I am reading Srednicki's book on QFT. Specifically, I am reading about the loop corrections to the fermion propagator (Chapter 62). The relevant expression representing the one-loop and counterterm corrections to the fermion propagator in spinor electrodynamics, after using on-shell renormalization scheme, is denoted by $\Sigma(p\llap{\unicode{x2215}})$ and it is given by $$-ie^2\int \frac{d^4l}{(2\pi)^4} \frac{1}{(l^2-i\epsilon)}\gamma^{\mu} \frac{(-p\llap{\unicode{x2215}}-l\llap{\unicode{x2215}}+m)}{(l^2+2p\cdot l-i\epsilon)} \gamma_{\mu}\tag{62.28}$$ where $p$ and $m$ is the momentum and the mass of the electron, and $l$ the momentum of the additional virtual photon.

At the same time, I am reading Weinberg's book on QFT, specifically, vol. 1 Chapter 13. At some point in Chapter 13, Weinberg derives the contribution of virtual soft photons to scattering amplitudes. Specifically, he states that adding a vritual soft photon to a diagram representing a scattering amplitude results in supplying the aforesaid amplitude with a factor of $$\frac{1}{2}\sum_{ij}e_ie_j\eta_i\eta_jJ_{ij}$$ where $e_i$ the $i-$th fermion's charge, $\eta_i=\pm1$ depending on whether or not the $i-$th fermion is outgoing (+1) or incoming (-1), and $$J_{ij}=-i(p_i\cdot p_j)\int_{\lambda\le|\textbf{q}|\le\Lambda} \frac{d^4q}{(2\pi)^4}\frac{1}{(q^2-i\epsilon) (p_i\cdot q-i\eta_i\epsilon)(-p_j\cdot q-i\eta_j\epsilon)}$$ Normally, if I restrict myself, let's say to an incoming fermion and the term from the sum that concerns the latter emitting and re-absorbing the additional virtual photon, then I shall have something of the form $$-\frac{i}{2}e_1^2 (p_1\cdot p_1)\int_{\lambda\le|\textbf{q}|\le\Lambda} \frac{d^4q}{(2\pi)^4}\frac{1}{(q^2-i\epsilon) (p_1\cdot q+i\epsilon)(-p_1\cdot q+i\epsilon)}$$ multiplying the original amplitude.

Suppose I wish to obtain Weinberg's result for adding a single virtual soft photon to a Feynman diagram, from the electron self-energy amplitude. Then, I would start with (62.28), and using some Dirac matrix properties and neglecting the $l^2$ term in the denominator, since we would like to study the case in which the additional virtual photon is soft, would yield $$-ie^2\int_{\lambda\le|\textbf{q}|\le\Lambda} \frac{d^4l}{(2\pi)^4} \frac{1}{(l^2-i\epsilon)} \frac{[-2(p\llap{\unicode{x2215}}+l\llap{\unicode{x2215}})-4m]}{(2p\cdot l-i\epsilon)} $$ My thoughts for deriving the result of Weinberg include supplying the amplitude one additional propagator. However, according to the self-energy diagram, this fermion propagator should be on-shell, since the additional soft photon is reabsorbed by the fermion and only then the self-energy diagram is connected to the remaining interaction (with a fermion propagator).

To describe my attempt better, let $\Gamma(p)u(p)$ be the amplitude describing an electron going into some interaction. Then, adding one virtual photon to the external electron results in $\Gamma(p)u(p)\rightarrow i\mathcal{M}=\Gamma(p)S_F(p)i\Sigma(p)u(p)$, where $$i\mathcal{M}= -e^2\Gamma(p)\int_{\lambda<|\vec{l}|<\Lambda} \frac{d^4l}{(2\pi)^4}\frac{1}{l^2-i\epsilon} \frac{(-p\llap{\unicode{x2215}}+m)}{p^2+m^2-i\epsilon} \frac{(p\llap{\unicode{x2215}}-l\llap{\unicode{x2215}})}{(p\cdot l-i\epsilon)}u(p)$$

How should I proceed in order to derive Weinberg's result?



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.