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

An operator $\mathcal{O}(w,\bar{w})$ is a primary operator of conformal weights $(h,\tilde{h})$ when its OPEs with the stress tensors arewith no more singular terms. Its scaling dimension and two-dimensional spin are

Normal ordering defines a composite field by subtracting all singular self-contractions before bringing its constituent points together. For example,This makes $T$ and $\bar{T}$ well-defined local operators and is equivalent in oscillator language to placing creation operators to the left of annihilation operators.

Only the fermionic part of $T$ contracts with $\psi$. Applying Wick theorem to $T_{\psi }(z)=-\frac{1}{2}:\psi \partial \psi :(z)$ and using $\psi (z)\psi (w)\sim 1/(z-w)$ givesThere is no singular OPE with $\bar{T}(\bar{z})$. Thus $\psi$ is a primary operator with

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 $T(z)T(w)$ giveSince the fourth-order coefficient is $c/2$,

The free equations of motion are $\bar{\partial }\psi =0$ and $\partial \bar{\partial }X=0$, henceThus the superconformal current $G=i\psi \partial X$ is holomorphic. Differentiating the boson OPE gives $\partial X(z)\partial X(w)\sim -1/(z-w{)}^2$. The double contraction in $G(z)G(w)$ is consequently $1/(z-w{)}^3$, while the single contractions combine into twice the full stress tensor. ThereforeThe leading coefficient equals $2c/3=1$ for $c=3/2$.

Extract a mode usingIn the radially ordered double contour for the anticommutator, the simple pole $2T(w)/(z-w)$ gives $2L_{m+n}$. Expanding $z^{m+1/2}$ about $w$, the third-order pole contributes one half of its second derivative,The remaining contour is nonzero only for $m+n=0$. Restoring the general leading OPE coefficient $2c/3$ givesThis is the fermionic relation in the N=1 super-Virasoro algebra.