Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 46 / 4 / i / Solution

Take metric signature $(+---)$ and the Fourier transform convention $e^{-ipx}$. Introduce an even source $J$ for the gauge field and odd sources $\eta ,\bar{\eta }$ for the Faddeev-Popov ghost field:At tree level the two-point functions come from the quadratic action. Integrating the gauge kinetic term by parts gives its momentum-space kernelIts inverse follows from transverse and longitudinal projectors: $K=-p^2(P_T+{\zeta }^{-1}{P}_L)$, so $K^{-1}=-p^{-2}(P_T+\zeta P_L)$. Completing the bosonic Gaussian square gives $Z_0[J]=\exp [-iJ{K}_F^{-1}J/2]$, with the Feynman i-epsilon prescription in the inverse. Two source derivatives and their insertion factors give the gauge-boson propagator $D_F=i{K}_F^{-1}$. HenceFor $\zeta =1$ this reduces to the Feynman-gauge adjoint propagator. The longitudinal double pole is understood with the compatible Feynman boundary prescription.