# Majorana Propagator

1. Dec 4, 2011

### ryanwilk

The Dirac propagator (e.g. for an electron) is given by the inverse of the field equation in momentum space i.e. ($\displaystyle{\not} p - m)\psi$ = 0, which gives:

$\frac{i}{(\displaystyle{\not} p - m)}$ = $\frac{i(\displaystyle{\not} p + m)}{(p^2-m^2)}$.

So is the propagator for a Majorana particle just the inverse of the Majorana equation: $\displaystyle{\not}p \psi + m \psi_{C}=0$?

But then this just leads to the Dirac equation if the particle is a Majorana spinor, so is the propagator just the same? If so, where does the difference come into effect in e.g. Feynman integrals?

Thanks.

Last edited: Dec 4, 2011
2. Dec 4, 2011

### Bill_K

I think it is legitimate to treat a Majorana particle as simply a Dirac particle that is subject to the constraint ψ = ψC.

3. Dec 5, 2011

### Parlyne

Majorana Feynman rules tend to be a little tricky. If you're not careful, you end up with charge conjugation operators floating all over the place. But, there are ways of taming them. You may find the treatment by Gluza and Zralek from Phys. Rev. D, vol. 45, num. 5 (march 1992), p. 1693 to be useful.

4. Dec 5, 2011

### Bill_K

Parlyne, For the benefit of those who don't have paper access to Phys Rev, could you indicate a little more what the issues are? It's clear that Majorana particles can't have interactions which violate C symmetry, e.g. they have to be electrically neutral. What else?

5. Dec 5, 2011

### Parlyne

There are a number of generically subtle issues. First, unlike Dirac fermions, a Majorana fermion can't, generically, absorb a phase. So, in fact, even the defining equation $\psi = \psi^C$ may need to be modified by a phase.

In terms of the Feynman rules, there's an issue that the exact form of what you get from the usual treatment is dependent on the basis you choose for the gamma matrices. The paper I cited discusses a method to eliminate this dependence systematically by looking at the forms that amplitudes involving the Majorana particles take, rather than just reading off rules from the Lagrangian.