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.
 A: 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 $dN$.
