Does neutrino mass imply right-handed neutrinos exist?

[Jim Kata]

Does that fact that it has been shown that neutrinos have mass in any way imply that there must be right handed neutrinos?

 

[Jim Kata]

Follow up question. If there are right handed neutrinos is there an explanation as to why they haven't been observed?

 

[mjsd]

Does that fact that it has been shown that neutrinos have mass in any way imply that there must be right handed neutrinos?

not necessarily, because you can have a majorana mass term consisting of only one type of neutrino field: $$(\nu_L)^c \nu_L \Delta$$ but to make this term (after spontaneous symmetry breaking) to be invariant under the unbroken gauge group (SU(3) color and U(1) electric charge), then you need $$\Delta$$ to be a triplet field under weak-SU(2). So you can do without the right handed neutrino $$\nu_R$$ but have to introduce a new scalar field $$\Delta$$ to the Standard Model instead to give neutrino a mass.

having said that there are reasons to favor the existence of the right-handed neutrino to explain neutrino mass (eg. see-saw mechanism)

Follow up question. If there are right handed neutrinos is there an explanation as to why they haven't been observed?

One explanation (the typical one) is that they are too heavy to be seen at colliders at current operating energies. The see-saw mechanism where light neutrino masses are related to heavy right-handed neutrino masses via
$$M_\text{light} \simeq \frac{\langle\phi\rangle^2}{M_\text{heavy}}$$
where $$\langle\phi\rangle$$ is the electroweak breaking VEV which is about 174GeV, tells us that if $$M_\text{light} \sim 0.1 \,\text{eV}$$ then that implies a $$M_\text{heavy}$$ of the order of $$10^{14}\,\text{GeV}$$ which is 100,000,000,000 times higher in energy than the current colliders can reach.

 

[blechman]

not necessarily, because you can have a majorana mass term consisting of only one type of neutrino field: $$(\nu_L)^c \nu_L \Delta$$ but to make this term (after spontaneous symmetry breaking) to be invariant under the unbroken gauge group (SU(3) color and U(1) electric charge), then you need $$\Delta$$ to be a triplet field under weak-SU(2). So you can do without the right handed neutrino $$\nu_R$$ but have to introduce a new scalar field $$\Delta$$ to the Standard Model instead to give neutrino a mass.

Actually, if you allow for non-renormalizable operators in the SM (coming from a GUT theory, for example) then you immediately get a Majorana mass without adding anything (no new scalars)! In fact, the UNIQUE(!) dimension-5 operator will do it:

$$\mathcal{L}_5=\frac{c}{M}(HL)^2$$

where L is the lepton doublet and H is the Higgs doublet, and c is some dimensionless coupling, and M is the UV scale where the SM breaks down (GUT scale, for instance). When you set H equal to its vev, then this will become a Majorana mass for the left handed neutrino, whose value is the same as what you'd expect from the see-saw mechanism. No RH neutrinos necessary. No new scalars necessary.

 

[mjsd]

Actually, if you allow for non-renormalizable operators in the SM (coming from a GUT theory, for example) then you immediately get a Majorana mass without adding anything (no new scalars)! In fact, the UNIQUE(!) dimension-5 operator will do it:

yes, perhaps it would be a good idea to point this out to the OP as well.
hope I didn't confuse anyone.