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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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SRS
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  • see Nambu–Goldstone fermions – AccidentalFourierTransform Jan 17 '17 at 17:40
  • @AccidentalFourierTransform Can Goldstone bosons be anything other than spin-0 bosons? – SRS Jan 17 '17 at 17:46
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    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 (https://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. – Elliot Schneider Jan 17 '17 at 19:27
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    @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. – Cosmas Zachos Jan 17 '17 at 22:12
  • 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. – Cosmas Zachos Jan 17 '17 at 22:21
  • @CosmasZachos I know that the field with nontrivial VEV is not the Goldstone. But consider the case where I define a shifted field: $\eta+i\chi=\phi(x)-v$ (say, in the broken complex $\phi^4$-theory) where $v=\langle 0|\phi|0\rangle$. Since $\phi$ must be a spin-0 particle to preserve Lorentz invariance, $\eta,\chi$ should also be spin-0 because both sides should have the same Lorentz transformation property. This $\chi$ degree of freedom turns out to be the Goldstone particle. – SRS Jan 18 '17 at 07:35
  • ?!? Considering this case injects your unwarranted preconception, as the partner of the η which really gets the v.e.v. is a boson, χ. I invited you to consider a superfield, where the effective (super) partner of the viv'd particle is a fermion, eqn (3.10), instead. This then is a fermionic goldston, accounting for a SSBroken fermionic (super)symmetry. – Cosmas Zachos Jan 18 '17 at 11:53
  • You may improve your question by insisting that you are not inclined to consider fermionic symmetries, nor spacetime (Lorentz) violating symmetries, in which case you are stuck with the dull, obvious answer, yes. But you must understand you already rigidly enforced the answer you are getting. – Cosmas Zachos Jan 18 '17 at 13:47

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No. Magnons are spin-1 Goldstone bosons.

Everett You
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    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-matter for some reason; I just think that it is important to emphasise the role of the Lorentz symmetry). – AccidentalFourierTransform Jan 21 '17 at 10:05