1
$\begingroup$

On P. 208 of the book An Introduction of Quantum Field Theory by Peskin and Schroeder, the probability of production $n$ soft photos all with with energies between $E_- < E < E_+$ is discussed (Eq. (6.86)). It reads

$$\text{Prob}(n\gamma \text{ with } E_-<E<E_+)=\frac{1}{n!}\left[\frac{\alpha}{\pi}f_{IR}(q^2)\log\left(\frac{E_+^2}{E_-^2}\right)\right]^n\times \exp\left[-\frac{\alpha}{\pi}f_{IR}(q^2)\log\left(\frac{E_+^2}{E_-^2}\right)\right] $$

What I understand is the following. The probability to reproduce $n$ soft photons up to the detector's energy thread $E_l$ is

$$\text{Prob}(n\gamma \text{ with } \mu<E<E_l)=\frac{{d\sigma}/{d\Omega}(n\gamma \text{ with } \mu<E<E_l)}{\sum_{n'=0,1,2,\cdots}{d\sigma}/{d\Omega}(n'\gamma \text{ with } \mu<E<E_l)}\\ =\frac{\left(\frac{d\sigma}{d\Omega}\right)_0\frac{1}{n!}\left[\frac{\alpha}{\pi}f_{IR}(q^2)\log\left(\frac{E_l^2}{\mu^2}\right)\right]^n\times \exp\left[-\frac{\alpha}{\pi}f_{IR}(q^2)\log\left(\frac{-q^2}{\mu^2}\right)\right]}{\left(\frac{d\sigma}{d\Omega}\right)_0\exp\left[-\frac{\alpha}{\pi}f_{IR}(q^2)\log\left(\frac{-q^2}{E_l^2}\right)\right]}$$

The denominator consists of three factors. The first factor, $\left(\frac{d\sigma}{d\Omega}\right)_0$, corresponds to the vertex together with some given hard process. The second factor, $\frac{1}{n!}\left[\frac{\alpha}{\pi}f_{IR}(q^2)\log\left(\frac{E_l^2}{\mu^2}\right)\right]^n$, describes the emission of $n$ soft photons (the $Y$ term Eq.(6.82)), while the last factor, $\exp\left[-\frac{\alpha}{\pi}f_{IR}(q^2)\log\left(\frac{-q^2}{\mu^2}\right)\right]$, takes into account the contribution from the virtual photons (the $X$ term Eq.(6.79)).

This seems to be "reasonable", since by summing $n'$ of the numerator, one obtains

$$\sum_{n'=0,1,2,\cdots}{d\sigma}/{d\Omega}(n'\gamma \text{ with } \mu<E<E_l)\\ =\left(\frac{d\sigma}{d\Omega}\right)_0\sum_{n'}\frac{1}{n'!}\left[\frac{\alpha}{\pi}f_{IR}(q^2)\log\left(\frac{E_l^2}{\mu^2}\right)\right]^{n'}\times \exp\left[-\frac{\alpha}{\pi}f_{IR}(q^2)\log\left(\frac{-q^2}{\mu^2}\right)\right]\\ =\left(\frac{d\sigma}{d\Omega}\right)_0\exp\left[\frac{\alpha}{\pi}f_{IR}(q^2)\log\left(\frac{E_l^2}{\mu^2}\right)\right]\times \exp\left[-\frac{\alpha}{\pi}f_{IR}(q^2)\log\left(\frac{-q^2}{\mu^2}\right)\right]\\ =\left(\frac{d\sigma}{d\Omega}\right)_0\exp\left[-\frac{\alpha}{\pi}f_{IR}(q^2)\log\left(\frac{-q^2}{E_l^2}\right)\right]$$

Therefore, it restores Eq.(6.84) as mentioned in the textbook and implies $\sum_n\text{Prob}(n\gamma)=1$.

However the above expression is not consistent with Eq.(6.86) when the energy range of the undetected soft photons is $E_- < E < E_+$, since one cannot replace $-q^2$ by $E_+$ in the virtual photon term. What am I missing? Many thanks!

$\endgroup$
1
  • $\begingroup$ This turned out to be a silly question, one simply divides the numerator by the denominator and the two terms $-q^2$ cancel each other and one obtains $\log(E_l^2/\mu^2)$. Then some arguments lead to replacing $E_l$ by $E_+$ and $\mu$ by $E_-$. $\endgroup$
    – gamebm
    Jul 7, 2016 at 2:51

0

Your Answer

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