I realized that I do not really understand the way we define and calculate scattering cross-sections starting from transition probabilities in QM. I will explain my reasoning in a moment but the TLDR of my question is:
- How can scattering cross-sections be defined to be finite quantities if particles of definite momenta aren't localized?;
- Why does the usual way we calculate them work?.
In classical mechanics
I'll briefly go through how I think the scattering cross-section is defined in classical mechanics. We assume to have some potential $V(\vec r)$ and to throw at it particles with a fixed momentum but uniformly distributed in space. What we are describing is a stream of particles of constant volumetric number density $n$ and velocity $v$, giving rise to a flux of particles $J = nv\; [1/TL^2]$. Depending on the impact parameter ($b$) a thrown particle will take some trajectory and fly away in some direction. We can then choose a portion of solid angle and count how many particles fly in that direction per unit time; we denote this quantity with $f\; [1/T]$. Thanks to the fact that $J$ is known we can relate any number of particles per unit time to an area, the scattering cross section $\sigma = f/J\;[L^2]$. $\sigma$ is finite, although infinitely many particles are thrown to the target, because particles are finite and the bigger the impact parameter is the smaller is the interaction felt. A nice property of $\sigma$ is that, despite its definition, is independent of $J$ ($f$ is actually proportional to $J$). So $\sigma$ is only a function of the impact velocity and the shape of $V(\vec r)$.
In quantum mechanics
Now, we want to define and calculate scattering cross-sections in the context of quantum mechanics. We immediately have to face the problem that particles of definite momenta aren't localized. As a consequence, talking about impact parameters doesn't make sense and every one of the infinitely many particles we throw for each unit of time has the same probability of scattering towards the solid angle we selected. This makes $f$ infinite and therefore $\sigma$ as well. Also, the concept of the particle coming from far away is meaningless if the particle's wave-function permeates all of space. So, it's not clear how, starting from transition probabilities, we can define a scattering cross-section independent of $J$, finite and measurable (POINT 1). Note that this problem goes away if we only want to predict branching ratios (the infinities cancel), but this is not the case${}^*$.
I practically know how scattering cross-sections are calculated: they assume a finite universe of volume $\mathcal{V}$ (or periodic boundary conditions which also apply to the potential $V$) so that they can normalize plane waves, they calculate transition probabilities in first order perturbation theory $\text{d}P = W\text{d}t$, they set $\sigma = W/J$ and $J = v/\mathcal{V}$. Since both the initial and final waves are normalized with $1/\sqrt{\mathcal{V}}$ the dependency of $J$ from $\mathcal{V}$ cancels with that of $W$ and $\sigma$ is independent of it. If we take this procedure as a definition for $\sigma$, it's not clear how this scattering cross-section is related to experiments and reduce to the original definition in the classical limit (POINT 2).
${}^*$There's also a time-independent way of doing scattering in QM but, as far as I know, it allows to predict branching ratios but not scattering cross-sections (I'm probably wrong about this but I've never seen it being used to find $\sigma$).
Edit (my personal attempt)
Unable to find an explanation that I can understand, I rolled up my sleeves and gave a shot at reinventing scattering theory (notice the arrogance here). Confident that I could tackle the problem in a general setting first and do approximations only later when they would became necessary, I considered a generic wave packet $\vert\vec x_0, \vec p_0\rangle$. This ket is meant to represent a particle whose position and momentum, albeit not fixed, are centered around $\vec x_0$ and $\vec p_0$, with small enough uncertainties. By defining $\vert\Psi_0\rangle = \vert \vec 0, \vec p_0 \rangle$ the following identities should follow. $$ \langle \vec x \vert \vec x_0, \vec p_0\rangle = \langle \vec x - \vec x_0 \vert\Psi_0\rangle \tag{1} $$ $$ \langle \vec p \vert\vec x_0, \vec p_0 \rangle = \langle \vec p \vert\Psi_0\rangle e^{-ipx_0/\hbar} =: \tilde\Psi_0(\vec p) e^{-i\vec p\vec x_0/\hbar} \tag{2} $$ Let's say that a wave packet $\vert\vec x_0, \vec p_0\rangle$ is generated at time $t_0$ by my source of particles. After that, the wave packet travels towards the target and scatters. At time $t$ the state vector is $U(t, t_0)\vert\vec x_0, \vec p_0\rangle$. We apply Born's rule to determine the probabilities related to the measurement of the momentum of this particle. $$ p_\text{single}(\vec p; \vec x_0, \vec p_0) = \left\vert \langle \vec p \vert U(t, t_0) \vert\vec x_0, \vec p_0\rangle \right\vert^2 = \left\vert \int \text d^3 q\; \tilde\Psi_0(\vec q) e^{-i\vec q\vec x_0/\hbar} \langle \vec p \vert U(t, t_0) \vert\vec q\rangle \right\vert^2 \tag{3} $$ At this point, I considered a generic source of particles, that shoots particles with $\vec p_0 = (0,0,p_{0z})$ and $\vec x_0 = (x_0, y_0, 0)$. The particles-source is characterized by $J(x,y)$ which describes the density of particles generated at $(x,y,0)$ per unit time; so that its integral over the xy-plane gives the number of particles produced per unit time. With all of this definitions a transition rate $W(\vec q; \vec p_0)$ is easily introduced. $$ W(\vec q; \vec p_0) = \int \text d x_0 \text d y_0J(x_0, y_0)\; p_\text{single}(\vec p; \vec x_0, \vec p_0) \tag{4} $$ $$ W(\vec q; \vec p_0) = \int \text d x_0 \text d y_0J(x_0, y_0)\;\left\vert \int \text d^3 q\; \tilde\Psi_0(\vec q) e^{-i\vec q\vec x_0/\hbar} \langle \vec p \vert U(t, t_0) \vert\vec q\rangle \right\vert^2 \tag{5} $$
Now, if the interaction distance is small compared to the distance over which the beam density starts to fade, we can regard $J(x, y)$ as being constant $J(x, y) \equiv J_0$. Under this approximation, we fall back to the case of a universe uniformly filled with particles, in which case the scattering cross-section can be defined as $\sigma=W/J_0$ making it independent of $J_0$, as desired. $$ \boxed{ \sigma(\vec q; \vec p_0) = \int \text d x_0 \text d y_0\;\left\vert \int \text d^3 q\; \tilde\Psi_0(\vec q) e^{-i\vec q\vec x_0/\hbar} \langle \vec p \vert U(t, t_0) \vert\vec q\rangle \right\vert^2} \tag{6} $$ Urrah!... not so fast. Hoping I didn't make any error in the calculations, there's still one problem: the scattering cross-section is not meant to be dependent on the shape of the wave packets. One may think this is an easy fix: just send $\tilde\Psi_0(\vec q)$ to $\delta^{(3)}(\vec q - \vec p_0)$ but doing so makes $\sigma$ go to infinity, as predicted by all the above reasoning. It's clear: either I say $J$ is constant or I say that the particles have definite momentum, both assumptions together cannot give a finite answer.
I don't know how to fix this problem: the definition of $\sigma$ must be changed, but how?
