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The important point is the fact that such a mass term breaks the gauge symmetry (Edit: I am assming that you want to build the Majorana mass term using SM available fields -- no extension considered -- of which there is only $\nu_L$). Namely, the desired term is (one generation suffices): $$\frac{1}{2}\,M\, \nu_L^T \,\mathcal{C}^\dagger\,\nu_L\, +\, ...


3

Lets analyse the Majorana condition and the Majorana mass term. A massive Majorana neutrino $\chi_j$ (a Majorana spin $1/2$ fermion) having mass $m_j>0$ can be described in a local quantum field theory (eg. the standard model) by a four component spin $1/2$ field $\chi_j(x)$ which satisfies the Dirac equation and the Majorana condition which reads: $$ ...


2

The interaction term disappears because when you integrate out the right handed neutrinos you insert in a linear combination of the other fields in their place. For example I expect that we would have something of the form, \begin{equation} G _{ ij} \sigma \bar{\nu} _L ^i \nu _R ^j \rightarrow G _{ij} G _{jk}\sigma ^2 \bar{\nu} _L ^i \nu _L ^k ...


0

The real reason is in following. Let's assume Majorana field: $$ \Psi_{M} = \Psi_{L} + \hat{C}\bar{\Psi}^{T}_{L}, \quad \hat{C} = i\gamma_{2}\gamma_{0}, \quad \Psi_{L} = \begin{pmatrix} \psi_{L} \\ 0 \end{pmatrix}. $$ By using this notation it's not hard to see that kinetic term is equal to $$ \bar{\Psi}_{M}\gamma^{\mu}\partial_{\mu}\Psi_{M} = ...


0

The short asnwer to your question is that the overall factor $\frac{1}{2}$ from the Lagrangian of a Majorana field (in the 4-component notation) $$\mathcal{L}=\frac{1}{2}(\bar{\psi}i\gamma^{\mu}\partial_{\mu}\psi -m\bar{\psi}\psi)$$ compared to the general Dirac Lagrangian is usual for self-conjugate fields and it is introduced to ensure a consistent ...



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