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In QED, one can calculate the correction to the form factor $F_2$. To the lowest order, $F_1=1$ and $F_2=0$. At one loop, it is found that $F_2(0)$ receives a non-zero finite correction which is related to the anomalous magnetic moment of the electron. However, the correction to charge form factor $F_1$ turns out to be divergent and one gets rid of the divergence by renormalization.

$\bullet$ Is there any fundamental reason to believe that $F_1$ should not receive any finite correction?

$\bullet$ The vertex correction $\delta\Gamma_\mu$ is divided into two parts (in the following, I used notation of L. H. Ryder): $$\delta\Gamma_\mu=\Gamma_\mu-\gamma_\mu=\Lambda_\mu^{(1)}+\Lambda_\mu^{(2)}$$ $\Lambda_\mu^{(1)}$ contributing to $F_1$ and is divergent while $\Lambda_\mu^{(2)}$ contributes to $F_2$ and is convergent. It turns out that, $\Lambda_\mu^{(1)}$ itself has a divergent and a finite part. Renormalization gets rid of total $\Lambda_\mu^{(1)}$ including its finite part.

Does it not mean that we don't want to have any quantum correction to $F_1$?

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  • $\begingroup$ Hi! $F_1(0)$ is the electric charge of the electron (in units of $e$), and this is the reason why it must be equal to one at all orders. This happens to be true because the vertex correction is exactly compensated by the field strength renormalization as a consequence of the Ward identities. I might write a complete answer if I have time, but I think that this calculation can be found in most QFT books. Note that at $q^2\neq 0$ there is indeed a quantum correction. $\endgroup$
    – DelCrosB
    Commented Oct 18, 2016 at 20:59

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