4

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?

SRS
  • 26,333
  • The minima of your potential is at 0. Right? Then there's no spontaneously broken symmetry. – Ari Apr 01 '20 at 04:49
  • @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 01 '20 at 04:52
  • 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 01 '20 at 05:16
  • 1
    @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 01 '20 at 05:18

2 Answers2

2

FWIW, Goldstone's argument just yields that there is a massless mode. However, it could be an unphysical massless mode, cf. the BRST formalism. See also this & this related Phys.SE posts.

Qmechanic
  • 201,751
1

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, $\frac{1}{2}v^2 A_\mu(x) A^\mu(x)$.

Luke Pritchett
  • 7,195
  • I am aware of that. My question was whether Goldstone theorem is applicable here. I have asked a similar question here https://physics.stackexchange.com/q/532858/ – SRS Apr 04 '20 at 14:51