# Bare mass versus the mass form spontaneous symmetry breaking

Consider renormalization in $$\phi^4$$ theory $$\mathscr{L}=(\partial\phi)^2-\frac{1}{2}m^2\phi^2+\frac{\lambda}{4}\phi^4$$ where $$m$$ and $$\lambda$$ are respectively the unobservable bare mass and bare coupling.

Now suppose we consider a theory where this $$m^2<0$$. In that case, the Lagrangian after the spontaneous breakdown of symmetry is given by $$\mathscr{L}=\frac{1}{2}(\partial_\mu\eta)(\partial^\mu\eta)-\lambda v^2 \eta^2-\lambda v \eta^3 -\frac{\lambda}{4}\eta^4+\frac{\lambda v^4}{4}$$ where $$\phi=v+\eta$$. Here too, I think that the term $$\lambda v^2\eta^2$$ still represents unobservable bare mass because, in general, have nothing to do with the pole of the two-point function.

But in the Standard Model, we say that can measure the VEV ($$\langle \Phi\rangle=v$$) of the Higgs field $$\Phi$$ from muon decay. So is it saying that the VEV do not get renormalized but only $$\lambda$$ gets?