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.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-354/1/c/i/solution