The holomorphic free boson contributes <central charge> $c_X=1$, while one real chiral free fermion contributes $c_\psi=1/2$. Equivalently, the double contractions in $T(z)T(w)$ give
$$
T(z)T(w)\sim\frac{3/4}{(z-w)^4}
+\frac{2T(w)}{(z-w)^2}+\frac{\partial T(w)}{z-w}.
$$
Since the fourth-order coefficient is $c/2$,
$$
\boxed{c=1+\frac12=\frac32}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 306 / 3 / iv / Solution

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In a two-dimensional supersymmetric conformal field theory, the superconformal current has weight $3/2$ and its self-OPE closes on the stress tensor and the <central charge>.

Superconformal current

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The central charge is the coefficient of the fourth-order pole in the stress-tensor operator-product expansion, $T(z)T(w)\sim c/[2(z-w)^4]+\cdots$. A free real boson contributes one and a free real chiral fermion contributes one half.


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