# Ignoring SSB in massless phi^4 theory

1. Jul 10, 2014

### geoduck

If a theory is spontaneously broken, you usually make calculations by expanding the field about the ground-state expectation value $\nu$, so that you define a new field $\rho$ such that $\phi=\nu+\rho$ such that $\langle \rho \rangle =0$.

What if you just calculate perturbation theory with $\phi$ instead of the new field $\rho$?

It seems that massless $\phi^4$ is spontaneously broken at the 1-loop level. So do you have to do perturbation theory with the field $\rho$ instead of $\phi$?