Consider 4d spacetime, it seems that for massless particles, we can easily change

  • the left-handed complex Weyl spinor basis (2 component in complex $\mathbb{C}$ for Euclidean spacetime Spin(4))


  • the real Majorana spinor basis (4 component in real $\mathbb{R}$ for Euclidean spacetime Spin(4))

So naively, we can change N left-handed complex Weyl spinors to N real Majorana spinors.

However, the internal flavor symmetry of the N left-handed complex Weyl spinors is $G_{Weyl}=$ U(N).

Puzzle 1: What are the internal flavor symmetry of N real Majorana spinors? $G_{Majorana}=?$ Is that O(N) or O(2N)?

Puzzle 2: Why the internal flavor symmetry of the N left-handed complex Weyl spinors different from N real Majorana spinors?


1 Answer 1


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) $U(N)$.


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