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?

  • $\begingroup$ see Nambu–Goldstone fermions $\endgroup$ Jan 17, 2017 at 17:40
  • $\begingroup$ @AccidentalFourierTransform Can Goldstone bosons be anything other than spin-0 bosons? $\endgroup$
    – SRS
    Jan 17, 2017 at 17:46
  • 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. $\endgroup$ Jan 17, 2017 at 19:27
  • 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. $\endgroup$ Jan 17, 2017 at 22:12
  • $\begingroup$ 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. $\endgroup$ Jan 17, 2017 at 22:21

1 Answer 1


No. Magnons are spin-1 Goldstone bosons.

  • 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-matter for some reason; I just think that it is important to emphasise the role of the Lorentz symmetry). $\endgroup$ Jan 21, 2017 at 10:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.