= Solution
At order $e^2$, scalar particle–<antiparticle> scattering has a $t$-channel photon exchange and an $s$-channel annihilation and recreation diagram. They are shown below, with incoming momenta $p,q$ and outgoing momenta $p',q'$. The <seagull vertex> has only two scalar legs and does not yield a four-scalar tree diagram at this order.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-50-scalar-qed-tree.png]
{title=Leading scalar particle–antiparticle scattering diagrams with exchanged and annihilation-channel photons}
For the exchange graph, let $k=p-p'=q'-q$. Apart from overall charge signs, the two scalar currents are $J=p+p'$ and $K=q+q'$. The external <mass shell> conditions give
$$
k\cdot J=p^2-p'^2=0,\qquad k\cdot K=q'^2-q^2=0.
$$
For the annihilation graph, $k=p+q=p'+q'$, with currents $J=p-q$ and $K=p'-q'$, again up to vertex signs. Their contractions are $p^2-q^2=0$ and $p'^2-q'^2=0$. Thus each graph couples the photon to <conserved currents>, the relevant <scalar quantum electrodynamics Ward identity>.
Write $k=(\omega,\mathbf k)$ and $\kappa^2=|\mathbf k|^2$. <conserved current> means $\mathbf k\cdot\mathbf J=\omega J^0$ and $\mathbf k\cdot\mathbf K=\omega K^0$. For $\kappa\ne0$, contraction with the <Coulomb-gauge photon propagator> gives, displaying the rational identity before the pole boundary limit,
$$
\begin{aligned}
J^\mu D^{\mathrm C}_{\mu\nu}K^\nu
&=\frac{iJ^0K^0}{\kappa^2}
+\frac{i}{k^2}\left(\mathbf J\cdot\mathbf K-\frac{(\mathbf J\cdot\mathbf k)(\mathbf K\cdot\mathbf k)}{\kappa^2}\right)\\
&=\frac{i\mathbf J\cdot\mathbf K}{k^2}
+\frac{iJ^0K^0}{\kappa^2}\left(1-\frac{\omega^2}{k^2}\right)\\
&=\frac{i}{k^2}\bigl(\mathbf J\cdot\mathbf K-J^0K^0\bigr)
=-\frac{i}{k^2}J\cdot K.
\end{aligned}
$$
We used $k^2=\omega^2-\kappa^2$. Restoring the common <Feynman i-epsilon prescription> yields the <Coulomb-gauge propagator between conserved currents> result
$$
\boxed{J^\mu D^{\mathrm C}_{\mu\nu}K^\nu
=J^\mu\left(\frac{-i\eta_{\mu\nu}}{k^2+i0}\right)K^\nu.}
$$
Hence each on-shell tree amplitude can use the Lorentz-invariant propagator. This is equality after contraction, not equality of the gauge-dependent tensors. At $\mathbf k=0$, the separate Coulomb-gauge terms are ill-defined; combine them first and take the conserved-current limit. The combined expression has only the physical photon pole.
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