# What causes here an apparent violation of Elitzur's theorem?

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?

• 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.
– Buzz
Apr 5, 2020 at 5:40

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.
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.