Consider the Lagrangian for the type-I seesaw given by $$-\mathcal{L}=\bar{\nu}_{L}m_DN_{R}+\frac{1}{2}\overline{(N_{R})^c}M_R N_{R}+\text{h.c.}.$$

$\bullet$ In this Lagrangian, what is the nature of $N_R$ field? I think it cannot be Majorana even though we write a Majorana mass for it because we don't take $(N_R)^c=N_R$ (the defining condition for a field to be Majorana).

$\bullet$ I know that after diagonalization, the light, and heavy neutrinos will be linear combinations of the fields $\nu_L$ and $N_R$, and are all Majorana particles. But my question is what can we say about the fields $N_R$?


A Dirac spinor can be written as the sum of two Weyl (chiral) spinors, eigenstates of $\gamma_5$, $$ \psi= \psi_L + \psi_R,\tag{1} $$ but also, alternatively, as the sum of two Majorana (self-conjugate) spinors, eigenstates of C (and $i\gamma_2$) $$ \psi= \chi + i\omega= \chi^c + i\omega^c,\tag{2} $$ where each Majorana component is real in the Majorana representation, in which $\gamma_5= \sigma^3\otimes \sigma^2$, imaginary, $i\gamma_2= \sigma^2\otimes\sigma^2$, real, and $C=-i\sigma^1\otimes \sigma^2$, real. Note C and $i\gamma_2$ do not commute with $\gamma_5$ (a feature of 4 dimensions).

Consequently, a Majorana spinor cannot be Weyl, and a Weyl spinor cannot be Majorana. The Majorana and Weyl bases are mutually exclusive bases for the components of a Dirac spinor.

$N_R$ is a Weyl spinor, and combines with $\nu_L$ to resolve into two Majorana components. Don't worry about N vs ν, they are meant to recall EW quantum numbers.

A comprehensive treatment is in section 13.2 of ISBN-13: 978-0198506218, Gauge Theory of Elementary Particle Physics: Problems and Solutions, 1st Edition by Ta-Pei Cheng & Ling-Fong Li.

Edit as per references request : M Schwartz's notes, authoritative alright, illustrate the link in the Weyl, not Majorana basis, where everything is complex, instead, so contrast eqns (9) to (34).

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The answer sits in the definition of charge conjugation as:

$$\psi^c = C \bar{\psi}^T$$

where $C$ is some ''book dependent matrix''. In many notes, it is chosen to be $i\gamma_2$ but this will be irrelevant for our discussion.

With this in mind, we can now rewrite the action from your question as:

$$-\mathcal{L} = \bar{\nu_L} m_D N_R + \frac{1}{2}C (N_R)^T M_R N_R + h.c.$$

At which point is becomes very clear that $N_R$ is indeed a Majorana neutrino.

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  • $\begingroup$ You can also define $\psi^c$ for a Dirac field $\psi$. It's just that $\psi\neq \psi^c$ for a Dirac field. Therefore, the presence of $\psi^c$ in the Lagrangian doesn't ensure that $\psi$ is a Majorana neutrino. @gertian $\endgroup$ – SRS May 18 '17 at 9:35
  • $\begingroup$ I agree, my point was that $\bar{\psi}^c \sim \psi^T$ it is just a strange way to write down the transpose of your field. If you look in this paper you will find that they use the Lagrangian that I obtained after the substitution: arxiv.org/pdf/hep-ph/9911364.pdf $\endgroup$ – gertian May 18 '17 at 9:47
  • $\begingroup$ As soon as you write down a Majorana mass term for it you have implicitly assumed that it must obey $\psi^c = \psi$ otherwise you would lose the symmetries in your theory. $\endgroup$ – gertian May 18 '17 at 10:35
  • $\begingroup$ I'm not convinced (although you may be correct). Nowhere you need to assume $(N_R)^c=N_R$. And it appears to me that you write the Majorana mass term because it is compatible with Lorentz invariance and gauge invariance. @gertian $\endgroup$ – SRS May 18 '17 at 10:44
  • $\begingroup$ Yes indeed you write the Majorana term becouse it is compatible with your gauge (and Lorentz) invariance if and only if $(N_R)^c = N_R$. But to be fairly honest the exact reason why that is the case sits way to far back in my head...(and I am nearing deadline on my thesis so I can't really take the time to look it up.) $\endgroup$ – gertian May 18 '17 at 14:31

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