Past exam of the advanced quantum field theory course of the University of Oxford / 2022 / 3 / b / Solution

In unitary gauge the quadratic action is that of a Proca field. Use Minkowski metric $(+---)$ and add a source $J_{\mu }{A}^{\mu }$. Integration by parts, dropping the boundary term, givesWith Fourier transform convention $A(x)=\int d^{4}k{e}^{-ik\cdot x}A(k)/(2\pi {)}^4$, its kernel isFor $k^2\ne 0$, introduce the transverse and longitudinal momentum projectorsThey obey $P_T{P}_L=0$ and $P_T+P_L=1$ as mixed-index linear operators. HenceThe final expression is regular at $k^2=0$; the apparent singularities of the individual projectors have cancelled.

The normalized Gaussian functional integral is evaluated by completing the square with $A\mapsto A+K^{-1}J$:The source-independent functional determinant cancels in this ratio. Two source derivatives insert two fields, so the covariant Feynman propagator is $i{K}^{-1}$, with the Feynman i-epsilon prescription at the physical pole:This is the Proca propagator. Its pole residue has the three physical particle polarizations of a massive spin-one gauge boson. No additional gauge-fixing term is needed to invert this already gauge-fixed massive quadratic action. In a full interacting Higgs mechanism calculation, any field-dependent unitary gauge measure belongs to the interacting theory; the free quadratic gauge boson propagator above is unaffected.