How is the probability density for the deflection angle calculated from the differential cross-section?

Question

In a scattering process, the probability density for the deflection angle of the incident particle is proportional to the differential cross-section of that scattering process. Given the differential cross-section, how is the probability density for the deflection angle of the particle calculated?

For example, the differential cross-section for a Moller scattering process (collision of two electrons) is given in https://en.wikipedia.org/wiki/Møller_scattering (end of the article). What would be the probability that the electron is deflected by an angle between θ and 90 degrees, after colliding another electron?

Answer 1

Well, you know that $$ p(x) \propto \frac{d\sigma}{dx} $$ from which it is easy to find $$ p(x) = \frac{1}{\sigma} \frac{d\sigma}{dx} $$ which follows from $\int p(dx) dx = 1$.

Answer 2

A probability density can be obtained by integrating the differential cross section over $d\Omega$.

For Moller scattering we have $$ \frac{d\sigma}{d\Omega} =\frac{\alpha^2}{8E^2} \left( \frac{1+\cos^4(\theta/2)}{\sin^4(\theta/2)} +\frac{8}{\sin^2\theta} +\frac{1+\sin^4(\theta/2)}{\cos^4(\theta/2)} \right) $$

Let $$ I(\xi)=2\pi\int_\alpha^\xi\frac{d\sigma}{d\Omega}\,\sin\theta\,d\theta, \quad\alpha\le\xi\le\pi-\alpha $$

for some $\alpha>0$. The support range is restricted because $d\sigma$ is undefined at $\theta=0$ and $\theta=\pi$.

The cumulative distribution function is $$ F(\theta)=\frac{I(\theta)}{I(\pi-\alpha)}, \quad\alpha\le\theta\le\pi-\alpha $$

The probability density is $$ f(\theta)=\frac{dF(\theta)}{d\theta} =\frac{2\pi}{I(\pi-\alpha)} \left(\frac{d\sigma}{d\Omega}\right) \sin\theta, \quad\alpha\le\theta\le\pi-\alpha $$

Here is a graph of $f(\theta)$ for $\alpha=\pi/180$.