The symmetry depends on the Lagrangian.

Let $\gamma^\mu$ be a real representation of the Dirac matrices for 4d spacetime, and define $\Gamma := \gamma^0\gamma^1\gamma^2\gamma^3$. Then $\Gamma$ is also a real matrix, and $\Gamma^2=-1$.

If $\psi$ is a Majorana spinor field (with self-adjoint components), then the corresponding left-handed Weyl spinor is
$$
\newcommand{\pl}{\partial}
\newcommand{\opsi}{\overline\psi}
\newcommand{\cL}{{\cal{L}}}
\psi_L := \frac{1+i\Gamma}{2}\psi.
$$
So yes, the Weyl spinor $\psi_L$ may be written in terms of the Majorana spinor $\psi$ and conversely, but the two Lagragnians
$$
\cL \propto \opsi_L \gamma^\mu\pl_\mu\psi_L
\hskip2cm
\cL' \propto \opsi \gamma^\mu\pl_\mu\psi
$$
are not the same. (I'm suppressing the flavor index.) In particular, they have different flavor symmetries. If we start with $\cL$ and rewrite it in terms of the Majorana spinor, we get
$$
\cL \propto \opsi \gamma^\mu\pl_\mu\frac{1+i\Gamma}{2}\psi,
$$
which is different than $\cL'$. The flavor symmetry of $\cL$ is still $U(N)$. To see this, use the identity
$$
i\psi_L = -\Gamma\psi_L
$$
to see that multiplying the Weyl spinor $\psi_L$ by $i$ is the same as multiplying the Majorana spinor $\psi$ by $-\Gamma$, after which its components are still self-adjoint. This shows that every $U(N)$ flavor transformation of the original version of $\cL$ can be re-written as an equivalent flavor transformation of the new version of $\cL$, using $-\Gamma$ in place of $i$, so the flavor symmetry group is still (isomorphic to) $U(N)$.

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user Chiral Anomaly