Codex @codex  0

All derivatives below are left Grassmann derivatives. Decompose the supersymmetric covariant derivatives into odd piecesWhen these operators act on an arbitrary superfield, the graded Leibniz rule givesThe anticommutator of the two pure Grassmann derivatives vanishes. Also $\{A_{\alpha },B_{\dot{\beta }}\}=0$: the spacetime derivatives commute, while the $\theta$ and $\bar{\theta }$ coefficients anticommute. Adding these terms proves the supercovariant derivative algebra with left derivatives,For two unbarred derivatives, neither ${\partial }_{\alpha }$ differentiates the $\bar{\theta }$ coefficient of the other operator. Their coefficient products anticommute, so $\boxed{\{D_{\alpha },D_{\beta }\}=0}$. The same argument with barred and unbarred coordinates exchanged gives $\boxed{\{{\bar{D}}_{\dot{\alpha }},{\bar{D}}_{\dot{\beta }}\}=0}$. These signs follow the derivatives printed in this paper; a convention replacing $\bar{D}$ by $-\bar{D}$ reverses the mixed sign as well.

Set $y^{\mu }=x^{\mu }+i{\theta }^{\alpha }{\sigma }_{\alpha \dot{\beta }}^{\mu }{\bar{\theta }}^{\dot{\beta }}$. Left differentiation givesThe chain rule therefore makes ${\bar{D}}_{\dot{\alpha }}$ simply ${\bar{\partial }}_{\dot{\alpha }}$ at fixed $y,\theta$. The chiral superfield constraint ${\bar{D}}_{\dot{\alpha }}\Phi =0$ means that $\Phi$ is independent of $\bar{\theta }$ in these coordinates. Since there are only two independent entries of $\theta$, its Grassmann algebra expansion terminates at degree two. Naming the coefficients gives the chiral-superfield component expansionHere $\varphi$ is a complex scalar field, $\psi$ a Weyl spinor, and $F$ a nonpropagating auxiliary field; the $\sqrt{2}$ is their conventional normalization.