# Why are we allowed to spontaneously break the Higgs field symmetry?

In the lepton Lagrangian there are no mass terms allowed for the gauge bosons, due to gauge symmetry. To fix the problem of existing gauge boson masses, we introduce another field (Higgs) and say that here we

1) have self-interactions

(Why do we have self-interactions for the Higgs field but not in the lepton case? Is the Higgs "charged" like the self-interacting gluon is?)

2) just flip the sign of the $|\phi|^4$ term

(Why are we allowed do to that?)

3) hence break the gauge symmetry and get gauge boson masses

(Why are we allowed to do break the symmetry here but not for the lepton field?)

→ We have similar initial complex fields (lepton and Higgs) with similar couplings to the other gauge bosons, so why are we allowed to treat both fields in such a different way?

Essentially, Higgs self-interactions are allowed because they don't violate any laws or symmetries (i.e. the $\phi^4$ term is gauge invariant, Lorentz invariant etc). Informally, a Lagrangian can (and possibly should) consist of any/all combinations of fields, derivatives of fields etc that respect the symmetries of the theory. In the Standard Model the Higgs is not charged.
The sign of the $\phi^4$ term is flipped by convention, because the coupling constant is a parameter of the theory, to be determined by experiment (the theory does not offer a prediction for the coupling constant) so yes, you are allowed to choose the sign.
• Thank you for your answer, it already helps me a lot. But why do we need the Higgs when 1) $\phi^4$ is gauge invariant and 2) we are allowed to break any symmetries we wish? We could 1) just add a self-interaction term to the lepton Lagrangian and 2) break the symmetry there, so that gauge boson masses arise without having a Higgs field. – Thomas Apr 3 '15 at 14:27
• Lepton self interaction terms like $(\overline{\psi}\psi)^2$ are not renormalizable. – Robin Ekman Apr 3 '15 at 14:38