In a four-dimensional supersymmetric conformal field theory, a <chiral primary operator in four-dimensional N=1 supersymmetry> obeys $\Delta=3R/2$. The meson has <R-charge> $R[M]=2R[\Phi]$, hence
$$
\Delta(M)=3R[\Phi]=3\left(1-\frac{N_c+1}{N_f}\right).
$$
At $N_f=3(N_c+1)$ this gives
$$
\boxed{\Delta(M)=2}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 307 / 4 / iii / Solution

The fermion in a chiral multiplet has <R-charge> $R[\Phi]-1$, while the gluino has charge one. Cancellation of the $Sp(N_c)^2U(1)_R$ anomaly gives
$$
I(\mathrm{adj})+2N_fI(\square)(R[\Phi]-1)=0.
$$
Using $I(\mathrm{adj})=2(N_c+1)$ and $I(\square)=1$ yields
$$
\boxed{R[\Phi]=1-\frac{N_c+1}{N_f}
=\frac{N_f-N_c-1}{N_f}}.
$$

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 307 / 4 / i / Solution

The magnetic superpotential $W\sim qMq$ must have <R-charge> two, so
$$
R[M]=2-2R[q]=2\left(1-\frac{N_c+1}{N_f}\right).
$$
Therefore
$$
\boxed{\Delta(M)=\frac32R[M]
=3\left(1-\frac{N_c+1}{N_f}\right)},
$$
exactly matching the R-charge and dimension of the electric meson $\Phi^i\Phi^j$.

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 307 / 4 / vi / Solution

{c}
{title2=$R$}

R-charge is the charge under a continuous R-symmetry. In four-dimensional N=1 supersymmetry, the fermion in a chiral multiplet has R-charge one less than its scalar, and every superpotential term has R-charge two.


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