# 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.

Think of scattering events like this; the blue and red wavepackets in some handwavy way represent the wavefunctions of two particles moving toward each other):

for some time, the two particles' wavefunctions overlap, so in some region of space they can be approximated as two overlapping plane waves. This is the framework in which we calculate scattering cross sections.

Indeed, as you point out, if the wave packets are spread out infinitely in the direction transverse to the collision, there is 0 probability of a collision event. And the collision probability is proportional to the cross section divided by the area of the wavepacket in the transverse direction.

Allow me to be a bit handwavy with the exact numbers and try to get the concept right (for example is a gaussian's "width" its standard deviation or it's 90% confidence interval or what... I don't care)

However, we typically aren't talking about two particles with tailored wavepackets moving toward eachother (to my knowledge... actually no experiment works this way). Rather, we might be talking about $$N$$ particles confined in a beam whose cross sectional area has been measured as $$A$$. Each individual particle is somewhere within that beam, and it's wavefunction is at least confined to be smaller than $$A$$ in the transverse direction. Let's say this beam is shot at a single particle target, then the probability of a single scattering event is $$\sigma N/A$$ ($$\sigma$$ is the cross section calculated with quantum mechanics).

In the end, the particulars of each of the $$N$$ particles' wavefunctions don't matter. Suppose for example that the particles in the beam are all identical overlapping gaussian wavepackets of cross sectional area $$A$$, then we trivially get the answer. Suppose on the contrary that our particles have tiny wavepackets of area $$a$$ located in random positions throughout the beam (this is... not the case). Then the probability that one particle's wavefunction overlaps with the target particle is $$Na/A$$, and if there is an overlap the cross section is $$\sigma/a$$. So we're left with the same result $$N\sigma/A$$.

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 $$dN$$.

To get a rate due to a physical process whose rate is proportional to a flux, you have to multiply that flux by an area. That's all there is to it.