Ignoring SSB in massless phi^4 theory

[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 $\phi$?

 

[azntoon]

It is possible to do perturbation theory with the field \phi, however it is not necessarily the most accurate or efficient way to approach the problem. Expanding the field about the ground-state expectation value \nu by defining a new field \rho allows you to find the exact ground-state energy of the system as well as calculate the mass of the particle created due to the spontaneous breaking. This method is often thought of as the more effective way to approach the problem.