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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.

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  • $\begingroup$ Could you explain what you mean by "renormalized/bare scattering amplitudes"? Strictly speaking, Weinberg only defined one notion of "amplitude", namely the physical one $S_{\beta\alpha}$, right? Do you mean his $M$, as defined in eq.(10.2.3) or equivalently the second line of eq.(10.3.2)? $\endgroup$
    – chaostang
    Mar 29, 2023 at 10:38
  • $\begingroup$ Weinberg only defines one kind of amplitude, yes. But, in principle, no one is stopping me from writting down a "bare" Lagrangian (one that contains only the bare fields) and deriving S-matrix elements from it... Those matrix elements are going to be related to the physical ones in the way I describe above (I believe this is what he means by that...) $\endgroup$
    – schris38
    Mar 29, 2023 at 11:14
  • $\begingroup$ I beg to differ. Weinberg didn't define an "unphysical S-matrix", and I don't agree when you say "derive S-matrix elements from (the Lagrangian)". I think logically speaking there is no direct connection between the Lagrangian and the S-matrix. Somewhere along the line, we must employ the LSZ formula, which would yield the same answer whether we use bare or renormalized fields to compute Green's functions. $\endgroup$
    – chaostang
    Mar 29, 2023 at 13:24
  • $\begingroup$ I'm also struggling with the discussions at the end of section 13.2. Here's what I'm able to understand so far. If $b$ denotes the bare field and $r$ denotes the renormalized field, we have the wavefunctions $\langle\Psi_0|b(0)|\Psi_{p,\sigma}\rangle=\sqrt{Z}\langle\Psi_0|r(0)|\Psi_{p,\sigma}\rangle=\sqrt{Z}u(p,\sigma)$ and the full propagators which schematically reads $\langle bb\rangle=Z\langle rr\rangle=Z/(p^2-m_P^2)$. Then, LSZ tells us that, for $N$-particle scattering, schematically, $S/(p^2-m_P^2)^N=\langle r^N\rangle=\langle b^N\rangle/\sqrt{Z^N}$. [to be continued] $\endgroup$
    – chaostang
    Mar 29, 2023 at 13:29
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    $\begingroup$ I agree that one can write down Feynman rules and compute diagrams, however, I prefer to call these Green's functions, because it is not clear how they relate to on-shell physical processes without LSZ. As for the $m=n$ terms, I fully agree that "one needs counterterms to deal with self-energy diagrams". Could you elaborate on how the cancellation works? I'm confused because it seems self-energy diagrams sum up to full propagators, which are proportional to $Z^N$ instead of the $\sqrt{Z^N}$ needed in $\langle r^N\rangle^A=\langle b^N\rangle^A\sqrt{Z^N}$. $\endgroup$
    – chaostang
    Mar 29, 2023 at 14:15

2 Answers 2

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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.

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  • $\begingroup$ Hi and thank you for your answer. I am not sure I follow your line of reasoning though... You seem to suggest that the divergences associated with soft emission negate the ones coming from virtual soft photons. If that is what you are trying to say, you are probably refering to chapter 13 of Weinberg's book, right?? However, there I can not see how the electromagnetic form factor enters the calculation... Can you shed some light please on how the form factors are related to the virtual divergences presented on chapter 13 and how these are associated with the external field strength $Z_2$??. $\endgroup$
    – schris38
    May 30, 2022 at 12:04
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I think I can now answer the question in a satisfactory manner. Weinberg claims that there is some sort of relationship between what I will be calling the renormalized amplitude and the bare amplitude. To support this relationship, one can write down the bare Lagrangian $$\mathcal{L}_B=-\frac{1}{4}F_{B\mu\nu}F_B^{\mu\nu}-\bar{\psi}_B(\gamma_{\mu}\partial^{\mu}+m_B)\psi_B- ie_B\bar{\psi}_B\gamma_{\mu}A_B^{\mu}\psi_B$$ and performing the following transformations $$\psi_B=Z_2^{1/2}\psi\\ A_B^{\mu}=Z_3^{1/2}A^{\mu} \\e_B=Z_3^{-1/2}e\\m_B=m-\delta m$$ one obtains the renormalized Lagrangian $$\mathcal{L}_R=-Z_3\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-Z_2\bar{\psi}(\gamma_{\mu}\partial^{\mu}+m_B)\psi- iZ_2e\bar{\psi}\gamma_{\mu}A^{\mu}\psi$$

This supports that the bare amplitude $S_{\beta\alpha}^{(B)}$ is related to the renormalized one, $S_{\beta\alpha}^{(R)}$, in the following way $$S_{\beta\alpha}^{(B)}=Z_2^{-E/2}Z_2^{V/2}Z_2^{I} S_{\beta\alpha}^{(R)}\label{*}\tag{*}$$ where $E$ denote the fermion external lines, $I$ denote the fermion propagators and $V$ denote the QED vertices.

Now, the key to understanding what is going on here is to recall that both the bare Lagrangian and the renormalized one yield scattering amplitudes that do not suffer from UV divergences. In the "on-shell" renormalization scheme, $Z_2$ is chosen such that it cancels with the corrections to the propagator exactly, when those corrections are located on an external leg. So, the usual procedure of "on-shell" renormalization fixes $Z_2$. I think this what the author means when he says that

virtual soft photons produce infrared divergences, not only directly, but also through their effect on the renormalization constants $Z_2$.

It is because the renormalization scheme has fixed $Z_2$ to be both IR and UV divergent. Since the quantum corrections to the external leg cancel exactly with the counter-term to the external leg, one might be tempted to suggest that we can neglect the $n=m$ contributions to the amplitude describing the virtual soft photon insertion to a generic interaction, on the grounds that it is to be negated by the relevant counter-terms.

However, Weinberg supports neglecting the $n=m$ contributions along with the relevant counter-terms would be wrong, because the divergences introduced by the virtual soft photons through $Z_2$ (from internal lines and vertices) would not have been cancelled by the factors of $Z_2$ introduced from the external lines (see eq.\ref{*}), which are present so that the corrections to the external line do not diverge.

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