# Why are there no radiative corrections to the electron charge for (an on-shell interaction)?

My question stems from the preliminary discussion of the electron vertex function found in section 6.2 of Peskin and Schroeder's Introduction to Quantum Field Theory.

Specifically, they calculate the amplitude for electron scattering from an external potential, taken to be electrostatic and slowly varying. The corresponding potential (by comparison with the Born approximation) is

$$V(\mathbf{x}) = e F_1(0) \phi(\mathbf{x}) \tag{1} \label{eq1}$$

where $$\phi$$ is the electrostatic potential and $$F_1(q^2)$$ is one of the form factors appearing in the vertex function ($$q^\mu$$ is the momentum of the photon line in the relevant diagrams).

The book goes on to say (emphasis added)

Thus $$F_1(0)$$ is the charge of the electron, in units of $$e$$. Since $$F_1(0) = 1$$ already in the leading order of perturbation theory, radiative corrections to $$F_1(q^2)$$ should vanish at $$q^2 = 0$$.

Why must radiative corrections to $$F_1(q^2)$$ vanish on-shell? Is it because (1) is the definition of the electron's charge, and thus by definition $$F_1(0) = 1$$?

Here $$q^2=0$$ means "no interaction" limit - the transferred momentum from/to an external field is zero. $$q$$ is not an integration (out-of-shell) variable in some loop. In other words, it is sort of a classical limit; thus, the definition of the electric charge.
• Thanks for pointing out that the limit is not merely that of an "on-shell" transfer: I checked the text and the limit taken is that of all components of $q^\mu$ approaching $0$. – Styg Jan 21 '19 at 13:20
• If $(1)$ is in fact the definition of the electron charge, is it in practice measured and standardized via such a nonrelativistic interaction with a macroscopic electrostatic field? – Styg Jan 21 '19 at 13:21