Renormalization and virtual soft divergences I am reading Weinberg's book on QFT. Specifically, chapter 13.2. The author calculates the effect of including infrared quantum corrections (i.e. associated with soft virtual photons) to amplitudes. While doing so, he also calculates the effect of Feynman graphs in which a soft photon is emitted and reabsorbed by the same fermion. However, this is a self-energy effect, which we know that must not be included in our renormalized amplitudes.
The author justifies (if I understand correctly) by relating the renormalized scattering amplitude with the bare one: namely
$$S_{\beta\alpha}^{(R)}=Z_2^{-E/2}Z_2^{V/2}Z_2^{I} S_{\beta\alpha}^{(B)}$$
where $S_{\beta\alpha}^{(R)}$ is the renormalized amplitude, $S_{\beta\alpha}^{(B)}$ the bare one and $Z_2$ are the factors that relate the renormalized fermion field with the bare one. Above, we have $E$ external fermion lines, $I$ internal ones and $V$ vertices. He then proceeds in saying that the counter terms that cancel the external line radiative corrections (i.e. self energy effects) are now cancelled by the $Z_2$ factors arising from vertices and internal lines.
Can someone elaborate on that? I am not sure I fully understand it.
 A: There are two distinct divergent pieces associated with external fermion lines: soft photon emissions and self-energy effects.
Because of the QED Ward Identity, the external line's field strengths $Z_{2}$ must get renormalized the same way as the vertex does so as not to upset gauge invariance in one-loop order calculations. A consequence of this is that divergent soft photon emissions get canceled by the divergence in the electromagnetic form factor $F_{1}(q^2)$. At the same time, the Ward identity fixes $Z_{1} = Z_{2}$ and that takes care of the self-energy part you mentioned (i.e, the photons that were not radiated away get renormalized by the same factor as the vertex). In more general diagrams, the internal lines also contribute to canceling the divergences in soft photon emissions, and Weinberg shows how.
A: I think I can now answer the question in a satisfactory manner. If any of the following are not correct, please anyone who sees it, leave a comment or offer a better answer. Weinberg calculates the contribution of the virtual soft photons to the unrenormalized scattering amplitude (using bare fields). Then, according to
$$S_{\beta\alpha}^{(R)}=Z_2^{-E/2}Z_2^{V/2}Z_2^{I} S_{\beta\alpha}^{(B)}$$
where $S_{\beta\alpha}^{(R)}$ is the renormalized amplitude and $S_{\beta\alpha}^{(B)}$ the one constructed using bare fields, one can see that  adding a virtual photon the the bare amplitude $S_{\beta\alpha}^{(B)}$ introduces some self-energy diagrams as well. We know that these diagrams are not present in the renormalized amplitudes. Those diagrams are cancelled by the $Z$ factors from the external lines (i.e. $Z_2^{-E/2}$). Then, those $Z$ factors will have introduced some additional divergences on the renormalized amplitude. They will cancel with the divergences coming from the $Z$ factors from the vertices and the internal lines.
As an example, one can think of an electron-muon scattering. At the tree level, there is no $Z$ factor, coming from internal lines. So, the diagrams contributing to the bare amplitude contain self-energy diagrams, as well. Those are cancelled by counter-term diagrams and the remaining diagrams comprise the renormalized scattering amplitude.
Therefore, one needs to include the self-energy diagrams (cases $n=m$ according to the book, the $n-$th fermion emitting the virtual photon and the $m-$th fermion absorbing it back), such that this nice cancellation takes place. Otherwise, it can not.
P.S.: Note that the $Z$ factor coming from the external lines are equal to the $Z$ factor coming from the vertex due to the Ward identity.
