Renormalization Conditions of QED

[Gaussian97]

TL;DR Summary

What is exactly the renormalization condition for the QED vertex?

Hello, I'm studying the renormalization of QED. I have the Lagrangian
$$\mathscr{L}_{QED}=\mathscr{L}_{physical}+\mathscr{L}_{counterterms}$$
where
$$\mathscr{L}_{physical}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\bar{\psi}(i\gamma^\mu\partial_\mu - m)\psi - e \bar{\psi}\gamma^\mu\psi A_\mu$$
$$\mathscr{L}_{counterterms}=-\frac{1}{4}\delta_3 F_{\mu\nu}F^{\mu\nu}+\bar{\psi}(i\delta_2\gamma^\mu\partial_\mu - \delta_m)\psi - e \delta_1\bar{\psi}\gamma^\mu\psi A_\mu$$

with $\delta_k$ the counterterms, fixed by the renormalization conditions.

I don't have problems with the mass and field-strengths renormalization conditions, my problem is with the condition for the electric charge renormalization.
If $-ie\Gamma^\mu(p', p)$ is the amplitude for the 1PI vertex diagrams with $p'$ and $p$ the electron momenta, the renormalization condition is usually stated by imposing that, when the photon is on-shell ($q^2=0$) then this amplitude must reduce to $\Gamma^\mu = \gamma^\mu$.

My question is: In general, the momenta of the electrons in the vertex don't need to be on-shell, right? Then, do I need to impose on-shell electrons to determine the value of $\delta_1$? Because, if I understand this properly, $\delta_1$ should be independent of $p$ and $p'$ and I don't see how this is possible if I don't fix them.

My question arises because all the calculations that I've seen assume on-shell electrons, and I don't understand if:
1- By definition $\Gamma^\mu$ must have on-shell electrons or
2- The renormalization condition imposes on-shell electrons in addition to the on-shell photon.

Thank you very much!

 

[vanhees71]

Since QED is renormalizable in the Dyson sense the counter terms necessary to renormalize the theory look precisely as the ones in the original Lagrangian. If you use dim. reg. and minimal subtraction this will be automatically fulfilled and you also don't run into trouble with introducing additional IR problem when using a "physical renormalization scheme". Also note that there's a Ward identity, implying $\delta_1=\delta_2$. This is important, because it's necessary to keep the counter-term Lagrangian gauge invariant as it must be.

Of course, the finite part of the vertex renormalization leads to important additional terms as the contribution to the magnetic moment of the electron, which was one of the early radiative-correction calculations (first done by Schwinger) in agreement with experiment.

 

[Gaussian97]

Since QED is renormalizable in the Dyson sense the counter terms necessary to renormalize the theory look precisely as the ones in the original Lagrangian. If you use dim. reg. and minimal subtraction this will be automatically fulfilled and you also don't run into trouble with introducing additional IR problem when using a "physical renormalization scheme". Also note that there's a Ward identity, implying $\delta_1=\delta_2$. This is important, because it's necessary to keep the counter-term Lagrangian gauge invariant as it must be.

Of course, the finite part of the vertex renormalization leads to important additional terms as the contribution to the magnetic moment of the electron, which was one of the early radiative-correction calculations (first done by Schwinger) in agreement with experiment.

Ok thanks, but I'm not sure how to extract the answer to my question from here, can you be a little more specific?

 

[vanhees71]

For a nice discussion of renormalization of QED (and also QCD), see

https://arxiv.org/abs/hep-ph/0508242

There both the mass-independent $\overline{\text{MS}}$ renormalization scheme as well as the socalled "on-shell" renormalization scheme are described in detail.