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Elitzur's theorem [Ref. Andreas Wipf, Statistical approach to quantum field theory] states that

A local gauge symmetry cannot break spontaneously. The expectation value of any gauge non-invariant local observable must vanish.

Consider a spontaneously broken ${\rm U(1)}$ gauge theory of a charged scalar field coupled to the electromagnetic field $$\mathscr{L}=(D_\mu\phi)^*(D^\mu\phi)-\mu^2\phi^*\phi-\lambda(\phi^*\phi)^2-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}\tag{1}$$ with $\lambda>0$ and $\mu^2<0$. When the field $\phi$ with the polar parametrization $$\phi=\frac{1}{\sqrt{2}}\big(v+h(x)\big)\exp{[i\zeta/v]}\tag{2}$$ plugged into Eq.$(1)$, the field $\zeta$ disappeaears from the theory upon making a suitable gauge transformation. Therefore, there is no Goldstone boson.

In this popular derivation, the gauge non-invariant field operator $\phi$ acquires a nonzero VEV in violation to Elitzur's theorem.

Question What is the reason for this apparent violation of Elitzur's theorem?

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  • $\begingroup$ Elitzur's "Theorem" is misleading, at best. It dates from a time when spontaneous breaking of gauge symmetry was only coming to be undersood. It's not wrong, exactly, but it uses terminology differently than we do today. $\endgroup$
    – Buzz
    Apr 5, 2020 at 5:40

2 Answers 2

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Gauge symmetries are not real symmetries but redundancies. Any two states related by a gauge transformation are physically equivalent. So if there isn’t a symmetry to begin with, it cannot be spontaneously broken.

More concretely, the vev of a gauge non-invariant operator such as $\phi$ is gauge dependent and not physically meaningful. People often get confused and talk about the “global part” of a gauge symmetry, which is still a redundancy. But if we pretend for a minute that this $U(1)$ is a real symmetry and act on the vev, we can always undo it by a gauge transformation.

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For normal symmetry breaking, $\phi$ acquires a nonzero vev, which in turn breaks some symmetry.

For gauge symmetry breaking, $\phi$ is not a gauge invariant. By Elitzur's theorem, its vev vanishes. Instead, $|\phi|$ acquires a nonzero vev, which however doesn't break any symmetry.

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