At spacelike separation $x_{12}^2>0$, conformal symmetry fixes the <scalar-primary two-point function> to
$$
\langle\mathcal O(x_1)\mathcal O(x_2)\rangle
=\frac{C_{\mathcal O}}{(x_{12}^2)^\Delta}.
$$
Applying the <d'Alembertian> at $x_2$ gives
$$
\boxed{F_0
=\frac{2C_{\mathcal O}\Delta(2\Delta+2-d)}
{(x_{12}^2)^{\Delta+1}}}
$$
away from contact terms. Thus, up to a real multiplicative constant, $F_0\propto(x_{12}^2)^{-\Delta-1}$. The scalar <conformal unitarity bound> makes the coefficient nonnegative and makes it vanish when the bound is saturated, as expected for the null free-field equation of motion.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 354 / 1 / b / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-354/1/b/i/solution