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  Understanding Weinberg's soft-photon theorem

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2206 views

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
asked Dec 9, 2013 in Theoretical Physics by prahar21 (545 points) [ no revision ]
It depends on the charge $q$, and you indeed expect that the coupling of a photon to some matter (or other particles) depends only on the charge, and not on something else, right ? So I'm afraid don't really understand what is your puzzle.

This post imported from StackExchange Physics at 2014-03-17 06:43 (UCT), posted by SE-user Bru
I'm asking why the amplitude factorizes universally? As in, why is the form of the factor always $\frac{p\cdot\epsilon}{p\cdot p_\gamma}$??

This post imported from StackExchange Physics at 2014-03-17 06:43 (UCT), posted by SE-user Prahar
@Prahar Would you mind giving a reference for the statement of the theorem?

This post imported from StackExchange Physics at 2014-03-17 06:43 (UCT), posted by SE-user joshphysics
@Prahar : There is a nice discussion in this paper, Chapter $4$ pages $17-24$, while the interesting formula is $109$, page $20$

This post imported from StackExchange Physics at 2014-03-17 06:43 (UCT), posted by SE-user Trimok
@Trimok the epsilon in the denominator is different from the epsilon in the numerator (look closely). The epsilon in the denominator has to do with the Feynman prescription for integrating around poles. (I would guess you were being facetious.)

This post imported from StackExchange Physics at 2014-03-17 06:43 (UCT), posted by SE-user NowIGetToLearnWhatAHeadIs
@joshphysics - One can look this up in Weinberg Vol. 2 Eq (13.1.3)

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

1 Answer

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The universality of the coupling of the photon to charged particles exhibited by this formula is only valid in the limit of ultrasoft photons. This is also known as the eikonal approximation, in which the photon couples only to the charge x velocity of the charged particle.

This post imported from StackExchange Physics at 2014-03-17 06:43 (UCT), posted by SE-user user35997
answered Dec 18, 2013 by user35997 (0 points) [ no revision ]

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