# Are majorana masses technically natural?

I understand that Dirac masses for fermions are "technically natural" (in the 't Hooft sense of technical naturalness) because they break chiral symmetry. In limit where the Dirac mass goes to zero, chiral symmetry is restored, so one expects any loop corrections to the Dirac mass to be proportional to the Dirac mass itself. That is to say, there is no power law UV sensitivity in a Dirac mass parameter.

Is the same true for Majorana mass terms? Because the Majorana fermions are real, is it still true that the mass term breaks chiral symmetry? It seems so, since one may write a 4-component Majorana fermion in terms of a single Weyl fermion:

$$\Psi_D = \begin{pmatrix} \psi_\alpha \\ \bar\psi^{\dot{\alpha}}\end{pmatrix}$$,

where the $$\alpha$$ and $$\dot{\alpha}$$ are Weyl ($$SL(2,C)$$) indices. Then a Majorana mass term takes the form:

$$M\bar\Psi_M \Psi_M = M(\psi^2 + \bar\psi^2)$$,

where the Weyl spinor contractions use the $$\epsilon$$ tensor. These terms appear to break chiral symmetry as well.

Is this the correct picture?

Because of the Majorana condition $$\psi=\psi^C$$, Majorana fermions are singlets with respect to gauge symmetries, including $$SU(2)$$. Furthermore, no chiral symmetry forbids a Majorana mass.
The right-handed neutrino could be Majorana, with interesting implications. See the see-saw mechanism; the difference between the Planck and the electroweak scales results in the neutrino mass scale, $$M_W^2/M_P$$.
• I'm not so sure about this. I agree that the Majorana fermion is self conjugate and thus must be a gauge singlet, but chiral symmetry is a separate issue. Is it not true that in the absence of any mass term, the Majorana fermion has a U(1) symmetry, $\psi \to e^{i\alpha}\psi$? And Isn't it further true that this symmetry is broken by the Majorana mass term? Hence the mass term should be technically natural, no? (I should note that I'm writing $\psi$ as a Weyl spinor.) – Pengpeng Xu Jun 20 '13 at 1:32