Majorana fermions If you write the Majorana spinors as
$$\chi = \begin{pmatrix}\psi_L\\ i\sigma_2\psi_L^* \end{pmatrix} \tag1$$
It satisfies the Dirac equation that leads you to the Majorana equation $$i\bar{\sigma}^\mu\partial_\mu\psi_L = im\sigma_2\psi_L.$$
But if $\chi$ satisfies Dirac, is Dirac's Lagrangian the Lagrangian for $\chi$? My questions arises from one of my QFT classes where the professor said that Majorana fields like Eq. (1) don't have $U(1)$ symmetry. Nevertheless, if it is satisfying Dirac's equation, its Lagrangian should be Dirac's and therefore, it should have $U(1)$ symmetry.
Moreover, how do you know that Majorana's are self-conjugated? Do you impose it or it's a result from the Lagrangian or Eq. (1) or somewhere? I've been trying to understand this but I'm really stuck.
 A: Perhaps a simpler example will help show the issue. Consider the complex scalar field Lagrangian,
$$\mathcal{L} = (\partial_\mu \phi^*)(\partial^\mu \phi) - m^2 \phi^* \phi.$$
This Lagrangian has a $U(1)$ symmetry by phase rotations.
Now consider a real field $\varphi$. We already know how to deal with them, but out of laziness, we could choose to write the real field as a complex field $\phi$ that happens to be its own conjugate, $\phi^* = \phi$. This is useful because we can just use the very same complex scalar field Lagrangian,
$$\mathcal{L} = (\partial_\mu \phi^*)(\partial^\mu \phi) - m^2 \phi^* \phi.$$
However, this Lagrangian does not have a $U(1)$ symmetry like the original one, even though it looks the same, because $\phi$ is actually real. You simply can't rotate its phase in the first place.
The same logic holds for the Majorana spinor field. We start with a Weyl spinor field $\psi_L$. If we only know about the Dirac Lagrangian, then we don't know how to write down a Lagrangian for this Weyl spinor alone. So we choose to write it as a full Dirac spinor $\chi$ which is constrained to be its own conjugate, and just use the Dirac Lagrangian for $\chi$. However, this Lagrangian does not have a $U(1)$ symmetry like the original one, even though it looks the same, because $\chi$ is self-conjugate. You simply can't rotate its phase in the first place.
