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Mass scales in See-saw mechenism

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Are both types of Majorana masses $$\mathcal{L}^L_M=-\frac{m_L}{2}[\overline{(\psi_L)^c}\psi_L + \overline{\psi_L}(\psi_L)^c]$$ and $$\mathcal{L}^R_M=-\frac{m_R}{2}[\overline{(\psi_R)^c}\psi_R + \overline{\psi_R}(\psi_R)^c]$$ $SU(2)\times U(1)$ invariant? If not, then both the masses must arise by $SU(2)\times U(1)$ symmetry breaking and both the mass scales are fixed by electroweak symmetry breaking scale. Right? Then why does one take $m_R>>m_L$ in see-saw mechanism? Why should we assume $m_R$ is around the GUT scale?

This post imported from StackExchange Physics at 2014-04-13 14:45 (UCT), posted by SE-user Roopam
asked Apr 5, 2014 in Theoretical Physics by Roopam (145 points) [ no revision ]
retagged Apr 19, 2014 by dimension10

1 Answer

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In this context, the right-handed neutrino is a singlet under the Standard Model gauge groups. Only the right-handed neutrino is allowed a Majorana mass. The left-handed term is not gauge invariant. If the SM and right-handed neutrino fields were embedded into a Grand Unified gauge group, the right-hand term would break that symmetry. It is expected therefore that $m_R\sim m_{GUT}$.

This post imported from StackExchange Physics at 2014-04-13 14:45 (UCT), posted by SE-user innisfree
answered Apr 5, 2014 by innisfree (285 points) [ no revision ]
Why do you say that the left-handed term is not gauge invariant but the right-handed is? @innisfree

This post imported from StackExchange Physics at 2014-04-13 14:45 (UCT), posted by SE-user Roopam
The left-hand is an SU(2) doublet with nonzero U(1) hypercharge. The right hand is a singlet. The Major an a masses don't contain $\sim\psi^*\psi$, but $\sim\psi\psi$, so they are not invariant.

This post imported from StackExchange Physics at 2014-04-13 14:45 (UCT), posted by SE-user innisfree

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