For a boundary-independent source, the <AdS scalar-field boundary asymptotics> reduce to
$$
\phi(z)=J+z^5A+o(z^5).
$$
One may therefore extract the source and response directly:
$$
\boxed{J=\lim_{z\to0}\phi(z)},
\qquad
\boxed{A=\lim_{z\to0}z^{-5}\bigl(\phi(z)-J\bigr)
=\frac15\lim_{z\to0}z^{-4}\partial_z\phi}.
$$
The operator expectation value is proportional to $A$; choosing the boundary-operator normalization $\mathcal O=A$ gives the requested expressions. A conventional action normalization can instead multiply this relation by the fixed factor $2\Delta-d=5$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 354 / 1 / b / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-354/1/b/ii/solution