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A Majorana particle is a fermion whose antiparticle is itself. In other words, a Majorana fermion has only two components, rather than 4 components for a Dirac fermion. Neutrinos are alleged as possibly Majorana fermions in most text books.

However with the observation of neutrino oscillations, it's known that neutrinos do possess masses, either Dirac or Majorana masses (or most likely both, required by the seesaw mechanism to account for the smallness of neutrino masses). Either case indicates the existence of right-handed (RH) neutrinos in addition to the left-handed (LH) neutrinos (hence there are 4 independent components for neutrinos, invalidating the Majorana neutrino hypothesis), since:

  • Dirac mass couples LH neutrinos with RH neutrinos.
  • Majorana mass is only allowed for sterile fermions, such as RH neutrinos. Whereas LH neutrinos participate in weak interactions, thus could not directly carry Majorana mass (unless one is willing to introduce the non-renormalizable Weinberg operator to beget the Majorana mass term via the VEV of the Higgs field, as noted by Cosmas Zachos), which would explicitly violate the electroweak symmetries. In the case of Majorana neutrino, the antiparticles of LH neutrinos are still involved in weak interactions, even though they are purported as RH. The Wikipedia page on Neutrino Oscillation espouses the false claim (is introducing non-renormalizability the smallest modification?) that "The smallest modification to the Standard Model, which only has left-handed neutrinos, is to allow these left-handed neutrinos to have Majorana masses".

So with the observation of neutrino oscillations since 20 years ago (dated back to 1998), why is the notion of neutrinos as possiblly Majorana fermions still promoted?

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    $\begingroup$ The Weinberg operator is gauge invariant, alright. You are using "Majorana particles" in a nonstandard fashion. "Majorananess" today is tantamount to the existence of lepton-number-violating Majorana mass terms. $\endgroup$ – Cosmas Zachos Sep 17 '18 at 16:51
  • $\begingroup$ @CosmasZachos is correct. As far as I know, the experiments that talk about possible Majorana neutrinos today are those that are looking for lepton-number violation (e.g. neutrinoless double-beta decay experiments). $\endgroup$ – probably_someone Sep 17 '18 at 17:12
  • $\begingroup$ Thank you @CosmasZachos! So the "standard definition" of Majonana fermion is keeping on changing with time! The latest "standard definition" would amount to the existence of sterile neutrino, i.e. the RH neutrino. Man, physicists are the sneaky bunch! $\endgroup$ – MadMax Sep 17 '18 at 17:31
  • $\begingroup$ So the latest "standard definition" of Majonana neutrino is tantamount to requiring 4 independent component (2 for LH and 2 for RH ) neutrinos, right? What an irony! given the original definition of 2 components only. $\endgroup$ – MadMax Sep 17 '18 at 17:39
  • $\begingroup$ So the "subtlety" in Wikipedia's definition of "left-handed neutrinos" (as quoted in the question body) actually means right-handed neutrinos (sterile neutrinos )? Hmm, physicists are so artful with "matter of language". $\endgroup$ – MadMax Sep 17 '18 at 18:47
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In 4 dimensions, if you are a fermion, you cannot be simultaneously Majorana and Weyl. So a 4-component spinor can be written as a composite of L and R chirality components, but also as two different Majorana (self-conjugate) components, $$ \psi=\psi_L + \psi_R =(\chi +\omega)/2 +\gamma_5 (\omega-\chi)/2, $$ where $$ \chi=\chi^c=\psi_L+ \psi_L^c, \qquad \omega=\omega^c=\psi_R+\psi_R^c. $$

The most general mass matrix (consistent with Lorentz invariance) is $$ D\overline{\psi_L}\psi_R +A \overline{\psi_L} \psi_L^c +B \overline{\psi_R} \psi_R^c + \mathrm{h.c.} ~~~~. $$

The first term (D) is the standard Dirac mass term preserving lepton (fermion) number, and merely changing chirality of the fermion; but the other two (B,A) violate lepton number, as they convert fermions into antifermions and vice-versa, and appear in the Majorana equation for self-conjugate fermion fields.

The general mass eigenstates are then self-conjugate linear combinations of $\psi_L, \psi_L^c,\psi_R, \psi_R^c$.

Now, in the Standard Model, you only have D, where $\psi_L$ is a weak isodoublet (active), while $\psi_R$ is an isosinglet (sterile); and likewise for their opposite chirality antiparticles. But, through the magic of Higgs saturation (the second job of the Higgs) $\psi_L$ is effectively a singlet, as it "really" is the gauge-invariant term $\langle \Phi\rangle \sigma_2\cdot \psi_L$ inside the Yukawa coupling term. This is dimension 5/2, so saturating it with a dimension 3/2 sterile $\psi_R$ yields a dimension 4 term, thus renormalizable!

However, Weinberg further noticed that the square of this composite, corresponding to A, is also gauge invariant, but of dimension 5, so unrenormalizable; it could be induced by extraneous BSM interactions at higher energies.

The conventional see-saw sets it =0, but if renormalizability were mooted for a moment, it could be there. The B term never threatened gauge invariance, and most people consider it enormous.

So the eigenstate of the mass matrix is mostly χ with a tiny admixture of sterile ω , provided B is enormous and D "conventional", and A small. The other superheavy eigenvector is mostly sterile ω with a small contamination of active neutrino. This is not really necessary, as it could be that A=B=0 and lepton number is conserved, but a consensus of expectation doesn't bet much on this dull possibility.... and theorists like dreaming big...

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  • $\begingroup$ Will the observation of neutrinoless double-beta decay be the confirmation of non-zero A only, B only, A or B, or A and B? $\endgroup$ – MadMax Sep 17 '18 at 22:26
  • $\begingroup$ Hmm... a combination of A and B, not sure which... Vogel 2000 discusses such, but the 3 generations substantially complicate the picture... $\endgroup$ – Cosmas Zachos Sep 17 '18 at 22:52

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