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

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
  • $\begingroup$ I should also point out that, at present, D could vanish but A not; then we'd know little about B. Or A and B could vanish, in which case D should not. But, for the lightest neutrino, D and A could vanish, in which case we'd know nothing about B, also. $\endgroup$ – Cosmas Zachos Mar 12 at 23:18

Cosmas Zachos's answer is completely correct, but it took me a while to fully understand it, so I'll give a slightly different more basic perspective on the same idea.

In my opinion, it's not particularly useful to frame the question as whether neutrinos "are" Dirac or Majorana fermions: as Srednicki points out, the "old" Standard Model with massless neutrinos can be equivalently formulated with the neutrino physics being expressed in terms of either Dirac or Majorana bispinors, so that's really just a question of semantics. The better way to frame the question is whether neutrinos have Majorana mass terms, Dirac mass terms, both, or neither.

In the "old" Standard Model (considering only a single lepton generation for simplicity), neutrinos are characterized by a single two-spin-component (left-chiral) Weyl field $\nu$. (Before electroweak symmetry breaking, this field is $SU(2)$-symmetric with another left-chiral Weyl field $e$. Electroweak symmetry breaking breaks this symmetry, and $\nu$ describes the massive neutrino while $e$ makes up the left-chiral half of the massive electron bispinor.) If we want to express $\nu$ in terms of a four-spin-component bispinor $\mathcal{N}$, then we have two options: we can either form a Majorana bispinor $$\mathcal{N} := \left( \begin{array}{c} \nu \\ \nu^\dagger \end{array} \right),$$ or a different bispinor $$\mathcal{N}_L := \left( \begin{array}{c} \nu \\ 0 \end{array} \right).$$ $\mathcal{N}_L$ (which Cosmas Zachos calls $\psi_L$ in his answer) is kind of like a Dirac field, but it's nonstandard one because it only consists of one independent Weyl field rather than the usual two (and there is no $U(1)$ symmetry). Which scheme you prefer is a matter of personal opinion; the latter is more common because it's more mathematically convenient, but the former is arguably conceptually clearer because it manifestly demonstrates that the field only has two independent spin components, rather than four as for a general Dirac bispinor.

The minimal way to modify the old Standard Model to incorporate neutrino masses is to introduce a completely new independent left-chiral Weyl field $\bar{\nu}$ which is uncharged under any of the gauge fields (i.e. in the singlet represtation). This new field simply isn't a part of the postulated matter content of the old Standard Model. It indirectly couples to the Higgs field via a three-point Yukawa coupling with the Higgs field and the $SU(2)$ doublet Weyl field $(\nu, e)$ (note that this ordered pair ranges over $SU(2)$ indices, not spin indices).

(I believe that Cosmas Zachos's answer is embedding $\bar{\nu}$ into a four-spin-component bispinor $$\psi_R := \left( \begin{array}{c} 0 \\ \bar{\nu}^\dagger \end{array} \right),$$ but I personally find it more straightforward to work directly with the two-spin-component Weyl spinor fields. Depending on your perspective, either $\bar{\nu}$ or $\psi_R$ "is" the sterile neutrino.)

From the two Weyl fields $\nu$ and $\bar{\nu}$, we can form three different times of quadratic combinations, which are the three different types of mass terms that Cosmas Zachos describes in his answer. The term that mixes $\nu$ and $\bar{\nu}$ (Cosmas Zachos's "$D$" term) is the Dirac mass term, which conserves lepton number. The $A$ and $B$ terms are Majorana masses that violate lepton number. (As Cosmas Zachos discusses, the $A$ term has some subtleties regarding renormalizability.) As Cosmas Zachos points out, the question of whether neutrinos "are" Majorana fermions is more precisely the question of the $A$ and $B$ terms are nonzero. The alternatives are either (a) $A$ and $B$ are fined-tuned to zero for some unknown reason, which is rare but not unheard of (see the theta angle), or (b) the entire premise of introducing a new sterile-neutrino $\bar{\nu}$ is wrong, and we need a more convoluted explanation for the neutrino masses.

My only quibble with Cosmas Zachos's answer is his claim that "in the Standard Model, you only have $D$". I would perhaps call perhaps the "new Standard Model", which accomodates neutrino masses with the mimimal number of additional new nonzero parameters. In the "old Standard Model", where neutrinos are completely massless, then all three of the terms in Cosmas Zachos's answer are zero, because $\psi_R$ simply isn't a part of the theory.

I've frankly never understood why the presence of Majorana mass terms is described as the neutrino "being" a Majorana fermion - there are still Dirac mass terms as well. I'd welcome any comments in this issue.

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