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

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?

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 $$\int p(dx) dx = 1$$.
• it;s the differential cross section with respect to some variable $x$. The $x$ should be some information about the phase space of the final state particles. In your case, you would probably want $x \to \theta$, the angle between outgoing electron and beam axis (in com frame) – innisfree May 29 at 9:40