This question is a spin-off from this related question: Why does the Born approximation for the scattering amplitude depend on the potential $V$ everywhere in space, unlike classical scattering? This question deals with a broadly similar topic, which is solving scattering problems "locally", i.e., without knowing the potential $V$ everywhere in space, which is possible in the case of classical scattering.
In the textbook derivation of the scattering cross section (e.g., Griffiths QM) we look for solutions to the Schrödinger equation which have the form
$$ \psi \propto e^{ikz}+f( \theta) \frac{e^{ikr}}{r}$$
where $f$ the scattering amplitude has the interpretation that $\lvert f \rvert^2 = \frac{d \sigma}{d \Omega} $. The way I understand cross sections is as they relate to the attenuation of a particle beam, but a particle beam is a localized in a way that a plane wave is not. If wave packets in a particle beam are written as a linear combination of plane waves $\psi_{\text{beam}} = \sum_{k \in K} A_k e^{ikz},$ wouldn't we find the appropriate scattering amplitude to be some $\tilde f = \sum_{k \in K} f_k \neq f$ from solving the Schrödinger equation for each plane wave? Is there some identity like $\frac{d\sigma}{d \Omega} = \lvert \tilde f \rvert^2 = \lvert f \rvert^2$ or $\sigma_{\text{tot}}=\int_0^{\pi} \lvert \tilde f \rvert^2 \sin \theta\, d\theta = \int_0^{\pi} \lvert f \rvert^2 \sin \theta\, d\theta$?
