More specifically, how does a wave-particle duality differ from a quasiparticle/collective excitation?
What makes a photon a gauge boson and a phonon a Nambu–Goldstone boson?
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asked Dec 30, 2012 at 22:43
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2$\begingroup$ For the latter question, did you research this up beforehand? Or do you just find wikipedia unsatisfying?: en.wikipedia.org/wiki/Photon#The_photon_as_a_gauge_boson . As to the first question, one striking difference is that the duality refers to actual particles whereas excitations are not real particles (hence the term 'quasi'). $\endgroup$– Chris GerigCommented Dec 31, 2012 at 6:55
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1$\begingroup$ Photons can be thought of as quasi-particles too: they have sources and absorbers and are involved in the corresponding source/absorber equations. In other words, they are not thought of as independent of matter, to tell the truth. $\endgroup$– Vladimir KalitvianskiCommented Dec 31, 2012 at 9:59
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2 Answers
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Not all phonons are Nambu-Goldstone bosons and not all Nambu-Goldstone bosons are phonons. Nambu-Goldstone bosons are (usually) gapless excitations that arise from spontaneous symmetry breaking. For instance, in a spinless Bose-Einstein condensate, the NG boson is indeed a phonon, with a linear dispersion at low energy. However, in a ferromagnet the NG boson is called a magnon. This magnon is gapless but has a quadratic dispersion relation, like a massive particle, and should generally not be called a phonon.
In a periodic crystal, for instance, phonon modes arise because of (discrete) translational symmetry but not spontaneous symmetry breaking -- they are not NG bosons. As someone pointed out, the periodicity is not necessary. In fact, in virtually all condensed matter systems have phonons because of translational symmetry (air has plenty!).
As you can probably tell, I'm much more inclined towards condensed matter than high energy theory, so I'm not sure if I can say anything useful about photons!
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$\begingroup$ Wow! That's really cool thanks for taking the time to explain that. $\endgroup$ Commented Dec 31, 2012 at 2:33
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1$\begingroup$ @emarti: I think phonons are NG bosons in periodic crystal too since spacetime boosts (and traslations, rotations) are spontaneously broken by the lattice. The residual discrete symmetries constraints the NG boson low-energy interactions. In fact, the theory of phonons is Galileo (or Poincare') invariant rather than be symmetric only under the discrete unbroken subgroup of the lattice. See my answer below. $\endgroup$ Commented Dec 31, 2012 at 10:51
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$\begingroup$ I beg your pardon, but the existence of phonons does not require periodicity or any other symmetry breaking. Every continuous solid possesses 3 branches of gapless acoustic phonons. $\endgroup$– SlaviksCommented Dec 31, 2012 at 10:55
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2$\begingroup$ @Slaviks: yes, the existence of photons doesn't require periodicity of the background (in fact they exist for fluids and other gelly solids too), but it does require instead the spontaneous breaking of 3 spacetime symmetries, namely the boosts (or translations). This is a very basic fact that boils down to the fact that a background breaks the invariance under Galilean or Poincare' boosts. Local boosts corresponds to gapless excitations because they cost no energy since the laws of physics are indeed Galilean (or Poincare') invariant. $\endgroup$ Commented Dec 31, 2012 at 13:13
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$\begingroup$ @argopulos +1; a nice perspective! Is that right that the difference essentially boils down to whether there is ether (hence, phonons) or not (hence, photons)? $\endgroup$– SlaviksCommented Dec 31, 2012 at 15:50
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Phonons are Goldstone bosons of a spontaneously broken spacetime symmetry (see e.g. http://arxiv.org/abs/hep-ph/9609466). Typically, one breaks Galileo's (or Poincare') boosts and traslations but the resulting number of Goldstones is less or equal than the number of broken generators (e.g. a spacetime dependent traslation is not independent than a boost). Phonons are spin-0 and become strongly coupled, as every Goldstone boson, to the scale (e.g lattice scale) where the underlying microscopic degrees of freedom can be excited.
Massless Spin-1 bosons are instead described by mean of gauge invariance which is a bookkeeping 'symmetry' which erase all interactions that would either give the photon a mass or lower the cutoff of the theory making it strongly coupled at quite large distances. One can try to connect spin-1 massless gauge bosons with Goldstone bosons by breaking spacetime symmetries with a spin-1 order parameter. These are very speculative ideas that have a tiny chance to work only in Poincare' broken spacetimes.
