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In flat four-dimensional N=1 supersymmetry, let ${\bar{D}}^2={\bar{D}}_{\dot{\alpha }}{\bar{D}}^{\dot{\alpha }}$, contracting the two spinor indices with the antisymmetric epsilon invariant tensor. The supercovariant derivative algebra with left derivatives has $\{{\bar{D}}_{\dot{\alpha }},{\bar{D}}_{\dot{\beta }}\}=0$. There are only two dotted components, so every product of three barred derivatives repeats one component and vanishes. Therefore ${\bar{D}}_{\dot{\alpha }}{\bar{D}}^{2}X=0$ for any superfield $X$: ${\bar{D}}^{2}X$ is a chiral superfield. This map is called chiral projection, although ${\bar{D}}^2$ is not an idempotent projector.

If $\Phi$ is a chiral superfield, holomorphic closure of chiral superfields also makes $W^{\prime }(\Phi )$ chiral. Thus an equation ${\bar{D}}^2{\Phi }^{†}=4W^{\prime }(\Phi )$ is consistent with the chiral constraint: applying any ${\bar{D}}_{\dot{\alpha }}$ gives zero on both sides. Both sides are scalar superfields, and the squared supersymmetric covariant derivative commutes with the supercharges, so the equation is Lorentz invariant and supersymmetry covariant.

The barred notation denotes the antichiral supersymmetric covariant derivative. Write barred spinor indices as dotted indices, and use the stated plus-sign convention with left Grassmann derivatives:Their supercovariant derivative algebra with left derivatives gives $\{{\bar{D}}_{\dot{\alpha }},{\bar{D}}_{\dot{\beta }}\}=0$ and $\{D_{\alpha },{\bar{D}}_{\dot{\beta }}\}=2i{\sigma }_{\alpha \dot{\beta }}^{\mu }{\partial }_{\mu }$. The constraint $\boxed{{\bar{D}}_{\dot{\alpha }}\Phi =0}$ defines a chiral superfield, also called an L-superfield.

To solve the constraint, put $y^{\mu }=x^{\mu }+i\theta {\sigma }^{\mu }\bar{\theta }$. For a left Grassmann derivative, ${\partial }_{{\bar{\theta }}^{\dot{\alpha }}}(\theta {\sigma }^{\mu }\bar{\theta })=-{\theta }^{\beta }{\sigma }_{\beta \dot{\alpha }}^{\mu }$. Therefore ${\bar{D}}_{\dot{\alpha }}{y}^{\mu }=0$ and ${\bar{D}}_{\dot{\alpha }}{\theta }^{\beta }=0$, so $\bar{D}$ becomes ordinary differentiation with respect to $\bar{\theta }$ at fixed $y$. Its general solution is the chiral-superfield component expansionHere $A$ is a complex scalar, $\chi$ a Weyl spinor, and $F$ a complex auxiliary field. Before imposing field equations of motion this is four real bosonic and four real fermionic components. Chirality means independence of $\bar{\theta }$ at fixed $y$, rather than at fixed $x$. The differential supercharges $Q_{\alpha }={\partial }_{{\theta }^{\alpha }}-i{\sigma }_{\alpha \dot{\beta }}^{\mu }{\bar{\theta }}^{\dot{\beta }}{\partial }_{\mu }$ and ${\bar{Q}}_{\dot{\alpha }}={\partial }_{{\bar{\theta }}^{\dot{\alpha }}}-i{\theta }^{\beta }{\sigma }_{\beta \dot{\alpha }}^{\mu }{\partial }_{\mu }$ anticommute with both covariant derivatives, so a supersymmetry transformation preserves the chiral constraint.

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.

For a real Abelian vector superfield, $W_{\alpha }=-{\bar{D}}^2{D}_{\alpha }V/4$ is chiral. The supercovariant derivative algebra with left derivatives gives $D^{\alpha }{\bar{D}}^2{D}_{\alpha }={\bar{D}}_{\dot{\alpha }}{D}^2{\bar{D}}^{\dot{\alpha }}$, proving the displayed reality condition. Its component equations contain the ordinary gauge-theory Bianchi identity, not the Maxwell equations of motion. For a non-Abelian gauge group, replace these by gauge-covariant derivatives and compare both field strengths in the same gauge frame; the identity follows from the graded Jacobi identity of the superconnection.