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Varying the displayed gauge-fixed Lagrangian density gives the kinetic operatorIts quantum field theory propagator should satisfyand inversion into transverse and longitudinal projectors gives the numerator

The paper instead prints $-{\eta }_{\mu \nu }+(\alpha -1)k_{\mu }{k}_{\nu }/k^2$. Except at $\alpha =1$, that is not the inverse of the displayed Lagrangian's kinetic operator. Taken literally, for $\alpha \ne 2$ it is the Green function ofand at $\alpha =2$ its longitudinal part is noninvertible. Thus the longitudinal sign in the printed propagator is a typographical error; the two forms coincide in Feynman gauge.

With all momenta directed into each vertex, the momentum-space Feynman rules are:

Each vertex also carries its momentum-conserving delta function; integrate every independent loop momentum and attach the appropriate external wavefunctions and photon polarization vectors.

Use dimensional regularization with $d=4-ϵ$ and choose Feynman gauge. Combining the electron and photon denominators with a Feynman parameter, shifting the loop momentum, and discarding the odd term leaves the numeratorThe pole of the rotationally symmetric integral consequently givesWith the inverse-propagator conventionthe minimal subtraction scheme chooses the counterterm pole to equal ${\Sigma }_{\mathrm{pole}}$. Hence

For fixed $\xi \ne 0$, the Lorentzian quadratic density $-F^2/4+(\partial \cdot A{)}^2/(2\xi )$ has momentum kernel $K^{ab}=-p^2{\eta }^{ab}+(1+1/\xi )p^a{p}^b$. At $p^2\ne 0$ its inverse, defined by $K^{ac}{G}_{cb}={\delta }_b^a$, is $G_{ab}=-{\eta }_{ab}/p^2+(1+\xi )p_a{p}_b/(p^2{)}^2$. A vacuum photon propagator is $iG$ with the appropriate Feynman i-epsilon prescription. The usual parameter in the negative-sign covariant gauge term is $\alpha =-\xi$, so Feynman gauge is $\xi =-1$.