The quadratic momentum-space operator is inverted by the <Proca propagator>. With the <Feynman propagator> boundary prescription,
$$
\Delta_{\mu\nu}(x-y)
=\int\frac{d^4p}{(2\pi)^4}
\frac{-i}{p^2-m^2+i\epsilon}
\left(\eta_{\mu\nu}-\frac{p_\mu p_\nu}{m^2}\right)
e^{-ip\cdot(x-y)}.
$$
Multiplication by the Fourier-space Proca operator gives $i\delta_\mu{}^\nu$, with the overall sign determined by the convention for the Green function.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 301 / 1 / d / Solution

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>.

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

Completing the square in the <Gaussian functional integral> and choosing
$$
\mathcal N^{-1}=\int\mathcal D\phi\,
\exp\left(i\int d^4x\,\mathcal L_b\right)
$$
normalizes $Z_0[0]=1$. The result is
$$
Z_0[J]=\exp\left[-\frac12\int d^4x,d^4y\,J(x)D_F(x-y)J(y)\right],
$$
where the <Feynman propagator> in the convention of the question is
$$
D_F(x-y)=\int\frac{d^4p}{(2\pi)^4}\frac{i,e^{-ip\cdot(x-y)}}{p^2-M^2+i\epsilon}.
$$

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 304 / 2 / a / Solution

{wiki}

The Feynman propagator is the time-ordered Green function selected by the $i\epsilon$ prescription.


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