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  Majorara mass and fermion number violation

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How can it be shown that the Majorana mass violates the fermion number by two units? Can even a Noether charge be defined in presence of Majorana mass term?

This post imported from StackExchange Physics at 2014-04-13 14:48 (UCT), posted by SE-user Roopam
asked Apr 4, 2014 in Theoretical Physics by Roopam (145 points) [ no revision ]

1 Answer

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In the Standard Model, fermion number is not conserved. Lepton number is conserved, because of an accidental symmetry. One cannot write down a renormalizable, gauge and Lorentz invariant operator that violates lepton number conservation in the Standard Model.

A Majorana neutrino would violate lepton number conservation by two units. To see this, consider, e.g. neutrinoless double beta decay. You can draw Feynman diagrams with two $W^-e^-v_e$ vertices in which two incoming $W$-bosons each decay into an electron and an electron-neutrino. The two electron-neutrinos annihilate (possible because they are Majorana particles), leaving a final state of two elecrons, violating lepton number conservation by $2$ units.

You can see that Majorana neutrinos violate lepton number conservation by $2$ units from the mass term. The mass term, $$ \mathcal{L} = \frac12 m \psi^T C^{-1}\psi, $$ is not invariant under the $U(1)$ lepton number symmetry, $\psi\to\exp(iL\theta)\psi$. It picks up a phase of twice the lepton number of the neutrino, i.e. $\Delta L=2$ rather than $\Delta L=0$. A Majorana neutrino cannot be charged under a $U(1)$ symmetry.

Because there is not a lepton number $U(1)$ symmetry, there is no conserved Noether charge corresponding to lepton number.

This post imported from StackExchange Physics at 2014-04-13 14:48 (UCT), posted by SE-user innisfree
answered Apr 4, 2014 by innisfree (295 points) [ no revision ]

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