# Bremsstrahlung and infrared divergences in Peskin & Schroeder 6.5

At the very end of chapter 6 in Peskin and Schroeder (page 208), there is a calculation of the expected number of soft photons radiated in a hard electron scattering process. The text claims that the cross-section for the process of producing $n$ soft photons is given by the cross-section for producing no soft photons in the process (or more precisely, for producing no photons with energy greater than the detector threshold energy), times the additional factor $$\frac{1}{n!} \left[ \frac{\alpha}{\pi} f_\mathrm{IR}(q^2) \log\left(\frac{E_+^2}{E_-^2}\right)\right]^n \times \exp\left[ - \frac{\alpha}{\pi} f_\mathrm{IR}(q^2) \log \left(\frac{E_+^2}{E_-^2}\right)\right] \,,$$ where the soft photons all have energy between $E_-$ and $E_+$. The first factor in this expression make sense to me, since the factor in square brackets is simply the contribution to the cross-section of producing one real soft photon, as in equation (6.82) on the previous page (which gives the contribution to the cross-section of emitting one real photon with energy between the infrared cutoff $\mu$ and the detector threshold energy $E_\ell$).

What I don't understand is where the exponential factor comes from. The cross-section for the hard process with no detectable soft photons (6.84) contains similar exponential factors, but these arise from summing over a) all possible number of virtual photon exchanges, and b) all possible number of undetectable real photon emissions. Since we are explicitly asking for $n$ real photons to be produced, I cannot see where such an exponential could arise from. Thank you.

The expression is just a Poisson distribution for getting $n$ events, with the rate parameter being $$\lambda = \frac{\alpha}{\pi} f_\mathrm{IR}(q^2) \log\left(\frac{E_+^2}{E_-^2}\right)$$
• Thanks for the answer. The book itself points out that this is a Poisson distribution, but it seems more that this is being deduced from the expression (6.86) (the expression in my original post) rather than being used to derive it. The quantity $\mathbf{Y}$ in (6.82) is the contribution to the cross-section for emission of one real photon, so I don't see why the expression (6.86) isn't merely $\mathbf{Y}^n/n!$. – gj255 Jan 16 '18 at 15:38