Why does SSB deal only with scalar fields and not with fermion or vector fields?

My professor told me that it's closely related to the Lorentz invariance of the theory, but I don't understand at all the meaning of that.

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    $\begingroup$ Great question! +1 $\endgroup$ Apr 29, 2016 at 13:04
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    $\begingroup$ 1. What renormalizable terms do you propose to add to the action to generate SSB for fermions and vectors? 2. E.g. $\langle A^\mu \rangle$ (for a massive vector field) would be the VEV we have to look at in SSB for a vector field, but that would distinguish a direction in the broken theory, which is certainly something we don't observe! $\endgroup$
    – ACuriousMind
    Apr 29, 2016 at 13:38
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    $\begingroup$ Possible duplicate of Why cannot fermions have non-zero vacuum expectation value? $\endgroup$
    – innisfree
    May 1, 2016 at 9:43
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    $\begingroup$ Hmm maybe not an exact duplicate as this questions asks about vector bosons and fermions. I'd like to see a definitive answer on this topic - I didn't follow the discussion about the answers to the above. $\endgroup$
    – innisfree
    May 1, 2016 at 9:47
  • $\begingroup$ Side note: From a condensed matter point of view this questions reads nonsensical – you might want to add more context. In condensed matter you do have spontaneous symmetry breaking with vector or tensor fields that have non-zero "vacuum" expectation values. (But of course, if Lorentz invariance shall not be broken spontaneously, then tensor (these would break rotation invariance) and fermion order parameters are obviously ruled out). $\endgroup$ May 1, 2016 at 14:11

1 Answer 1


Essentially you are asking why only scalars are allowed to develop a vacuum expectation value (VEV).

A scalar (as the name suggests) does not point to any direction -it has spin 0- therefore it can have a VEV without breaking the Lorentz symmetry. On the other hand, a boson with higher spin, e.g. a vector (spin 1) would spontaneously break Lorentz by singling out a 'direction', and this is experimentally very constrained as no sign of Lorenz violation has ever been observed.

For fermions, we can apply the same arguments although there could be a mathematical reason forbidding them from taking VEV, see for instance this discussion (it would be nice if someone could develop this further here).

However, the scalar field need not be fundamental. As long as you construct an object which transforms trivially under Lorentz (this could be formed by a pair of more fundamental fermions for instance), it can have a VEV and, eventually, spontaneously break some symmetry depending on its other charges.


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