Under a finite conformal coordinate change $z'=f(z)$ and $\bar z'=\bar f(\bar z)$, a <primary operator> of <conformal weight> $(h,\widetilde h)$ obeys
$$
\mathcal O'(z',\bar z')
=\left(\frac{df}{dz}\right)^{-h}
\left(\frac{d\bar f}{d\bar z}\right)^{-\widetilde h}
\mathcal O(z,\bar z).
$$
Equivalently, its <operator product expansion> with the <holomorphic stress-energy tensor> has singular part
$$
T(z)\mathcal O(w,\bar w)
\sim\frac{h\,\mathcal O(w,\bar w)}{(z-w)^2}
+\frac{\partial\mathcal O(w,\bar w)}{z-w}.
$$
The antiholomorphic stress tensor gives the analogous formula with $\widetilde h$.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 306 / 3 / a / Solution

For three identical operators, conformal invariance gives
$$
\langle O(x)O(0)O(y)\rangle
=\frac{C_3}{|x|^\Delta|y|^\Delta|y-x|^\Delta}.
$$
When $|x|\ll|y|$, $|y-x|=|y|[1+O(|x|/|y|)]$, so
$$
\langle O(x)O(0)O(y)\rangle
\sim\frac{C_3}{|x|^\Delta|y|^{2\Delta}}.
$$
Inserting the <operator product expansion>, the identity term has zero expectation with $O(y)$, while the $O(0)$ term gives
$$
\frac{C_{OOO}^{\rm OPE}}{|x|^\Delta}
\langle O(0)O(y)\rangle
=\frac{C_{OOO}^{\rm OPE}C_2}{|x|^\Delta|y|^{2\Delta}},
$$
which has precisely the required position dependence.

Solved by gpt-5.6-sol high.

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 354 / 1 / c / i / Solution

{title2=OPE}
{wiki}

An operator product expansion expresses the short-distance product of local operators as a sum of local operators multiplied by singular coefficient functions.


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