A <chiral superfield> is annihilated by the antichiral <supersymmetric covariant derivative>:
$$
\overline{\mathcal D}_{\dot\alpha}\Phi=0.
$$
Equivalently, it depends on $(y^\mu,\theta)$ with $y^\mu=x^\mu+i\theta\sigma^\mu\bar\theta$, and has the finite expansion $\Phi=\phi+\sqrt2\theta\psi+\theta^2F$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 307 / 1 / iii / Solution

{title2=$D_\alpha,\bar D_{\dot\alpha}$}

Supersymmetric covariant derivatives anticommute with the supersymmetry generators. Their antichiral member defines a chiral superfield through $\bar D_{\dot\alpha}\Phi=0$.


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 Incoming links: Supersymmetric covariant derivative