Elitzur's theorem, stating that spontaneous breakdown of a gauge symmetry is impossible, was originally proved for a lattice gauge theory. Is it valid in continuum field theory? Any ref?


Well, you have to specify what you mean by continuum gauge theory. The only way I know how to regulate gauge theories in the continuum directly is the perturbative way, which needs gauge fixing, which already breaks gauge symmetry. In this kind of contexts what can break is a global symmetry, not a local one (this is the Higgs mechanism, which is often sloppily called spontaneous gauge symmetry breaking).

I remember thinking about this question once. :) It particularly bothered me, because you would need symmetry breaking in an SU(2)-Higgs model in particle physics. And since you can define nonperturbatively such QFT by taking a continuum limit of the lattice gauge theory, you seem to have a problem, you seem to always have no symmetry breaking at all. It is easy to show by formal manipulations of the path integral that the Higgs VEV is always zero.

The only solution I could think of in this context was the same as for spontaneous symmetry breaking in general. I think the only way you can see symmetry breaking is to include an explicit breaking term, than do the infinite volume and continuum limit first, and than do the symmetry breaking term goes to zero limit. It is important that the order of the limits is not interchangeable. If you do it the other way you always get zero. Unfortunately I have not seen this kind of calculation carried out anywhere, but this is my best guess.

If anyone has a better understanding of this, I am also very interested.

  • $\begingroup$ You can can still have a "Higgs" phase in a lattice gauge theory, e.g. link. The ground-state remains gauge-invariant and doesn't explicitly break the symmetry, but otherwise it should exhibit the same properties as obtained from perturbative methods. The appearance of gauge-symmetry breaking simply comes from gauge-fixing. $\endgroup$ – Dominic Else Aug 19 '15 at 22:18
  • $\begingroup$ I am not sure what you mean. In the perturbative formulation, gauge-symmetry breaks because of gauge fixing, but there remains a global version of the gauge symmetry. Than you pick a vacuum to expand around for perturbation theory, and with this choice you also break the global symmetry, and the Higgs VEV will no longer be zero. To reproduce this behavior on the lattice, you have to introduce an external source, otherwise the Higgs VEV will always be 0. $\endgroup$ – evilcman Sep 1 '15 at 7:56
  • $\begingroup$ The Higgs VEV is not a gauge-invariant quantity, so its value has no physical content in itself, and in particular is sensitive to gauge-fixing. The fact that you see a nonzero Higgs VEV after gauge-fixing does not imply that it would be nonzero in a gauge-invariant treatment. $\endgroup$ – Dominic Else Sep 1 '15 at 15:27
  • $\begingroup$ The appearance that the residual global symmetry is spontaneously broken following gauge-fixing can, in fact, easily occur in a lattice model. I go into more detail in my answer to this question: link $\endgroup$ – Dominic Else Sep 1 '15 at 15:28

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