of producing massless scalar bosons as in the Goldstone situation, the mixing between the condensed fermions and the gauge bosons generates massive gauge bosons. In the case of superconductivity, one mixes photons to the plasmons (electron-phonon composite -- usually one considers those as electrons, and forget about the mechanism giving rise to superconductivity), and the photons got a mass.
So in the simplified picture for superconductivity, the electron (the fermions that are the elementary constituent of a metal) condense and the photons (the gauge bosons of the electromagnetic field) got a mass. This is the Meissner effect, when a magnetic field does not enter a superconductor. In a superconductor, one talks about the Anderson-Higgs mechanism rather than Higgs mechanism, since Anderson predicted the presence of a massive plasmon mode in superconductors as a consequence of the preservation of the gauge redundancy a few years before Higgs and co generalise this result to relativistic field theories.
answered Aug 8, 2016 at 8:40
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Please let me know if this is basic, childishly simple information, well known to you already, (I suspect it is) but just on the chance that it is not:
From Wikipedia Superconductity
The situation is different in a superconductor. In a conventional superconductor, the electronic fluid cannot be resolved into individual electrons. Instead, it consists of bound pairs of electrons known as Cooper pairs. This pairing is caused by an attractive force between electrons from the exchange of phonons. Due to quantum mechanics, the energy spectrum of this Cooper pair fluid possesses an energy gap, meaning there is a minimum amount of energy ΔE that must be supplied in order to excite the fluid. Therefore, if ΔE is larger than the thermal energy of the lattice, given by kT, where k is Boltzmann's constant and T is the temperature, the fluid will not be scattered by the lattice. The Cooper pair fluid is thus a superfluid, meaning it can flow without energy dissipation.
Due to the above process, the photon, in effect, has a mass within the superconductor.
Again, I would stress that I am happy to delete this answer if it is too basic for you.