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Adding $-(\partial_\mu A^\mu)^2/(2\alpha)$ gives a family of Lorentz-covariant gauges; $\alpha=1$ is <Feynman gauge>.

Covariant gauge

Varying the displayed gauge-fixed <Lagrangian density> gives the kinetic operator
$$
D^{\mu\rho}=\eta^{\mu\rho}\Box-\left(1-\frac1\alpha\right)\partial^\mu\partial^\rho.
$$
Its <quantum field theory propagator> should satisfy
$$
D_x^{\mu\rho}\Delta_{\rho\nu}(x-y)=i\delta^\mu_\nu\delta^{(4)}(x-y),
$$
and inversion into transverse and longitudinal projectors gives the numerator
$$
\Pi_{\mu\nu}^{\mathrm{correct}}
=-\eta_{\mu\nu}+(1-\alpha)\frac{k_\mu k_\nu}{k^2}.
$$

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 $\alpha\ne2$ it is the Green function of
$$
\left[\eta^{\mu\rho}\Box+\frac{\alpha-1}{2-\alpha}
\partial^\mu\partial^\rho\right]\Delta_{\rho\nu}
=i\delta^\mu_\nu\delta^{(4)},
$$
and 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>.

Solved by gpt-5.6-sol high.

Solution

Paper 301 / 3 / a / Solution

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

* scalar line of momentum $p$: $i/(p^2-m^2+i\epsilon)$;
* photon line in <Feynman gauge>: $-i\eta_{\mu\nu}/(k^2+i\epsilon)$;
* one photon of index $\mu$, scalar momentum $p$ and conjugate-scalar momentum $p'$: $ie(p-p')^\mu$;
* the <seagull vertex> with photon indices $\mu,\nu$: $2ie^2\eta^{\mu\nu}$.

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

Solved by gpt-5.6-sol high.

Paper 301 / 4 / b / Solution

Use <dimensional regularization> with $d=4-\epsilon$ 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 numerator
$$
(2-d)(1-x)\gamma^\mu p_\mu+dm.
$$
The pole of the rotationally symmetric integral consequently gives
$$
\Sigma_{\rm pole}(p)
=\frac{e^2}{8\pi^2\epsilon}
\int_0^1dx\,[-2(1-x)\gamma^\mu p_\mu+4m]
=\frac{e^2}{8\pi^2\epsilon}(-\gamma^\mu p_\mu+4m).
$$
With the inverse-propagator convention
$$
S^{-1}(p)=\gamma^\mu p_\mu-m+\delta_2\gamma^\mu p_\mu-\delta_m-\Sigma(p),
$$
the <minimal subtraction scheme> chooses the counterterm pole to equal $\Sigma_{\rm pole}$. Hence
$$
\delta_2=-\frac{e^2}{8\pi^2\epsilon},
\qquad
\delta_m=-\frac{me^2}{2\pi^2\epsilon}.
$$

Solved by gpt-5.6-sol high.

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 Incoming links: Feynman gauge