# Understanding quantum cross sections as areas

In scattering cross sections we deal with $$d\sigma/d\Omega$$, incident area per scattered solid angle. When a particle scatters into a small finite $$\Delta\Omega$$, the incident particle was in a small finite area $$\Delta\sigma$$. However, in QM the incident state is a plane wave / asymptotic momentum eigenstate, so it's totally delocalized in position space. Isn't the probability for the incident particle to be found in the area $$\Delta\sigma$$ therefore zero (a small area out of an infinite plane)? If we integrate $$d\Omega$$ we'd find the total probability to be zero, which is absurd. Where did this reasoning go wrong?

It seems to me it would make more sense to define $$dP/d\Omega$$ instead of $$d\sigma/d\Omega$$. In a scattering there is some probability the final momentum angle is in some $$d\Omega$$. This would then integrate to 1. But obviously this is not done and the cross-sectional area $$\sigma$$ is somehow necessary.

## 1 Answer

Here's my take on the thing, based on what is expressed in chapter 11 of N. Zettili's Quantum mechanics: Concepts and applications. The scattering cross section is defined as the number of particles $$d\sigma$$ scattered into an element of solid angle $$d\Omega$$ defined by the angles $$(\theta, \varphi)$$. This is related with the incident flux of particles $$J_{inc}$$ as $$\frac{d\sigma(\theta, \varphi)}{d\Omega}=\frac{1}{J_{inc}}\frac{dN(\theta, \phi)}{d\Omega}$$ where $$dN$$ is the number of particles scattered into an element of solid angle. The incident flux can be calculated as $$J_{inc}=\vert A\vert^2\frac{\hbar k_0}{\mu}$$ where $$\mu$$ is the reduced mass of the system, $$A$$ is a normalization constant and $$k_0$$ is the wave number for the incident wave.

Now, what you're saying isn't exactly wrong, you're just forgetting that position and probability are actually bound in a quantum system. So, this is information is stored in the wave number somehow, but specially in the probability amplitude, and surely in the number of particles $$dN$$.