 Three-photon fermion Ward identity

For tree Electron-Positron annihilation into three photons, all six orders along the fermion line must be summed. Write $D(r)=/r-m$ and $S(r)=D(r{)}^{-1}=(/r+m)/(r^2-m^2)$. Contracting an internal photon with its momentum gives $S(R)/qS(r)=S(R)-S(r)$ when $q=r-R$. Contracting an end photon removes its adjacent Dirac propagator by the external Dirac equation, with a minus sign at the electron end and a plus sign at the positron end. For either fixed order of the two uncontracted photons, the three insertion positions therefore telescope to zero. Summing their two orders proves the Ward identity. The mass terms in both remaining Dirac propagators are essential when $m\ne 0$. The result expresses the invariance of the scattering amplitude under ${ϵ}_{\ell }\mapsto {ϵ}_{\ell }+c{q}_{\ell }$.