For $n=2$, the coupling is a mass squared, $m^2=g$. After integration by parts, the quadratic action has kernel $K=-(\Box+g-i0)$. Completing the square in the <Gaussian integral> and choosing the source-independent normalization so that $Z_2[0]=1$ gives
$$
\boxed{Z_2[J]=\exp\!\left[-\frac i2\int d^dx\,d^dy\,
J(x)K^{-1}(x-y)J(y)\right]}.
$$
Equivalently, in momentum space,
$$
\boxed{Z_2[J]=\exp\!\left[-\frac i2\int\frac{d^dp}{(2\pi)^d}
\frac{J(-p)J(p)}{p^2-g+i0}\right]}.
$$
The $i0$ prescription selects the <Feynman propagator>.

Solution

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

 Solution

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