# Internal flavor symmetry of the $N$ left-handed complex Weyl spinors v.s. $N$ real Majorana spinors: ${\rm U}(N)$ vs. ${\rm O}(2N)$ or ${\rm O}(N)$

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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))

to

• 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?

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

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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
answered Apr 5, 2020 by (70 points)

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