# Why is the derivative necessary to connect left and right-hand spinors?

I am studying Weyl and Dirac spinors. Suppose we have two Weyl fermions $$\eta, \chi$$ transforming under $$(1/2,0)$$ representation of the Lorentz group. I learned that to construct Lorentz invariant term in the Lagrangian, we have terms like $$i\eta^\dagger\bar\sigma^\mu\partial_\mu\chi$$ or $$i\eta\sigma^\mu\partial_\mu\chi^\dagger$$. I understand that $$\sigma^\mu$$ maps left-handed spinors to right-handed and it is Lorentz invariant, therefore sandwiched between the two spinors, but I am still not quite sure how can I justify that $$\partial^\mu$$ is needed, and we don't need higher-derivative terms?

Also, why for a vector field $$A_\mu$$, the possible interaction is similar to the derivative term: $$i\eta^\dagger\bar\sigma^\mu A_\mu\chi$$ ?

Index notation is rather useful here.

Left-handed spinors have $$\psi_\alpha$$ and right-handed spinors have $${\bar \psi}_{\dot \alpha}$$ (or the other way around, I can never remember what the conventions are).

If you want a Lagrangian with only left-handed spinors, we need to write a term of the form $$\psi_\alpha D^{\alpha\beta} \psi_\beta$$. The only tensorial object with this tensor structure is $$( \sigma^{\mu\nu} )^{\alpha\beta}$$. This has extra $$\mu\nu$$ indices so we need to contract them. However, it is anti-symmetric in $$\mu\nu$$ and there is no other such tensor quantity we can contract with. So a kinetic term involving two $$\psi_\alpha$$'s is impossible.

If you follow the same sort of argument with $${\bar \psi}_{\dot \alpha} D^{{\dot \alpha} \beta} \psi_\beta$$ then you will see that at the lowest derivative order, the Dirac Lagrangian is the only possibility. The same argument follows for the gauge field as well.

• Thanks for the answer! Could there be a kinetic term looks like $\psi^\gamma(\sigma^{\mu\nu})^{\alpha}_\gamma(\sigma_{\mu\nu})_{\alpha}^\delta\psi_\delta$ which involves two left-hand spinors?
– IGY
Feb 26 at 2:19
• Or that term looks more like an interaction term instead?
– IGY
Feb 26 at 2:32
• Firstly, that term has no derivatives so it cannot be a kinetic term. That term simplifies to $-3\psi^\alpha \psi_\alpha$ and such a term is totally allowed. Feb 26 at 10:55