Electron electromagnetic vertex form factors ($F_1(q^2),F_2(q^2)$)

For external electrons on shell, factor out signed electric charge in the parity-preserving interaction vertex:The $u$ are Dirac spinors; ${\gamma }^{\mu }$ are gamma matrices. With on-shell renormalization scheme normalization $F_1(0)=1$, the Gordon identity gives $g_s/2=F_1(0)+F_2(0)$ and $a_e=F_2(0)$.^[ref]

This explains a subtlety: the Dirac equation already predicts a spin magnetic dipole moment with $F_2=0$. Rewriting its Dirac electromagnetic current with the Gordon identity produces a spin-dependent term even before any loop order correction. $F_2$ measures the additional response after that baseline is included. The limit $q\to 0$ is taken after extracting its coefficient; setting $q=0$ in the interaction vertex at the outset would hide it. This electromagnetic anomalous magnetic moment is not a failure of gauge invariance of the kind called a quantum anomaly.

Write the QED vertex as $-ie({\gamma }^{\mu }+{\Lambda }^{\mu })$. In Feynman gauge and $d=4-2ϵ$, the one-loop ultraviolet pole has the local form ${\Lambda }_{\mathrm{UV}}^{\mu }=e^2{\gamma }^{\mu }/(16{\pi }^2ϵ)$. Its coefficient follows from the large-momentum gamma matrices contraction ${\gamma }^{\alpha }/k{\gamma }^{\mu }/k{\gamma }_{\alpha }\mapsto (d-2{)}^2{k}^2{\gamma }^{\mu }/d$. It is independent of external kinematics, whereas the full finite vertex requires additional structures off shell. Its cancellation verifies the Ward identity for QED renormalization constants.

For the QED proper Feynman vertex with both external four-momenta zero, a massive internal fermion and a massless internal photon in Feynman gauge, the even numerator reduces under Lorentz covariance to ${\gamma }^{\mu }[\alpha k^2+\beta m^2]$. The gamma matrices identity ${\gamma }^{\nu }{\gamma }^{\mu }{\gamma }_{\nu }=(2-D){\gamma }^{\mu }$ and the integrated replacement $k_{\rho }{k}_{\sigma }\mapsto g_{\rho \sigma }{k}^2/D$ prove the coefficients. The term $m^2/[k^2(k^2-m^2+i0{)}^2]$ has neither an infrared divergence nor an ultraviolet divergence for $m>0$; its integral in four dimensions is $-i/(16{\pi }^2)$. Only the remaining squared massive denominator supplies the ultraviolet pole.