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Massless vector bosons have only two independent degrees of freedom, while massive ones have three. In spontaneous symmetry breaking, the massless vector belonging to the broken group becomes massive and absorbs the Goldstone boson degree of freedom as its third independent (longitudinal) component.

Breaking a symmetry explicitly does not supply an additional field that can serve as third degree of freedom. From this I gather two possibilities:

  1. Vector bosons that belong to explicitly broken symmetries remain massless or
  2. there is some mechanism that I'm not aware of that can supply the required degree of freedom.

Which option is it? Or is there a third option that I'm not considering?

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Explicit symmetry breaking, i.e. adding a mass term for a particle simply implies that this particle is massive. In the case of a vector boson, there will be three degrees of freedom. The "additional" degree of freedom does not have to come from somewhere else by a certain mechanism, it is simply there from the beginning. An example would be Proca theory, which contains a massive spin one boson. Another example would be QCD, where explicit quark mass terms break chiral symmetry. This is nothing more than the statement that quarks are massive, without reference to any kind of mechanism.

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  • $\begingroup$ What if I write down a symmetry-breaking term that is not a mass term? Can that always be rewritten as a mass term plus a different interaction term? $\endgroup$ – Neuneck Nov 5 '14 at 13:14
  • $\begingroup$ @Neuneck: Then you have a theory with massless vector bosons that do not correspond to a particular symmetry. $\endgroup$ – Frederic Brünner Nov 5 '14 at 14:25

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