For $N_f=2$, the gauge-invariant chiral fields are the four mesons $M_i{}^j=\widetilde\Phi_i\Phi^j$ and the baryons
$$
B=\epsilon_{ab}\Phi_1^a\Phi_2^b,
\qquad
\widetilde B=\epsilon^{ab}\widetilde\Phi_{1a}\widetilde\Phi_{2b}.
$$
Classically they obey
$$
\boxed{\det M-B\widetilde B=0}
$$
up to the sign chosen in the definitions. No ordinary superpotential has the required R-charge. Nonperturbative dynamics instead produces the <quantum-deformed moduli space>
$$
\boxed{\det M-B\widetilde B=\Lambda^4}.
$$
The right-hand side has exactly the axial charge, dimension and other symmetry properties required to deform the classical relation.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 307 / 3 / ii / Solution

When $N_f=N_c$, SQCD has no generated superpotential but its classical constraint is shifted by the strong-coupling scale. For $SU(2)$ with two flavors, one convention gives $\det M-B\widetilde B=\Lambda^4$.


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 Incoming links: Quantum-deformed moduli space