EDIT: Bosonic fields with spin $s>0$ transform non-trivially under Lorentz transformation. Hence, if any of them acquires a VEV, that would violate Lorentz invariance as I learnt from the posts 1 , 2 , and 3. Does it mean that Goldstone bosons, obtained after spontaneous symmetry breaking, are necessarily spin-0 particles in a theory which respects Lorentz invariance? If yes, does it mean that one can have $s>0$ bosonic Goldstone particles in non-relativistic field theories such as condensed matter systems?
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$\begingroup$ see Nambu–Goldstone fermions $\endgroup$– AccidentalFourierTransformJan 17, 2017 at 17:40
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$\begingroup$ @AccidentalFourierTransform Can Goldstone bosons be anything other than spin-0 bosons? $\endgroup$– SRSJan 17, 2017 at 17:46
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2$\begingroup$ Ordinary continuous symmetry breaking leads to spin-0 Goldstone bosons. Supersymmetry breaking leads to spin-1/2 Goldstone fermions ("Goldstinos"). More generally, $p$-form symmetry breaking leads to $p$-form Goldstones (arxiv.org/abs/1412.5148v2). For example, U(1) gauge theory has a spontaneously broken 1-form global symmetry, whose Goldstone is the photon itself. $\endgroup$– Elliot SchneiderJan 17, 2017 at 19:27
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1$\begingroup$ @SRS. It does not mean anything of the sort! You appear stymied by a a misconception that the field with the nontrivial v.e.v. is the Goldstone particle. Studying the SSB mechanism reminds you that this simply is not so. The goldstons of the σ-model are the πs, not the σ. Likewise in Susy-breaking models, you recall the vev particle is a scalar, but the goldstons are its former super partner fermions. $\endgroup$– Cosmas ZachosJan 17, 2017 at 22:12
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$\begingroup$ To have a vector goldston, however, you'd have to work harder, as you cannot transfer a scalar v.e.v. to the transform of a vector. So, you'd have to break spacetime symmetries, as is well-known. $\endgroup$– Cosmas ZachosJan 17, 2017 at 22:21
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1 Answer
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No. Magnons are spin-1 Goldstone bosons.
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5$\begingroup$ But magnons only appear in a non-relativistic context, and OP is asking about "a theory which respects Lorentz invariance". This might seem to be a trivial requirement, but symmetry breaking and Lorentz symmetry are interrelated. (I'm not saying that the answer is bad; after all, OP chose to include the tag
condensed-matterfor some reason; I just think that it is important to emphasise the role of the Lorentz symmetry). $\endgroup$ Jan 21, 2017 at 10:05