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 $F_1(0) = 1$?


1 Answer 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.

  • $\begingroup$ 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 $0$. $\endgroup$
    – Styg
    Jan 21, 2019 at 13:20
  • $\begingroup$ 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? $\endgroup$
    – Styg
    Jan 21, 2019 at 13:21
  • $\begingroup$ Yes, to a great extent, for example, in electron mass spectrometers, etc. $\endgroup$ Jan 21, 2019 at 13:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.