The action from which Weyl equation can be derived is


where $\bar{\sigma}^\mu=(1,\vec{\sigma})$. Imposing $\delta S=0$ we arrive at

$$\bar{\sigma}^\mu \partial_\mu\psi_{L/R}=0$$

But this, in turn, means that the action is zero for every spinor that satisfies $\delta S=0$. Maybe there is nothing special about it always being zero, but why does it always have the same value? Is there any interesting mathematical reason for this, or maybe any physical consequence to it? Thank you!



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