Weinberg's virtual soft photon addition to Feynman diagrams

I am reading Weinberg's book on QFT, especially Chapter 13.2, talking about the effect of adding virtual soft photons to a Feynman graph in the theory of QED. Weinberg supports that adding $$N$$ virtual soft photon lines to a Feynman diagram corresponds to supplying the amplitude with a factor of $$\frac{1}{N!2^N}\bigg[\frac{1}{(2\pi)^4}\sum_{n,m} e_ne_m\eta_n\eta_mJ_{nm}\bigg]^N\tag{13.2.2}$$ where $$e_n,\ e_m$$ are the charges of the $$n-$$th and $$m-$$th charged particles, $$e_n$$ and $$e_m$$ are equal to $$+1$$ if the $$n-$$th or the $$m-$$th particle is outgoing and $$-1$$ if they are incoming, and $$J_{nm}=-i(p_n\cdot p_m)\int_{\lambda\le|\textbf{q}|\le\Lambda} \frac{d^4q}{(q^2-i\epsilon)(p_n\cdot q-i\eta_n\epsilon) (-p_m\cdot q-i\eta_m\epsilon)}\tag{13.2.3}$$ with $$p_n$$ and $$p_m$$ being the momentum of the $$n-$$th and the momentum of the $$m-$$th particle, respectively. The momentum of the virtual photon, $$q$$, is restricted such that the spatial component of it, $$\textbf{q}$$ is greater than some IR regulator, but not less than an energy scale beyond which the virtual photon is no longer considered to be soft.

My question is very specific: Weinberg claims that we have divided the contribution of the $$N$$ virtual photons to the "elastic" amplitude (i.e. the one without any soft virtual photons)

by a factor of $$2^NN!$$, because the sum over all places to which we may attach the two ends of the soft photon lines includes spurious sums over the $$N!$$ permutations of the photon lines and over interchanges of the two ends of these lines.

I understand what the spourious sum over the $$N!$$ permutations of the $$N$$ additional virtual soft photons is, but I would like to ask what are the spourious sum over the interchanges of the two ends of the soft photon lines? I suspect that the spourious sum over the interchange of the two ends of the soft photon line

1. implies including a sum over the Lorentz indices of the additional photon propagator, i.e. something like $$\frac{-i\eta_{\mu\nu}}{q^2-i\epsilon}\Big(ie\gamma^{\mu}S_F(p_n-q)\Big) \Big(ie\gamma^{\nu}S_F(-p_m-q)\Big)+\mu\leftrightarrow\nu$$ where $$S_F(p)$$ is the fermion propagator in momentum space, and $$\mu$$ and $$\nu$$ are the Lorentz indices of the additional vertices that have been formed upon attaching the additional photon line to the Feynman diagram without the photon.

2. implies including a sum over the transformation $$q\rightarrow-q$$, i.e. $$\frac{-i\eta_{\mu\nu}}{q^2-i\epsilon}\Big(ie\gamma^{\mu}S_F(p_n-q)\Big) \Big(ie\gamma^{\nu}S_F(-p_m-q)\Big)+q\leftrightarrow-q$$

So, which of the two is it? Any help will be appreciated!

EDIT: I guess my question can be reduced to "If I add a virtual soft photon to a Feynman diagram that connects the external line $$n$$ to the external line $$m$$, why do I divide with a symmetry factor of 2?"

• You are supposed to sum over all pairs of external particles $(i,j)$. We instead sum over all $i$ and all $j$ separately and then divide by $2$. Jul 14, 2023 at 12:54
• Let us restrict ourselves to a solid example: a $2\rightarrow2$ scattering. Let $1,\ 2$ be the initial state particles and $3,\ 4$ be the final state ones. If I were to sum over all pais of external particles I would be summing photon exchange diagrams between $\{(1,1),\ (1,2),\ (1,3),\ (1,4), (2,2),\ (2,3),\ (2,4),\ (3,3),\ (3,4),\ (4,4)\}$. That is correct, right? And you say that instead of that, we sum over the combinations $\{(1,1),\ (1,2),\ (1,3),\ (1,4),\ (2,1),\ (2,2),\ (2,3),\ (2,4),\ (3,1),\ (3,2),\ (3,3),\ (3,4),\ (4,1),\ (4,2),\ (4,3),\ (4,4)\}$ and divide by two. Correct? Jul 14, 2023 at 13:06
• The $i=j$ diagrams get divided by two because of a symmetry factor in the individual Feynman diagram. Note that you need to deal with the $i=j$ case separately from the $i\neq j$ case as the former corresponds to field renormalization. It turns out to give EXACTLY the same contribution as the general case EXCEPT that it has a different $i\epsilon$ prescription. Jul 14, 2023 at 13:27
• Yes, to all your comments. For the last comment - yes, they are completely different diagrams, but if you evaluate the Feynman diagram for the $m \neq n$ and $m=n$ case, you will see that the $m=n$ result obtained by taking the $m \neq n$ result and then setting $m=n$ EXCEPT for the $i\epsilon$. Jul 14, 2023 at 15:08
• yes, they are the same. It's the same Feynman diagram and it only depends on the pair $(m,n)$, not the ordering. To clarify once more, let $f(m,n)=f(n,m)$ be the value Feynman diagram where photon couples to particles $m$ and $n$. Let $f(m)$ be the field renormalization diagram for the $m$th particle. Based on what I said earlier, we have $f(m) = \frac{1}{2} f(m,m)$ (apart from the $i\epsilon$ prescription). The total sum is therefore $\sum_{m < n} f(m,n) + \sum_m f(m) = \frac{1}{2} \sum_{m \neq n} f(m,n) + \frac{1}{2} \sum_m f(m,m) = \frac{1}{2} \sum_{m,n} f(m,n)$. Jul 15, 2023 at 8:37