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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?
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?