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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 ...