# How to prove the equivalence of two definitions of the scattering cross section

I have noticed that there are two definitions of differential scattering cross section in non-relativistic quantum mechanics.

One of them is the most popular, particularly it is used in the book of Landau (text in italics meant for clarification):

The probability per unit time that the scattered particle will pass through a surface element $dS = r^2 d\Omega$ ... is $\left( \frac{v}{r^2} \right) |f|^2 dS = v |f|^2 d\Omega$ (f is the scattering amplitude). Its ratio to the current density in the incident wave is $$d\sigma = |f(\theta)|^2 d\Omega$$ This quantity has the dimensions of area, and is called the effective cross-section, or simply the cross-section, for scattering into the solid angle $d\Omega$.

Another definition was given in the book of Feynman Hibbs, Quantum Mechanics and Path integrals. It says

The cross section $\sigma$ is the defined as the effective target area (from classical point of view) of the atom that must be hit by an electron in order that the electron be scattered into a unit solid angle.

My question is how can I understand that these two definitions are equivalent?

• I don't have the time to sketch out the full argument, but I'm pretty sure it can be found in Chapter 11 of Griffith's" Introduction to Quantum Mechanics" (at least in the first edition). Apr 19 '18 at 17:30

First, it's important to keep in mind that particle scattering is an inherently quantum mechanical process. The "effective target area" description is nothing more than a suggestive classical metaphor to provide intuition for this non-classical process.

In this metaphor, we imagine our incoming particle beam to be composed of $dN_{\rm incoming}$ particles distributed uniformly over patches with cross-sectional area $A_{\rm beam}$ and depth $dl$ along the beam path. The flux of incident particles is therefore $v \,dN_{\rm incoming}/(A_{\rm beam} dl)$ where v is the particle velocity.

Some fraction of the particles in each incoming patch will be scattered into solid angle $d\Omega$. What fraction is that? To find out, we imagine that the entire scatterer is sub-divided into many component target types which determine the path of the particles that hit them. The rate at which particles hit a particular target type is then just the particle flux times the the total cross-sectional area of targets of this type, given by $d\sigma / d\Omega$. In other words,

$$\frac{dN_{\rm scattered}}{d\Omega dt} = \left( v \frac{dN_{\rm incoming}}{A_{\rm beam} dl}\right) \frac{d\sigma}{ d\Omega}$$

Given the way we've described the beam, $v = dl/dt$, and after multiplying both sides by the area of the beam $A_{\rm beam}$ and the time interval $dt$ for the particles in the beam patch to encounter the targets, we get

$$$$\frac{dN_{\rm scattered}}{d\Omega} \, \bigg / \, dN_{\rm incoming} = \frac{d\sigma}{d\Omega} \bigg / A_{\rm beam} \label{fraction}$$$$

In other words, the fraction of the particles in the beam patch scattered into solid angle d$\Omega$ is given by the ratio of the target's cross-sectional area $d\sigma/d\Omega$ to the beam area $A_{\rm beam}$.

Equating the RHS here to the quantum mechanical expression for the LHS, as given by Landau, establishes a definition for $d\sigma / d\Omega$ in terms of an effective scattering target area, as indicated in the book by Feynman and Hibbs.

Next we might want to quantify the extent to which particles in the incident beam are scattered in any direction. Particles that continue on their incident path without scattering contribute nothing to the scattering solid angle, because their "scattered" direction is a point of dimension zero on the unit sphere representing scattered directions. We integrate our equation for the fraction of scattered particles over solid angle to obtain

$$$$\frac{dN_{\rm scattered}}{dN_{\rm incoming}} = \frac{\sigma}{A_{\rm beam} }$$$$

where $\sigma = \int (d\sigma / d\Omega) \, d\Omega$. So again we have a way to think of scattering probability in terms of an effective target area.

Remember, in reality there aren't little targets of area $d\sigma/d\Omega$, there are quantum mechanical scattering amplitudes. What we've done here is built a classical analogy for $\sigma$ such that it ends up giving the same result as the quantum mechanical calculation.