 Fermi superfield

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 ${\sum }_a{f}_a{J}^a=0$.