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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?

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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.

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