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

The transformation of the holomorphic strong-coupling scale is fixed by the mixed $SU(2{)}^{2}U(1)$ anomaly. The baryon-number contributions of $\Phi$ and $\tilde{\Phi }$ cancel, so ${\Lambda }^{b_0}$ is neutral under $U(1{)}_B$. Under $U(1{)}_A$, all $2N_f$ doublets have charge one and Dynkin index one, giving axial charge $2N_f$. For the R-symmetry, each matter fermion has charge $(1-2/N_f)-1=-2/N_f$, so the matter contribution is $-4$; the gaugino has R-charge one and contributes $I(\mathrm{adj})=4$, giving zero total anomaly. Thus

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.

For $N_f=2$, the gauge-invariant chiral fields are the four mesons $M_i{}^j={\tilde{\Phi }}_i{\Phi }^j$ and the baryonsClassically they obeyup to the sign chosen in the definitions. No ordinary superpotential has the required R-charge. Nonperturbative dynamics instead produces the quantum-deformed moduli spaceThe right-hand side has exactly the axial charge, dimension and other symmetry properties required to deform the classical relation.

For $N_f=3$, the infrared fields at the origin are the nine mesons $M_i{}^j$ and three baryons plus three antibaryons, for fifteen chiral multiplets. Every elementary quark superfield has R-charge $1/3$, so its fermion has charge $-2/3$. In the ultraviolet there are twelve quark Weyl fermions and three adjoint gauginos. Hence the two anomalies areEach meson or baryon superfield contains two quarks and has R-charge $2/3$, so each composite fermion has charge $-1/3$. The unconstrained infrared fields therefore giveBoth results agree, establishing the requested 't Hooft anomaly matching. This smooth composite description is the $SU(2)$, $N_f=3$ example of s-confinement.