Gauge-boson propagator

For the Euclidean covariant gauge term $({\partial }_{\mu }{A}^{\mu }{)}^2/(2\xi )$, the free gauge-boson propagator is

With a mostly-plus metric signature, $e^{iS}$, and gauge-fixing term $-(\partial \cdot A{)}^2/2$, the free adjoint gauge-field kernel is $-p^2{\eta }_{\mu \nu }{\delta }_{ab}$. Inverting it gives the displayed Feynman propagator. This convention uses the same denominator and sign as a massless real scalar propagator.

At $\xi =1$, the Euclidean photon propagator is $D_{\mu \nu }(k)={\delta }_{\mu \nu }/k^2$. Its Lorentz-index contraction simplifies loop numerators in quantum electrodynamics.

At a nonzero Euclidean momentum $k$, the transverse projector of a vector field is $P_T^{ab}={\delta }^{ab}-k^a{k}^b/k^2$. In a Lorentzian Minkowski metric, its mixed-index form is $(P_T{)}^a{}_b={\delta }_b^a-p^a{p}_b/p^2$, defined for $p^2\ne 0$. It obeys $P_{T}p=0$ and $P_T^2=P_T$. A nonzero null four-momentum is insufficient for this particular decomposition because the denominator vanishes.

At a nonzero Euclidean momentum $k$, the longitudinal projector of a vector field is $P_L^{ab}=k^a{k}^b/k^2$. Its Lorentzian mixed-index form is $(P_L{)}^a{}_b=p^a{p}_b/p^2$ for non-null four-momentum. Together with the transverse projector of a vector field, it obeys $P_T+P_L=I$, $P_T{P}_L=0$, $P_T^2=P_T$, and $P_L^2=P_L$. This formula is not a pointwise decomposition at null four-momentum.