The soft-photon theorem is the following statement due to Weinberg:

Consider an amplitude ${\cal M}$ involving some incoming and some outgoing particles. Now, consider the same amplitude with an additional soft-photon ($\omega_{\text{photon}} \to 0$) coupled to one of the particles. Call this amplitude ${\cal M}'$. The two amplitudes are related by
$$
{\cal M}' = {\cal M} \frac{\eta q p \cdot \epsilon}{p \cdot p_\gamma - i \eta \varepsilon}
$$
where $p$ is the momentum of the particle that the photon couples to, $\epsilon$ is the polarization of the photon and $p_\gamma$ is the momentum of the soft-photon. $\eta = 1$ for outgoing particles and $\eta = -1$ for incoming ones. Finally, $q$ is the charge of the particle.

The most striking thing about this theorem (to me) is the fact that the proportionality factor relating ${\cal M}$ and ${\cal M}'$ is independent of the type of particle that the photon couples to. It seems quite amazing to me that even though the coupling of photons to scalars, spinors, etc. takes such a different form, you still end up getting the same coupling above.

While I can show that this is indeed true for all the special cases of interest, my question is: **Is there a general proof (or understanding) that describes this universal coupling of soft-photons?**

This post imported from StackExchange Physics at 2014-03-17 06:43 (UCT), posted by SE-user Prahar