# Behavior of derivative on Lorentz transformations of spinors

I'm currently working through Supersymmetry Demystified by Patrick Labelle and one passage in particular confuses me.

Specifically, if $$\eta$$ and $$\chi$$ are right and left Weyl spinors respectively, the Weyl equation: $$i\bar{\sigma}^\mu \partial_\mu \chi = m\eta$$ with $$\bar{\sigma}^\mu \equiv \left(1,-\vec{\sigma}\right)$$, shows that $$i\bar{\sigma}^\mu \partial_\mu \chi$$ transforms as a right chiral spinor under Lorentz transformations. The author then states that therefore the expression: $$\left( \partial_\mu \phi \right) \bar{\sigma}^\mu \chi$$ transforms in the same (right chiral) representation of the Lorentz group, regardless of the fact that the derivative acts on a complex scalar field $$\phi$$ rather then the spinor $$\chi$$. I can't find an argument anywhere justifying that statement. Why does the fact that the derivative acts on a scalar field instead of the spinor not influence its behavior under Lorentz transformations?

For a complex scalar $$\phi$$ and left-chiral Weyl spinor $$\chi$$, the quantity $$\partial_\mu (\phi \bar{\sigma}^\mu \chi)$$ transforms in some representation of the Lorentz group. Expanding out the product, $$(\partial_\mu \phi) \bar{\sigma}^\mu \chi \text{ and } \phi \partial_\mu \bar{\sigma}^\mu \chi$$ transform in the same representation of the Lorentz group. Since a scalar transforms trivially, the latter transforms in the same representation as $$\partial_\mu \bar{\sigma}^\mu \chi$$, which we established transforms like a right-chiral Lorentz spinor.