Can a Majorana field $\psi$ be charged under some $U(1)$ with a charge other than zero? I know Majorana particles have to be electrically neutral because electric charged is conserved.
My question, however, is whether at all a Majorana field $\psi$ be charged under any $U(1)$ (other than $U(1)$ of electromagnetism) with a nonzero $U(1)$ charge? If yes, how? Doesn't the defining condition $\psi=\psi^c$ always mathematically constrain all $U(1)$ charges to be zero? Let me explain why I think so. Suppose, we assign a nonzero U(1) charge $\alpha$ to the field $\psi$ so that $$\psi^\prime=e^{i\alpha}\psi.$$ What is the charge of the field $\psi^c$? It can be easily seen that $$\psi^{c\prime}=C\bar{\psi^\prime}^T=C\gamma^{0T}\psi^{\prime *}=e^{-i\alpha}C\bar{\psi}^T=e^{-i\alpha}\psi^c.$$
Therefore, for any $U(1)$ charge $\alpha$ assigned to the field $\psi$, the complex-conjugated field $\psi^c$ will have the charge $-\alpha$. What does it imply for a Majorana particle where $\psi=\psi^c$? It necessarily implies that $\alpha=0$ i.e., Majorana fields cannot be charged under any U(1) group. Therefore, it is not at all possible to assign a nonzero $U(1)$ charge for this fields. Am I missing something here?
If my conclusion is correct, what does it mean to say that a Majorana mass violates $U(1)$ quantum number (such as lepton number) by 2 units? I must be missing some caveat. What is that?
 A: Majorana field can not be charged under the U(1) group. However, it can be charged under the $\mathbb{Z}_2$ subgroup of U(1). Under the $\mathbb{Z}_2$ transformation, $\psi\to\psi'=-\psi$, any Majorana fermion Hamiltonian (such as the Majorana mass $m\psi\psi$) should be invariant under this $\mathbb{Z}_2$ symmetry, which is also called the fermion parity symmetry. Therefore we can assign a $\mathbb{Z}_2$ quantum number (i.e. the fermion parity) to the Majorana field, which is the lepton number modulo two. In other words, the lepton number is not conserved for the Majorana fermion, but it can only change by 2, so the parity of the lepton number is still conserved.
In the derivation, the charge conjugation condition $\psi'=\psi'^{c}$ requires:
$$e^{\mathrm{i}\alpha}\psi=e^{-\mathrm{i}\alpha}\psi^c=e^{-\mathrm{i}\alpha}\psi,$$
which implies
$$e^{\mathrm{i}\alpha}=e^{-\mathrm{i}\alpha}.$$
This equation has two solutions: $\alpha=0$ and $\alpha=\pi$. The solution of $\alpha=\pi$ corresponds to the non-trivial $\mathbb{Z}_2$ charge. In general, all composite fields made out of the Majorana field (such as $\psi_a\psi_b\psi_c\cdots$) can carry the $\mathbb{Z}_2$ charge. Let $q=0,1$ be the $\mathbb{Z}_2$ charge, then a charge-$q$ operator $O_q$ transforms under the $\mathbb{Z}_2$ symmetry as
$$O_q\to O'_q=e^{\mathrm{i}q\pi}O_q.$$
One can see the Majroana field $\psi$ itself has $q=1$, which has one unit of lepton number (modulo two). However it is meaningless to talk about the U(1) lepton number of the Majorana fermion, because the U(1) symmetry has been broken down to $\mathbb{Z}_2$. So more presicely, we should say the Majorana field is fermion parity odd (meaning that it carries one unit of the $\mathbb{Z}_2$ quantum number). The Majorana mass $m_{ab}\psi_a\psi_b$, the interaction $V_{abcd}\psi_a\psi_b\psi_c\psi_d$ and all other terms that appear in the Hamiltonian have $q=0$, which are fermion parity even (meaning that they do not carry the $\mathbb{Z}_2$ quantum number).
A: Let me write it down as an answer. Yes, Majorana neutrino cannot be charged under $U(1)$, including global $U(1)_L$ of Lepton number. What we mean by violation of $L$: the processes, like neutrinoless double beta decay, sensitive to the Majorana mass term will violate $L$ by two units, which for the given example corresponds to the production of two electrons, and (surprise!) no neutrinos.
