The question of whether a given particle "is" a Dirac or Majorana fermion is more subtle than is sometimes presented. For example, if we just consider the "old" Standard Model with massless neutrinos, then as Srednicki points out (pg. 550), each neutrino species can be described using either a Dirac or a Majorana bispinor field. That's because each neutrino only has two independent spin degrees of freedom and is (arguably) most naturally thought of as being represented by a Weyl field. As far as I can tell, it only makes sense to talk about a type of fermion "being" Dirac or Majorana if one formalism is overwhelmingly more natural than the other. And I don't see why this is the case for massive neutrinos.

If we extend the "old" Standard Model (considering only one lepton generation for simplicity) by introducing a new Weyl field $\bar{\nu}$ that is uncharged under all the gauge fields and represents a sterile neutrino, then the most general quadratic mass term we can write down for the neutrino fields is $$\mathcal{L}_\text{mass} = -\frac{1}{2} \left( \begin{array}{cc} \nu & \bar{\nu} \end{array} \right) M \left( \begin{array}{} \nu \\ \bar{\nu} \end{array} \right) - \frac{1}{2} \left( \begin{array}{cc} \nu^\dagger & \bar{\nu}^\dagger \end{array} \right) M \left( \begin{array}{c} \nu^\dagger \\ \bar{\nu}^\dagger \end{array} \right),$$ where the mass matrix $$M := \left( \begin{array}{cc} M_L & D \\ D & M_R \end{array} \right).$$ (Unfortunately, the $M$ without a subscript stands for "mass" and the $M$s with subscripts stand for "Majorana".)

The $D$ terms comprise a Dirac-type mass term that conserves lepton number, while the $M$ terms comprise Majorana-type mass terms that do not conserve lepton number. (As explained here, the $M_L$ terms raise subtle issues of gauge invariance and renormalizability; they are renormalizable, but the Higgs mechanism only gives rise to them if we temporarily allow non-renormalizable terms in the pre-symmetry breaking Lagrangian. For simplicity, we'll neglect these terms in this question.)

It seems to me that the generic case has both Dirac and Majorana mass terms, so I don't understand what people mean when they talk about neutrinos "being Dirac or Majorana fermions". Please correct me if I'm wrong, but as far as I can tell, when people talk about the possibility of neutrinos "being" Dirac fermions, they're referring to the case $D \neq 0,\ M_R = 0$, and when they talk about the possibility of neutrinos "being" Majorana fermions, they're referring to the case $D, M_R \neq 0$, where the seesaw mechanism provides a natural(-ish) explanation for the tiny neutrino masses.

But why does the latter case correspond to neutrinos being Majorana fermions? There are still two independent Weyl fields, four independent spin degrees of freedom, and a Dirac mass term. It seems to me that the legitimate way to describe this situation is that neutrinos are neither Dirac nor Majorana fermions, as there are two independent Weyl fields (unlike the purely Majorana case) and lepton number is not conserved (unlike the purely Dirac case). Are people just using extremely sloppy language, or is there a sense in which neutrinos actually are Majorana fermions?


You're completely correct: it's perfectly allowed to have both Dirac and Majorana mass terms. However, the presence of a Majorana mass term (whether or not a Dirac mass term is present) implies the violation of lepton number. When people say they're testing for whether a neutrino is Majorana, they just mean that they're looking for such violations. For a nice review of some simple neutrino mass models, phrased in the same terms that you used, see the relevant chapter in Burgess and Moore, The Standard Model.

I don't think this necessarily is sloppy language. I think that in condensed matter, whether a fermion is Majorana or not is a sharply defined, important thing. However, in particle physics, when we say that a particle is a Blah fermion (where Blah could be Weyl, Majorana, or Dirac), we mean that we have in mind a description for that particle in terms of Blah fermion fields.

For example, a given massless neutrino state could be created by a left-chiral Weyl field, a right-chiral Weyl field, or a Majorana field. None of this affects the physics; the fields are just a bookkeeping tool that help us write down interactions for the particles. As a more extreme example, Burgess and Moore go further and describe all the fermions in the Standard Model as Majorana fields (i.e. the electron corresponds to two separate Majorana fields, but with their Majorana mass terms each set to zero), solely because this allows them to use 4-component spinors and the associated computational tools.

Historically, the distinction between Weyl, Dirac, and Majorana fields was based on the fields' Lorentz transformation properties. However, these days this is becoming less important, so the same words are repurposed. In condensed matter, the words' original meanings can't matter because there's no Lorentz symmetry, so they seem to be used to denote properties of the spectrum, or of the (anti-)commutation relations describing the system. And in particle physics, the original meanings are less important in neutrino physics for the reasons I gave above, so they are adapted to pin down the only physical thing that varies between the possibilities -- namely, whether the particle number is conserved.

  • $\begingroup$ Hmm, so you're saying that particle people just use "Majorana" to mean "any fermion field that isn't a (pure) Dirac fermion coupled to a conserved current"? $\endgroup$ – tparker Mar 13 '20 at 1:35
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    $\begingroup$ I often see people say things like "if neutrinos turn out to be Majorana fermions then that means that there is no distinction between a neutrino and an antineutrino." So that's just wrong? $\endgroup$ – tparker Mar 13 '20 at 1:36
  • $\begingroup$ @tparker I think people don't use it entirely consistently, but yes, that's what Majorana means. If the Majorana mass term were sufficiently small, however, I could imagine people describing the neutrino as Dirac. $\endgroup$ – knzhou Mar 13 '20 at 2:46
  • $\begingroup$ @tparker Well, I think that usage just agrees with what I said. Ignoring the other generations for a moment: there are two electron neutrino degrees of freedom we see. How did we decide which one to call the neutrino and which to call the antineutrino? The neutrino is the one so that, if you have a neutrino in a box, and put in a lot of energy (but nothing carrying lepton number), the box could eventually contain one electron, plus other stuff, but no other leptons. (Ditto for antineutrino and the positron.) If you lose electron number conservation, this distinction doesn't work anymore. $\endgroup$ – knzhou Mar 13 '20 at 2:47
  • $\begingroup$ I've long since given up on condensed matter people ever using the term "Majorana fermion" correctly, but I thought I could still count on the particle people :'-( $\endgroup$ – tparker Mar 13 '20 at 4:00

I have to say that I do not completely agree with the answer given by knzhou as I think he misses a crucial point in his explanation.

Of course it is right that the most general mass term contains both Dirac and Majorana terms and the appearance of Majorana terms implies lepton number violation. We can summarize the mass term in matrix form as $$-\mathcal{L}_m = \frac{1}{2}n_L^TC\mathcal{M}n_L + h.c.$$ with $$n_L = \left(\begin{matrix}\nu_L\\(N_R)^c\end{matrix}\right)$$ and $$\mathcal{M}=\left(\begin{matrix}M_L & M_D \\ M_D^{T} & M_R\end{matrix}\right)\label{eq:neutrino_mass_matrix}$$ Here, $M_D,M_L$ and $M_R$ are $n\times n$ matrices (where n is the number of generations) and represent Dirac mass terms, left-handed Majorana mass terms and right-handed Majorana mass terms.

So far so good. But we should not miss one point. Here we are looking at neutrinos as flavor states. When talking about massive particles we have to diagonalize the mass matrix. Assuming $M_R$ to be invertible, we can block-diagonalize by a base transformation $$-\mathcal{L}_m\longrightarrow\frac{1}{2}\chi_L^TC\mathcal{M}_{\rm{diag}}\chi_L + h.c.$$ with $$ n_L=U\chi_L\\\mathcal{M_{\rm{diag}}}= U^T\mathcal{M}U = \left(\begin{matrix}\tilde{M}_L & 0 \\ 0 & \tilde{M}_R\end{matrix}\right)$$ now we are left with massive fields $\chi_L$ which have only a Majorana mass term.

You can do the whole calculation in the limit of 1 generation to check.

This is nicely explained in the lectures on neutrino physics by Evgeny Akhmedov.

  • $\begingroup$ Interesting. What is the exact requirement for diagonalizability? That both $M_L$ and $M_R$ be invertible? Or at least one of them? It can't just be that $M_R$ be invertible as you imply, because that treats $M_L$ and $ M_R$ asymmetrically. $\endgroup$ – tparker May 14 '20 at 13:39
  • $\begingroup$ You're right $M_R$ being invertible is just a nice assumption that makes the calculation easier in the limit $M_R>>M_{i\neq R}$. You get diagonalizability from $\mathcal{M}$ being hermitian. $\endgroup$ – Katermickie May 20 '20 at 16:10
  • $\begingroup$ So it seems to me that you're saying that the mass eigenstates always have solely Majorana mass terms (except in the trivial case $\mathcal{M} = 0$ where there are no mass terms at all). So if that's the case, then what exactly is the debate about whether "neutrinos are Dirac or Majorana fermions"? Isn't the answer always Majorana fermions? $\endgroup$ – tparker May 21 '20 at 1:28
  • $\begingroup$ If $M_L = M_R = 0$ you will end up with pairs of majorana fermions which have the same mass $M_D$. Now you should know that you can always describe a Dirac fermion (which has 4 degrees of freedom) in terms of two Majorana fermions (which each have two degrees of freedom). This is the situation you get in the case of vanishing majorana masses $M_L$ and $M_R$. Note that you need the two majorana fields to have the same mass and set of other quantum numbers to do so. $\endgroup$ – Katermickie May 22 '20 at 7:00
  • $\begingroup$ In other words. You can always describe fields as Majorana by decomposing a Dirac field. However you cannot always describe fields as Dirac. $\endgroup$ – Katermickie May 22 '20 at 7:02

There is an experimentally observable difference between the two. If neutrinos were Dirac fermions, we would never observe neutrinoless double beta decay. If neutrinos were Majorana fermions, they could never carry an additive charge, like U(1) electric charge, no matter how small. Since we do not observe neutrinoless double beta decay, and neutrinos are uncharged under electromagnetism, it is difficult to rule on the issue. If the converse happened in either direction, i.e., we observed a neutrinoless double beta decay process, or found out that the neutrino carried a tiny electric charge, the matter would be settled.

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    $\begingroup$ Can somebody explain why this was downvoted? (I'm way out of my area here, just reading all I can find) $\endgroup$ – Sebastián Vansteenkiste Mar 13 '20 at 14:33
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    $\begingroup$ @SebastiánVansteenkiste Because it doesn't answer the question at all. $\endgroup$ – tparker Mar 14 '20 at 1:23

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