# Does the differential cross section curve correspond to a probability density function?

Quick question here. If I know the differential cross section function for a given phenomenon, and if I normalize this function such as its integral over its domain is 1, can I interpret this normalized function as a probability density function?

Here's what I mean. Let's take Compton scattering as an example. The cross section for a scattering to produce an electron in energy interval ($k$, $k+$d$k$) is d$\sigma$/d$k$ (known). If I find the constant $C$ such as $\int{C \;d\sigma/dk\;\;dk}=1$, would the function $f(k) = C\; d\sigma/dk$ be a probability density function?

So if I wanted to get the average energy of and electron produced by Compton scattering, I could simply use $\int{k\;f(k)\;dk}$?

This function $f(k)$ would then be the (normalized) energy spectrum of the scattered electrons, is that correct? And does this work with any kind of differential cross section (angular, energy, etc.)?

Thanks a lot!

• The Rutherford scattering cross section is infinite, so that might be a problem (this is true for any $1/r^{2}$ potential). – Jon Custer Aug 2 '16 at 17:09
• Yeah, but in this case could I use new boundaries for the integral? Let's say I'm interested in the average scattering angle of every electron that is scattered with an angle greater than a certain limit $\theta>0$. Then I could integrate between $\theta$ and $\pi$, which would be a finite cross section if I remember the Rutherford scattering correctly. – Jasmeru Aug 2 '16 at 17:24
• $p (x)=\frac1\sigma \frac{d\sigma}{dx}$ is correct, might write more details if i have time later – innisfree Aug 2 '16 at 18:43
• Have a look at this tcm.phy.cam.ac.uk/~bds10/aqp/lec20-21_compressed.pdf . In a nutshell the differential crossection is proportional to the probability of scattering, and hence to the complex conjugate squared of the wavefunction in the problem. – anna v Aug 3 '16 at 5:46
• @innisfree I can see no way that your formula is correct. probability and scattering crossection have to be proportional. – anna v Aug 3 '16 at 5:51

The probability density over the Lorentz invarant phase space (LIPS) is proportinal to the matrix element squared and a Dirac function for energy momentum conservation, $$p(\text{LIPS}) \propto|\mathcal M|^2 \delta(p_i -p_f)$$ The right hand side appears in the formula for a cross section (NB I don't include the Dirac function in the $d\text {LIPS}$) $$\sigma = \frac1 {2I} \int |\mathcal M|^2\delta(p_i -p_f)\,d\text{LIPS}$$ such that $$p(\text{LIPS}) \propto \frac {d\sigma}{d\text {LIPS}}$$ By requiring $\int p(\text{LIPS})\,d\text{LIPS}=1$, we find that $$p(\text{LIPS}) = \frac {1}{\sigma} \frac {d\sigma}{d\text {LIPS}}$$
We find the pdf for a particular observable by marginalization over the full phase space, \begin {align} p(x) &= \int \delta(x - x(\text{LIPS})) p(\text{LIPS})\,d\text{LIPS} \\ &= \frac {1}{\sigma}\int \delta(x - x(\text{LIPS})) \frac {d\sigma}{d\text{LIPS}}\,d\text{LIPS} \\ &= \frac {1}{\sigma} \frac {d\sigma}{dx} \end {align} The last equality follows more or less from the meaning of a differential cross section (integration over the rest of the phase space is implicut in the ordinary notation for a differential cross section $\frac {d\sigma}{dx}$).