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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.

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The antineutrinos do indeed form a doublet. The particle-antiparticle conjugation operator is usually denoted by $\hat{C}$ and is defined through: \begin{equation} \hat{ C}: \psi \rightarrow \psi ^c = C \bar{\psi} ^T \end{equation} where $ C \equiv i \gamma _2 \gamma _0 $. So given a neutrino you can always get its complex conjugate with this operator: \begin{equation} \nu _L ^{\,\,c } = i \gamma _2 \gamma _0 ( \overline{\nu _L} ) ^T \end{equation} 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: \begin{equation} \left( \begin{array}{c} \nu _L ^{\,c } \\ e _L ^{ \, c} \end{array} \right) \end{equation}

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).

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  • $\begingroup$ 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. $\endgroup$
    – SRS
    Mar 14 '14 at 18:18
  • $\begingroup$ 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. $\endgroup$
    – innisfree
    Mar 14 '14 at 19:38
  • $\begingroup$ It is Majorana since $\nu = \nu_L + \nu_R^c = \nu^c$ $\endgroup$
    – innisfree
    Mar 14 '14 at 19:40

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