Is $N_R$ a Majorana field in the Seesaw Lagrangian?

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).

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.

• 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 – SRS May 18 '17 at 9:35
• 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 – gertian May 18 '17 at 9:47
• 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. – gertian May 18 '17 at 10:35
• 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 – SRS May 18 '17 at 10:44
• 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.) – gertian May 18 '17 at 14:31