# Neutrinos and anti-neutrinos in the Standard Model

In standard model neutrinos and the left handed electron forms SU(2) doublet.

1. What about the anti-neutrinos in the standard model? Do they also form some doublet?

2. If neutrinos have tiny masses will it not imply indirectly and conclusively that right-handed neutrinos must exist in nature?

EDIT : Neutrinos will have Majorana mass term if they are Majorana fermion. Is that right? Now, if neutrinos are Majorana fermions, will they have definite handedness? For example, does $\nu_M=\begin{pmatrix}\nu_L\\ i\sigma^2\nu_L^*\end{pmatrix}$ have definite handedness? Therefore, doesn't it imply that if neutrinos are massive then a right-handed component of it $\begin{pmatrix} 0\\ i\sigma^2\nu_L^*\end{pmatrix}$ must exist? Although we are not using $\nu_R$ to construct this column, does it imply $\nu_M$ do not have a right handed component? It is the column $\nu_M$ which we should call a neutrino. Then it has both the components. However, one can say that a purely right-handed neutrino need not exist if the neutrino is a Majorana fermion. Therefore, it seems that if neutrinos are massive a right handed component of it must exist (be it a Dirac particle as well as a Majorana particle). Correct me if I am wrong.

The antineutrinos do indeed form a doublet. The particle-antiparticle conjugation operator is usually denoted by $\hat{C}$ and is defined through: $$\hat{ C}: \psi \rightarrow \psi ^c = C \bar{\psi} ^T$$ where $C \equiv i \gamma _2 \gamma _0$. So given a neutrino you can always get its complex conjugate with this operator: $$\nu _L ^{\,\,c } = i \gamma _2 \gamma _0 ( \overline{\nu _L} ) ^T$$ Its easy to check this that this antineutrino is actually right handed, by applying a left projector onto it.
The antineutrino forms a doublet with the antileptons: $$\left( \begin{array}{c} \nu _L ^{\,c } \\ e _L ^{ \, c} \end{array} \right)$$
With regards to your second question, no having neutrino masses does not imply that there exist right handed neutrinos. This is because neutrinos could have Majorana masses ($\frac{m}{2} \nu _L \nu _L +h.c.$) as well as Dirac masses $m( \overline{\nu_L} \nu_R + h.c.)$. Majorana masses could arise if for example there exists a heavy Higgs which is a triplet under $SU(2)_L$ (which can be rise to what's known as a type 2 See-saw mechanism).
• Neutrinos will have Majorana mass term if they are Majorana fermion. Is that right? Now, if neutrinos are Majorana fermions, will they have definite handedness? For example, does $\nu_M=\begin{pmatrix}\nu_L\\ i\sigma^2\nu_L^*\end{pmatrix}$ have definite handedness? Therefore, doesn't it imply that if neutrinos are massive then a right-handed component of it must exist? Although we are not using $\nu_R$ to construct this column, does it imply $\nu_R$ do not have a right handed component? The column is what we should call a neutrino. Then it has both the components. Correct me if I am wrong.
• I'm not sure about your Majorana mass term, shouldn't it read $\bar\nu_L \nu_R^c$? Which for active, ordinary left-handeded neutrinos, could only originate from a triplet? For a sterile, we can write such a term straight away. Commented Mar 14, 2014 at 19:38
• It is Majorana since $\nu = \nu_L + \nu_R^c = \nu^c$ Commented Mar 14, 2014 at 19:40