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

Question

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: $i\eta^\dagger\bar\sigma^\mu A_\mu\chi$ ?

Answer 1

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.