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

For three identical operators, conformal invariance givesWhen $|x|\ll |y|$, $|y-x|=|y|[1+O(|x|/|y|)]$, soInserting the operator product expansion, the identity term has zero expectation with $O(y)$, while the $O(0)$ term giveswhich has precisely the required position dependence.

Solved by gpt-5.6-sol high.

Comparing the coefficients in the short-distance limit givesIf the operator is normalized so that $C_2=1$, its OPE coefficient equals the three-point normalization.

Solved by gpt-5.6-sol high.