= Solution
Use signature $(+,-,-,-)$ and <electron> charge $q=-e$, with $e>0$. Then $\mathcal L_{\rm int}=-q\bar\psi\gamma^\mu\psi A_\mu=+e\bar\psi\gamma^\mu\psi A_\mu$. Write the process as $e^-(p,s)+\gamma(k,\epsilon)\to e^-(p',s')+\gamma(k',\epsilon')$. Physical external states satisfy
$$
p^2=p'^2=m^2,\qquad k^2=k'^2=0,\qquad
p+k=p'+k',\qquad p^0,p'^0,k^0,k'^0>0,
$$
and the <photon polarization vectors> obey $k\cdot\epsilon=k'\cdot\epsilon'=0$. Choose unit physical polarizations with $\epsilon^*\cdot\epsilon=\epsilon'^*\cdot\epsilon'=-1$; representatives differing by a multiple of the corresponding null <momentum> describe the same physical polarization. The external spinors obey $(\not p-m)u(p)=0$, $\bar u(p')(\not p'-m)=0$, with $\bar u_su_r=2m\delta_{sr}$. Outgoing polarization appears complex conjugated, so this formula includes helicity states as well as real linear polarizations.
The needed <QED Feynman rules> are a vertex $-iq\gamma^\mu=+ie\gamma^\mu$, an internal <Dirac propagator> $i(\not q+m)/(q^2-m^2+i0)$, the external spinors, and the incoming/outgoing <photon polarization vectors>. There is no internal <photon> at this order and no elementary two-photon–Dirac seagull vertex. The two <photon> orderings along the <electron> line give internal momenta $q_s=p+k$ and $q_u=p-k'$. Defining the <matrix> element by $S_{fi}-\delta_{fi}=i(2\pi)^4\delta^4(p+k-p'-k')\mathcal M$, their sum is
$$
\boxed{\mathcal M^{(2)}=-e^2\bar u(p')
\left[
\not\epsilon'^*\frac{\not p+\not k+m}{(p+k)^2-m^2+i0}\not\epsilon
+\not\epsilon\frac{\not p-\not k'+m}{(p-k')^2-m^2+i0}\not\epsilon'^*
\right]u(p).}
$$
The denominators are $2p\cdot k+i0$ and $-2p\cdot k'+i0$. Both terms are essential; this is the <tree-level Compton amplitude>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/iii/paper-44-compton-diagrams.png]
{title=Compton tree diagrams and the four one-loop classes before exchanging the external photons}
To prove the incoming <Ward identity>, replace $\epsilon$ by $k$ and write $R(q)=(\not q+m)/(q^2-m^2)$, with the Feynman boundary prescription understood. The physical nonzero-photon kinematics keeps these tree denominators off the <electron> pole. Since
$$
\not k=(\not q_s-m)-(\not p-m),
\qquad
\not k=(\not p'-m)-(\not q_u-m),
$$
the external <Dirac equations> imply
$$
R(q_s)\not k\,u(p)=u(p),\qquad
\bar u(p')\not k\,R(q_u)=-\bar u(p').
$$
Therefore the two contracted terms are $\bar u(p')\not\epsilon'^*u(p)$ and its negative, and cancel. By linearity,
$$
\boxed{\mathcal M(\epsilon+\alpha k)=\mathcal M(\epsilon).}
$$
This is the <two-photon fermion Ward identity>. An analogous cancellation holds for the outgoing <photon>. Gauge-equivalent polarization representatives have the same scattering amplitude, so a longitudinal pure-gauge component is not an extra physical <photon> state. Individual diagrams do not generally satisfy this identity; their sum does.
For perturbative order, each elementary <QED> vertex has one <photon> leg. If $I_\gamma$ is the number of internal <photon> lines and $E_\gamma=2$, the <vertex parity of a QED amplitude> gives
$$
V=2I_\gamma+2.
$$
Thus \b[there is no third-order contribution]. At fourth order, $V=4$ and, counting all three legs per vertex with four external legs, $I=(3V-4)/2=4$. A connected graph has loop number $I-V+1=1$.
The nonzero one-loop classes on the open <electron> line are an internal-electron <self-energy> insertion, a vertex correction at either external-photon vertex, and a virtual <photon> spanning both <photon> vertices, giving the four-point box class. The first figure sketches each class for one <photon> ordering; interchange the two external <photon> attachments for its crossed partner. This yields two internal-self-energy graphs, four vertex graphs and two spanning-box graphs. They use the usual <photon> propagator, for example $-ig_{\mu\nu}/(\ell^2+i0)$ in <Feynman gauge>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/iii/paper-44-compton-bookkeeping.png]
{title=External-leg insertions, fourth-order counterterms and the odd-photon loops that cancel}
For completeness, <one-loop diagram classes for Compton scattering> also depend on whether one is drawing an unamputated correlation function or a physical amplitude. <Electron> self-energies can be attached to either external <electron> leg, and <photon vacuum polarization> can be attached to either external <photon> leg, for both <photon> orders. In an on-shell renormalized <S-matrix>, these are accounted for by <LSZ reduction> and external-field residues rather than counted again as independent amputated graphs. Photon-leg corrections are not vacuum-polarization insertions on an internal Born <photon>: there is no such Born line.
At the same fourth order, a renormalized calculation has one <mass> or electron-field <counterterm> on the internal <electron> line, or one vertex <counterterm> at either Born vertex; the crossed partners are included. Each <counterterm> is a relative order-$e^2$ correction to the order-$e^2$ tree amplitude. External-field normalization must follow the same convention.
A possible closed <electron> loop with three <photon> vertices, linked to the external <electron> line by a virtual <photon>, cancels between its two orientations by <Furry's theorem>. Transposing the loop and using $C^{-1}\gamma^\mu C=-(\gamma^\mu)^T$ gives an extra sign $(-1)^3$, so its orientation-reversed partner is its negative. A one-photon tadpole likewise vanishes. Disconnected vacuum bubbles are removed by vacuum normalization. \b[The first genuine radiative corrections are fourth order, with the internal <self-energy>, vertex and spanning-box classes and their consistently included <renormalization> terms.]
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