 Solution

For $N_f=1$, antisymmetry makes the would-be baryons vanish, and the only generator of the chiral ring of a supersymmetric gauge theory is the meson operator in supersymmetric quantum chromodynamics $M=\tilde{\Phi }\Phi$. It obeys no classical constraint. Holomorphy, mass dimension and all three U(1) charges uniquely permit the Affleck–Dine–Seiberg superpotentialup to a nonzero constant absorbed into $\Lambda$. Since ${\partial }_{M}W=-{\Lambda }^5/M^2$ never vanishes at finite $M$, there is no finite supersymmetric ground state. Instead the potential approaches zero as $|M|\to \infty$: the theory has a supersymmetric runaway vacuum at infinity.