# What causes the evasion of the Goldstone theorem here?

For simplicity, I'll consider perhaps the simplest possible example of a gauge theory.

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 $$\phi(x)=\frac{1}{\sqrt{2}}\big(v+h(x)\big)\exp{[i\zeta(x)/v]}\tag{2}$$ plugged into Eq.$$(1)$$, the field $$\zeta$$ disappeaears from the theory upon making a suitable gauge transformation. Therefore, there is no Goldstone mode.

Question What causes the Goldstone theorem not to be applicable here? I mean, is there a crucial assumption used in the derivation of Goldstone theorem fails here?

• The minima of your potential is at 0. Right? Then there's no spontaneously broken symmetry.
– Ari
Apr 1, 2020 at 4:49
• @Ari Why? With $\mu^2<0$ and $\lambda>0$, you have a Mexican hat potential with the maximum at $\phi=0$ while minimum along a circle $|\phi(x)|= -\mu^2/\lambda\neq 0$.
– SRS
Apr 1, 2020 at 4:52
• Sorry I misinterpreted the sign. Can you explain the 'suitable gauge transformation'? 'Cause $-|\mu|^2\phi^*\phi+\lambda(\phi^*\phi)^2$ looks like the classic example of mexican hat potential. Where there should be goldstone modes.
– Ari
Apr 1, 2020 at 5:16
• @Ari Please see the corrected Lagrangian and let me know. This is a gauge theory and I'm alluding to the usual discussion of the Higgs mechanism.
– SRS
Apr 1, 2020 at 5:18

This is exactly the Higgs mechanism, which is what happens you have spontaneous symmetry breaking of a gauge symmetry. You are seeing the Goldstone mode get "eaten" by photon. You gauge-transform away the $$\xi(x)$$ field but you are left with a mass term for the photon, $$\frac{1}{2}v^2 A_\mu(x) A^\mu(x)$$.