Can 1-loop corrections generate a neutrino mass? I'm looking at the Electroweak theory with massless neutrinos. The neutrino has an interaction with the $W$-boson and electron which gives 1-loop corrections to the propagator via the self-energy. I'm finding that the self energy evaluated on shell has a non-zero piece which indicates that quantum corrections are shifting the pole of the neutrino propagator.
Is there anything in the standard model which would protect this from happening - in the same way that gauge invariance protects the photon propagator from having corrections that shift the mass pole? Or is this something to be expected?
 A: If you actually calculate the one-loop correction, with a neutrino mass in the propagator as a placeholder for the moment. You will end up seeing the finite correction due to the resumed loop effect is only proportional to the mass. Now here comes the problem, the SM doesn't give the mechanism of neutrino masses in the first place. Hence the loop effect, which is exactly proportional to the fictitious mass term goes away eventually. So no mass stems from the quantum effect. Why is that? The essential reason comes all the way from chiral symmetry, where the non-zero fermion mass breaks this global symmetry. So if the mass of a fermion given by the SM is zero, we have the exact chiral symmetry in the lagrangian, such that the masslessness shall go to all orders in perturbation theory. If the fermion mass is non-zero, then the correction due to the quantum effect will be proportional to the mass itself so it keeps the light fermion mass from large quantum corrections. This is so-called custodial chiral symmetry.
A: I don't know how many details do you want, but a simple argument to understand why neutrinos are massless in the SM up to any order in perturbation theory is the following.
From a Lorentz perspective, fermion masses require both left- and right-handed spinor components to form the $m_D\,\overline\Psi\Psi$ mass term. In the case of neutrinos, they are present only as part of the $\rm SU(2)_L$ doublet, $(\nu_L,e_L)$, meaning that there are no right-handed neutrinos. This has been experimentally corroborated.
Regardless of the loop order, you cannot write down a mass term -in the SM- without $\nu_R$.
