Anomalous Magnetic Moment of the Electron, Muon, and Nucleon

Abstract

The anomalous magnetic moment of the electron, $\frac{1}{2}(g\ensuremath{-}2)$, is computed using dispersion theory. The analytic continuation is made in the mass of one of the external electron lines and only the one-electron one-photon states are retained in the absorptive amplitude. In this way we relate $g\ensuremath{-}2$ to the Compton amplitude which has a known exact threshold behavior. Our approximation is an expansion in the low-energy behavior rather than a perturbation expansion in powers of 1/137, and we are able to show that a major contribution to $g\ensuremath{-}2$ comes from the low-mass region of the electron-photon system near the threshold of the absorptive amplitude. First, in a purely nonrelativistic calculation, we find that a major part of the $\frac{\ensuremath{\alpha}}{2\ensuremath{\pi}}$ correction is accounted for by the Thomson limit. Further refining our calculation by including the exact residue of the pole terms in the Compton amplitude in accord with the low-energy theorem on Compton scattering, we find that electron-photon states below $2.5m{c}^{2}$ in the absorptive amplitude reproduce 90% of the $\frac{\ensuremath{-}0.328{\ensuremath{\alpha}}^{2}}{{\ensuremath{\pi}}^{2}}$ contribution and predict a value of $\ensuremath{\sim}\frac{+0.15{\ensuremath{\alpha}}^{3}}{{\ensuremath{\pi}}^{3}}$ for the sixth-order term. We also give a simple physical interpretation of the difference of the muon and electron $g\ensuremath{-}2$ values. Finally we calculate with this approach the anomalous magnetic moments of the proton and neutron, with the Kroll-Ruderman theorem on meson photoproduction providing the low-energy "anchor" in this case. Again retaining only the low-mass region of the absorptive amplitude, we obtain fair agreement with the magnitude and the isovector character of the moments, finding $\ensuremath{\Delta}{\ensuremath{\mu}}^{P}\ensuremath{\approx}0.7(\ensuremath{\Delta}{\ensuremath{\mu}}_{\mathrm{expt}})$ and $\ensuremath{\Delta}{\ensuremath{\mu}}^{N}\ensuremath{\approx}0.9(\ensuremath{\Delta}{\ensuremath{\mu}}_{\mathrm{expt}})$.