# Goldstone's theorem and massless modes for $\phi^4$ theory

Consider a scalar field doublet $(\phi_1, \phi_2)$ with a Mexican hat potential

$$V~=~\lambda (\phi_1^2+\phi_2^2-a^2)^2.$$

When $a=0$ this is a quartic potential and the symmetry is not spontaneously broken. However when the field acquire a VEV, the fields splits into a massive mode and a massless boson mode called the Goldstone boson.

I am wondering about the initial potential with $a=0$: does it have 2 massive modes?

-

AFAIR it has two massless modes, as there are no quadratic terms around the minimum.

-
Sounds about right. –  David Z May 17 '12 at 1:02
You should say that this requires fine-tuning, as the mass-term will renormalize away from zero. You need to fine-tune the $m^2$ to some negative value to get to the critical point. –  Ron Maimon May 17 '12 at 1:24