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

A scalar chiral primary operator at a four-dimensional N=1 superconformal fixed point saturates a shortening bound and obeys $\Delta=3R/2$.


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 Incoming links: Chiral primary operator in four-dimensional N=1 supersymmetry