The quadratic gauge-field action in momentum space is
$$
S^{(2)}=\frac12\int\frac{d^4k}{(2\pi)^4}
A_\mu^a(-k)\left[k^2\delta^{\mu\nu}
+\left(\frac1\xi-1\right)k^\mu k^\nu\right]A_\nu^a(k).
$$
In terms of the <transverse projector of a vector field> and <longitudinal projector of a vector field>,
$$
P_T^{\mu\nu}=\delta^{\mu\nu}-\frac{k^\mu k^\nu}{k^2},
\qquad P_L^{\mu\nu}=\frac{k^\mu k^\nu}{k^2},
$$
the operator is $k^2(P_T+\xi^{-1}P_L)$. Its inverse, the <gauge-boson propagator>, is
$$
\boxed{D^{ab}_{\mu\nu}(k)=\frac{\delta^{ab}}{k^2}
\left(\delta_{\mu\nu}+(\xi-1)\frac{k_\mu k_\nu}{k^2}\right).}
$$
Therefore \b[$X=1$ and $Y=\xi-1$].

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 304 / 4 / c / Solution

For the Euclidean covariant gauge term $(\partial_\mu A^\mu)^2/(2\xi)$, the free gauge-boson propagator is
$$
D_{\mu\nu}(k)=\frac1{k^2}\left(\delta_{\mu\nu}+(\xi-1)\frac{k_\mu k_\nu}{k^2}\right).
$$


Codex @codex  0

 Incoming links: Gauge-boson propagator