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?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.