Tree-Level QED Processes

Tree-level QED amplitudes follow from one vertex, but a physical prediction requires more than multiplying Feynman rules: fermion flow, relative diagram signs, spin averages, flux, phase space, and the observable’s angular or energy resolution must all agree. Distinguishable-fermion scattering gives a clean normalization benchmark; Compton scattering shows why every gauge-related diagram is indispensable.

Required background. The QED action and its rules fixes signs and normalizations. Fermion signs and closed loops supplies the ordering rules, and cross sections and decay rates supplies flux and phase space.

Helpful background. Mandelstam channels and tree-level crossing makes the analytic continuation between the examples explicit.

Distinguishable-fermion scattering

Consider massless $e^-(p_1)\mu^-(p_2)\to e^-(p_3)\mu^-(p_4)$, using $e=|q_e|=|q_\mu|$ and

$$s=(p_1+p_2)^2, \qquad t=(p_1-p_3)^2, \qquad u=(p_1-p_4)^2, \qquad s+t+u=0.$$

Because the species are distinguishable, there is one tree diagram: $t$-channel photon exchange. In Feynman gauge,

$$i\mathcal M_t= \bar u(p_3)(+iq_e\gamma^\mu)u(p_1) \frac{-i\eta_{\mu\nu}}{t+i0} \bar u(p_4)(+iq_\mu\gamma^\nu)u(p_2), \qquad q_e=q_\mu=-e.$$

An unpolarized initial state requires a factor $1/4$, not merely a sum over final spins. With $\sum_su_s(p)\bar u_s(p)=p\!\!\!/$,

$$\overline{|\mathcal M_t|^2} =\frac{e^4}{4t^2} \operatorname{tr}(p_3\!\!\!/\gamma^\mu p_1\!\!\!/\gamma^\nu) \operatorname{tr}(p_4\!\!\!/\gamma_\mu p_2\!\!\!/\gamma_\nu).$$

Using

$$\operatorname{tr}(a\!\!\!/\gamma^\mu b\!\!\!/\gamma^\nu) =4(a^\mu b^\nu+a^\nu b^\mu-\eta^{\mu\nu}a\cdot b)$$

gives the benchmark

$$\boxed{ \overline{|\mathcal M_t|^2} =2e^4\frac{s^2+u^2}{t^2}}$$

At $(s,t,u)=(5,-2,-3)$ this expression is exactly $17e^4$. The exact numerical anchor is especially good at catching a missing spin average or an incorrect trace factor.

For massless $2\to2$ scattering,

$$\frac{\mathrm d\sigma}{\mathrm dt} =\frac{\overline{|\mathcal M_t|^2}}{16\pi s^2} =\frac{2\pi\alpha^2}{s^2}\frac{s^2+u^2}{t^2}, \qquad \alpha=\frac{e^2}{4\pi}.$$

The cross section has mass dimension $-2$, while $\mathrm d\sigma/\mathrm dt$ has dimension $-4$. Its forward divergence as $t\to0$ is the long-range Coulomb singularity. A nonzero minimum scattering angle or a physical screening scale can make the integrated rate finite. Keeping the fermion masses does not by itself remove this forward pole.

The trace and phase-space calculation is developed in Schwartz 2014, §§ 13.1–13.3, pp. 224–236.

Crossing to pair annihilation

Analytically crossing one incoming and one outgoing fermion, while also replacing the corresponding $u$ spinor by the correct $v$ spinor, gives $e^-e^+\to\mu^-\mu^+$. It is not enough to exchange the symbols $s$ and $t$ in a squared formula; the external-state interpretation and physical region change as well. The result is

$$\overline{|\mathcal M_s|^2} =2e^4\frac{t^2+u^2}{s^2}.$$

For massless particles in the center-of-momentum frame,

$$t=-\frac{s}{2}(1-\cos\theta), \qquad u=-\frac{s}{2}(1+\cos\theta),$$

and therefore

$$\frac{\mathrm d\sigma}{\mathrm d\Omega} =\frac{\alpha^2}{4s}(1+\cos^2\theta), \qquad \sigma_{\rm tot}=\frac{4\pi\alpha^2}{3s}.$$

The $1/s$ scaling is the dimensional high-energy limit, while the angular dependence records spin-$1$ exchange.

Identical fermions and exchange signs

For Møller scattering,

$$e^-(p_1)e^-(p_2)\to e^-(p_3)e^-(p_4),$$

the same external state can be reached by $t$- and $u$-channel exchange. With a fixed ordering of the external spinors, Fermi statistics requires

$$\mathcal M=\mathcal M_t-\mathcal M_u,$$

where

$$\begin{aligned} \mathcal M_t&=\frac{e^2}{t} [\bar u(p_3)\gamma^\mu u(p_1)] [\bar u(p_4)\gamma_\mu u(p_2)],\\ \mathcal M_u&=\frac{e^2}{u} [\bar u(p_4)\gamma^\mu u(p_1)] [\bar u(p_3)\gamma_\mu u(p_2)]. \end{aligned}$$

An overall sign depends on the $S$-matrix convention and does not affect the rate; the relative minus sign is invariant. For massless external electrons, the spin-averaged result is

$$\overline{|\mathcal M|^2} =2e^4\left[ \frac{s^2+u^2}{t^2} +\frac{s^2+t^2}{u^2} +\frac{2s^2}{tu} \right].$$

The interference term is a sensitive exchange-sign check. A second, independent issue is phase-space counting: if the two final electrons are integrated over the same full labeled phase space, include $1/2!$. Equivalently, integrate only one non-overlapping half of the final-state phase space without the factor. Doing both undercounts the rate; doing neither counts each physical final state twice.

Compton scattering needs both diagrams

For $e^-(p)+\gamma(k,\varepsilon)\to e^-(p')+\gamma(k',\varepsilon')$, momentum conservation is $p+k=p'+k'$. The two electron-exchange diagrams give

$$\begin{aligned} \mathcal M={}&-e^2\bar u(p')\Bigg[ \gamma^\mu\varepsilon'^*_{\mu} \frac{p\!\!\!/+k\!\!\!/+m}{(p+k)^2-m^2+i0} \gamma^\nu\varepsilon_{\nu} \\ &\qquad\qquad+ \gamma^\nu\varepsilon_{\nu} \frac{p\!\!\!/-k'\!\!\!/+m}{(p-k')^2-m^2+i0} \gamma^\mu\varepsilon'^*_{\mu} \Bigg]u(p). \end{aligned}$$

Neither term is gauge invariant separately. Replace the incoming polarization by its momentum. The identity

$$k\!\!\!/ =S_0^{-1}(p+k)-S_0^{-1}(p), \qquad S_0^{-1}(r)=r\!\!\!/-m,$$

collapses the first propagator. Applying the on-shell equations to the external spinors leaves a boundary term. The crossed diagram produces the same term with the opposite sign, so

$$\mathcal M(\varepsilon\to k)=0.$$

The same holds for $\varepsilon'^*\to k'$. A failed replacement test almost always means that a diagram, momentum direction, or vertex ordering is missing.

In the electron rest frame, let $\omega$ and $\omega'$ be the incoming and outgoing photon energies and $\theta$ their angle. Energy–momentum conservation gives

$$\omega'=\frac{\omega}{1+(\omega/m)(1-\cos\theta)}.$$

After the spin and polarization sums, the Klein–Nishina result is

$$\frac{\mathrm d\sigma}{\mathrm d\Omega} =\frac{\alpha^2}{2m^2} \left(\frac{\omega'}{\omega}\right)^2 \left( \frac{\omega}{\omega'}+\frac{\omega'}{\omega}-\sin^2\theta \right).$$

Its low-energy limit is the Thomson distribution,

$$\frac{\mathrm d\sigma}{\mathrm d\Omega} \xrightarrow[\omega/m\to0]{} \frac{\alpha^2}{2m^2}(1+\cos^2\theta), \qquad \sigma_{\rm T}=\frac{8\pi\alpha^2}{3m^2}.$$

That limit checks both normalization and the interference sign between the two diagrams. The two-diagram amplitude, physical-polarization sum, and laboratory-frame phase space are developed in Schwartz 2014, § 13.5, pp. 238–242, especially Eqs. (13.108), (13.124), and (13.132). His external-particle ordering names the crossed electron channel $t$; with the ordering used here it is $u=(p-k')^2$.

A compact validation table

Calculation	Invariant check	Sensitive limit	What must be declared
$e^-\mu^-\to e^-\mu^-$	current contraction with exchanged momentum vanishes	$t\to0$ Coulomb enhancement	spin average, angular acceptance, masses
$e^-e^+\to\mu^-\mu^+$	crossed trace agrees after state relabeling	threshold and $s\gg m^2$	physical region, flux, species masses
$e^-e^-\to e^-e^-$	antisymmetry under $p_3\leftrightarrow p_4$	$t\leftrightarrow u$ exchange	relative minus sign, $1/2!$ or non-overlapping phase space
Compton scattering	both polarization-to-momentum replacements vanish	$\omega/m\to0$ Thomson limit	both diagrams, polarization sum, frame

This table is a static check record, not a substitute for the full amplitude. The corresponding check can be automated; the analytic results above remain complete without it.

Common pitfalls

Crossing is not a blind variable substitution. Momentum signs, particle versus antiparticle spinors, and the physical kinematic region must be crossed together.

Spin sums and spin averages are different. Sum over unobserved final spins, but divide by the number of equally populated initial spin states.

One Compton diagram is not a QED amplitude. Gauge invariance appears only after the $s$- and $u$-channel terms are combined.

A singular limit needs an observable definition. A fermion mass regulates a collinear singularity, but does not remove the forward Coulomb pole. Angular acceptance or screening controls the latter; soft-photon divergences require the inclusive or dressed construction developed later in the chapter.

Exercises

Annihilation with angular acceptance. A detector accepts the outgoing $\mu^-$ only when $|\cos\theta|<c$, with $0<c\leq1$. In the massless approximation, find the accepted $e^-e^+\to\mu^-\mu^+$ cross section and its fraction of the total. Evaluate that fraction at $c=1/2$.

Solution

The final particles are distinguishable, so there is no factor $1/2!$. Integrating over azimuth and $x=\cos\theta$ gives

$$\sigma(c)=\frac{2\pi\alpha^2}{4s} \int_{-c}^{c}(1+x^2)\,\mathrm dx =\frac{\pi\alpha^2}{s}\left(c+\frac{c^3}{3}\right).$$

Thus $\sigma(c)/\sigma_{\rm tot}=(3c+c^3)/4$. It approaches one at $c=1$, vanishes as $c\to0$, and equals $13/32$ at $c=1/2$. The calculation assumes $s\gg m_\mu^2$ and neglects radiative corrections and detector efficiencies within the accepted region.

The Compton cancellation. Show explicitly that replacing the incoming polarization by $k$ makes the two terms in the amplitude cancel. Keep $m\ne0$ and use only the on-shell equations and momentum conservation.

Solution

Write the reduced electron propagator as $R(r)=(r\!\!\!/+m)/(r^2-m^2)$, with its factor $i$ stripped. Away from its poles, $R(r)(r\!\!\!/-m)=1$. The first diagram then obeys

$$R(p+k)k\!\!\!/\,u(p) =R(p+k)\bigl[(p\!\!\!/+k\!\!\!/-m)-(p\!\!\!/-m)\bigr]u(p) =u(p).$$

For the second, $k=p'-(p-k')$ implies

$$\bar u(p')k\!\!\!/\,R(p-k') =\bar u(p')\bigl[(p'\!\!\!/-m)-(p\!\!\!/-k'\!\!\!/-m)\bigr]R(p-k') =-\bar u(p').$$

The bracket in the amplitude therefore reduces to $\bar u(p')\varepsilon'^{*}\!\!\!/\,u(p)-\bar u(p')\varepsilon'^{*}\!\!\!/\,u(p)=0$. No sum over electron spins or photon polarizations is required: this is an amplitude identity for each external state.

Recoil and the Thomson limit. Derive $\omega'/\omega$ from $(p+k-k')^2=m^2$ in the electron rest frame. Then integrate the low-energy angular distribution to recover the total Thomson cross section.

Solution

Using $p\cdot k=m\omega$, $p\cdot k'=m\omega'$ and $k\cdot k'=\omega\omega'(1-\cos\theta)$ gives

$$m(\omega-\omega')=\omega\omega'(1-\cos\theta), \qquad \frac{\omega'}{\omega}=\frac{1}{1+(\omega/m)(1-\cos\theta)}.$$

For $\omega/m\to0$ at fixed angle, the ratio tends to one. The Klein–Nishina bracket tends to $2-\sin^2\theta=1+\cos^2\theta$, and hence

$$\sigma_{\rm T}=\frac{\pi\alpha^2}{m^2} \int_{-1}^{1}(1+x^2)\,\mathrm dx =\frac{8\pi\alpha^2}{3m^2}.$$

This checks the low-energy normalization after the two diagrams have been combined; discarding either diagram does not satisfy the preceding Ward test.

References

• Matthew D. Schwartz, Quantum Field Theory and the Standard Model, Cambridge University Press (2014), §§ 13.1–13.5, doi:10.1017/9781139540940.