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A Fermi superfield ${\Lambda }_{-}^a$ obeys ${\bar{D}}_{+}{\Lambda }_{-}^a=f_a(\Phi )$. A supersymmetric $J$-type coupling $\int d{\theta }^{+}{\Lambda }_{-}^a{J}^a(\Phi )$ requires ${\sum }_a{f}_a{J}^a=0$.

Put $\theta ={\theta }^{+}$, $\bar{\theta }={\bar{\theta }}^{+}$ and $y^{+}=x^{+}+i\theta \bar{\theta }$. The general two-dimensional N=(0,2) supersymmetry chiral superfield isIn ordinary coordinates this is $\Phi =\varphi +\sqrt{2}\theta {\psi }_{+}+i\theta \bar{\theta }{\partial }_{+}\varphi$. A supersymmetric kinetic action iswhose component form, up to light-cone conventions and total derivatives, is

A Fermi superfield satisfying ${\bar{D}}_{+}{\Lambda }_{-}=f(\Phi )$ has expansionHere $G$ is a complex bosonic auxiliary field. With the normalization in the question, the component action isup to equivalent sign conventions for the fermions.

The chiral integral in $S_3$ is supersymmetric only if its integrand is chiral. Applying ${\bar{D}}_{+}$ gives the necessary and sufficient conditionwith every $J^a$ holomorphic and with gauge charges chosen so that each product ${\Lambda }_{-a}{J}^a$ is gauge invariant. Its component expansion couples $G_a$ linearly to $J^a$ and supplies the corresponding Yukawa term ${\lambda }_{-a}{\psi }_{+}^i{\partial }_i{J}^a$.

Eliminating each $G_a$ by its algebraic field equation gives the nonnegative scalar potential. In the normalization displayed above and in the question,If the conventional definitions ${\bar{D}}_{+}{\Lambda }_{-}^a=\sqrt{2}{E}_a$ and $S_J=-\int d\theta {\Lambda }_{-a}{J}^a/\sqrt{2}+h.c.$ are used instead, the same result is written $V={\sum }_a(|E_a{|}^2+|J^a{|}^2)$; the two forms differ only by the normalization of $f$ and $J$.

Two-dimensional $\mathcal{N}=(0,2)$ supersymmetry retains one complex right-moving supercharge. Its basic matter multiplets are chiral superfields containing a scalar and right-moving fermion, and Fermi superfields containing a left-moving fermion and an auxiliary scalar.