= Solution
Take the <muon> <mass> to be $m_\mu$, and use the <quantum electrodynamics> convention $\mathcal L_{\rm int}=-e\bar\psi\gamma^\mu\psi A_\mu$ with $e>0$. Let $p_1,p_2$ be the incoming <muon> and antimuon momenta and $k_1,k_2$ the outgoing <photon> momenta, so $p_1+p_2=k_1+k_2$. External particles are <on shell>: $p_1^2=p_2^2=m_\mu^2$ and $k_1^2=k_2^2=0$. The two tree-level <Feynman diagrams> are the two orders in which the <photon> legs attach to the <muon> line. There is no three-<photon> <QED> vertex, and therefore no single-<photon> annihilation diagram into two <photons> at this order.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2012/iii/paper-46-tree-diagrams.png]
{title=Two photon orders for muon annihilation and Compton scattering}
The <QED Feynman rules> needed here are the vertex $-ie\gamma^\mu$, the <Dirac propagator>
$$
\frac{i(\not q+m_\mu)}{q^2-m_\mu^2+i0},\qquad \not q=\gamma^\mu q_\mu,
$$
an incoming <muon> spinor $u(p_1)$, an incoming antimuon adjoint $\bar v(p_2)$, and an outgoing <photon> polarization $\epsilon_\mu^*(k)$. At a vertex enforce <four-momentum> conservation. Spinor products are ordered along the <fermion> line; exchanging the external <photons> introduces no relative minus sign. Define $R(q)=(\not q+m_\mu)/(q^2-m_\mu^2+i0)$ and $\not\epsilon=\gamma^\mu\epsilon_\mu$.
With the convention that a diagram contributes $i\mathcal M$, the two <muon>-antimuon annihilation amplitudes are
$$
\begin{aligned}
\mathcal M_t&=-e^2\bar v(p_2)\not\epsilon_2^*R(p_1-k_1)\not\epsilon_1^*u(p_1),\\
\mathcal M_u&=-e^2\bar v(p_2)\not\epsilon_1^*R(p_1-k_2)\not\epsilon_2^*u(p_1),\\
\boxed{\mathcal M_{\rm ann}}&\boxed{=\mathcal M_t+\mathcal M_u.}
\end{aligned}
$$
Their denominators are respectively $t-m_\mu^2+i0$ and $u-m_\mu^2+i0$, where $t=(p_1-k_1)^2$, $u=(p_1-k_2)^2$. A different consistent overall amplitude-phase convention has no physical effect.
A physical <photon polarization vector> obeys \b[$k\cdot\epsilon=0$], and represents an equivalence class $\epsilon\sim\epsilon+\alpha k$. For a real null momentum, adding $\alpha k$ preserves transversality. One can additionally choose transverse spatial vectors with $\epsilon^0=0$ and $\epsilon^*\cdot\epsilon=-1$; there are two independent physical polarizations. The equivalence class removes the unphysical longitudinal direction, rather than imposing four independent physical polarization states.
The invariance of the total amplitude is a <two-photon fermion Ward identity>. Replace $\epsilon_1^*$ by $k_1$ and put $q_t=p_1-k_1$, $q_u=p_1-k_2=k_1-p_2$. The external <Dirac equations> give
$$
(\not p_1-m_\mu)u(p_1)=0,\qquad \bar v(p_2)(\not p_2+m_\mu)=0.
$$
Away from propagator poles, $R(q)(\not q-m_\mu)=(\not q-m_\mu)R(q)=I$; the identity extends with the common Feynman prescription. Therefore
$$
\begin{aligned}
R(q_t)\not k_1u(p_1)&=R(q_t)(\not p_1-\not q_t)u(p_1)=-u(p_1),\\
\bar v(p_2)\not k_1R(q_u)&=\bar v(p_2)(\not q_u+\not p_2)R(q_u)=\bar v(p_2).
\end{aligned}
$$
The two terms in the bracket then give $-\bar v\not\epsilon_2^*u$ and $+\bar v\not\epsilon_2^*u$, which cancel. Exchanging <photon> labels proves the second <Ward identity>. Thus \b[the sum is unchanged under $\epsilon_i\mapsto\epsilon_i+\alpha k_i$ for either <photon>]. Neither diagram is generally <gauge-invariant> separately. Physically, longitudinal pure-gauge polarization does not couple to the observable <scattering amplitude>; only the two transverse <photon> degrees of freedom contribute.
For <Compton scattering> write the incoming momenta as $p,k$ and outgoing momenta as $p',k'$, with $p+k=p'+k'$. The outgoing <muon> contributes $\bar u(p')$, the incoming <photon> contributes $\epsilon(k)$ without conjugation, and the outgoing <photon> contributes $\epsilon'^*(k')$. The two <tree-level Compton amplitudes> are
$$
\begin{aligned}
\mathcal M_s&=-e^2\bar u(p')\not\epsilon'^*R(p+k)\not\epsilon\,u(p),\\
\mathcal M_u&=-e^2\bar u(p')\not\epsilon\,R(p-k')\not\epsilon'^*u(p),\\
\boxed{\mathcal M_{\rm C}}&\boxed{=\mathcal M_s+\mathcal M_u.}
\end{aligned}
$$
These are the lower two diagrams. Their intermediate <muon> momenta are $p+k$ and $p-k'$, with denominators $s-m_\mu^2+i0$ and $u-m_\mu^2+i0$. The first attaches the incoming <photon> before the outgoing <photon> along <fermion> flow; the second reverses that attachment order. Both contribute at order $e^2$ in the amplitude.
\b[<Compton scattering> has one incoming and one outgoing <muon> and one <photon> on each side; annihilation has an incoming particle-<antiparticle> pair and two outgoing <photons>.] Accordingly the external spinors/polarizations and physical Mandelstam channels differ, although both arise from the same two-vertex <QED> matrix element. <Crossing symmetry> relates them by continuing $p_2\mapsto-p'$ and one outgoing <photon> momentum $k_1\mapsto-k$, with the corresponding external wavefunction replacements. The annihilation $t$ channel becomes the Compton $s$ channel and the other channel becomes its $u$ channel. Identical final <photons> require a factor $1/2!$ in an annihilation phase-space integral over an otherwise double-counted full final-state region; there is no analogous final-state identical-particle factor for the <muon>-<photon> Compton final state. Gauge cancellation holds for the summed Compton amplitude as well.
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