For $n=2$, the operator formula becomes
$$
Z_2[J]=\left[1+\frac{ig}{2}\int d^dx\,
\frac{\delta^2}{\delta J(x)^2}+O(g^2)\right]Z_0[J].
$$
Two source derivatives of the free Gaussian produce a $J$-independent coincident propagator and a source-dependent insertion between two free propagators. The former is a vacuum bubble and disappears when $Z_2[0]=1$. Thus
$$
\frac{Z_2[J]}{Z_2[0]}
=Z_0[J]\left[1-\frac{ig}{2}
\int\frac{d^dp}{(2\pi)^d}\frac{J(-p)J(p)}{(p^2+i0)^2}
+O(g^2)\right]
$$
Expanding the exact denominator from part (b),
$$
\frac1{p^2-g+i0}=\frac1{p^2+i0}+\frac{g}{(p^2+i0)^2}+O(g^2),
$$
gives the same source-dependent correction.


Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 304 / 1 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-304/1/d/solution