# Spinors Under Spatial Reflection

How eq(4.4) is a solution of eq(4.3)

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First of all, the spatial reflection in even spacetime dimension maps left-handed spinors to right-handed spinors. So there's no way to define the reflection for Weyl spinors.

For Dirac spinors, it's possible.

(4.4) is the most general solution to (4.3) because $$P = e^{i\phi} \gamma^0$$ is the most general matrix that commutes with $\gamma^0$ but anticommutes with all spatial $\gamma^i$ matrices. That follows from $$\{\gamma^\mu,\gamma^\nu\} = 2g^{\mu\nu}\cdot{\bf 1}$$ So if we conjugate the gamma matrices, $$P (\gamma^0,\gamma^i) P^{-1} = (\gamma^0,-\gamma^i)$$ we get the same sign for the temporal gamma matrix but the opposite sign for other matrices.

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But why exponential term in the solution ? Can you show me the mathematics ? – Fahim Sharif Nov 8 '12 at 15:41
The exponential term is just a phase factor commuting with all the $\gamma^\mu$ and cancelling out with the phase factor of $P^{-1}$. – Claudius Dec 8 '12 at 11:32

Just check it directly, using the anti-commutation relations of the gamma matrices: $\{\gamma^\mu, \gamma^\nu\} = 2\eta^{\mu\nu}$. When $\sigma = 0$, the LHS is just $(\gamma^0)^3 = \gamma^0$, and when $\sigma = i = 1,2,3$, it is $\gamma^0 \gamma^i \gamma^0 = -\gamma^i$. This agrees with the RHS.

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