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The expression for the first-order Born approximation to the scattering amplitude goes like $$ f(\theta) \propto \int d^3 r e^{-i ( k - k_i)r} V(r) $$ and so in particular the value of $ V(r) $ everywhere is relevant. However, if we want to consider for example scatters at $ \theta \sim 0 $ it would seem that the only relevant parts of the potential are those far away from the origin. Classically, particles with high impact parameters scatter at $ \theta \sim 0, $ and it does not "see" the potential at the origin at all during its orbit. Perhaps this is because the Born approximation is only approximate, but isn't it effective in the regime where $ V$ is small, which is certainly when $ r$ is big?

I am looking for a more physical/intuitive explanation of this difference between classical and quantum scattering rather than a mathematical rederivation of the Born approximation (unless of course there is a very helpful derivation).

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The key difference between treatments of quantum mechanical scattering and classical scattering is the nature of the incident particle. In the quantum mechanical case, the incident particle is typically treated as a delocalized plane wave, whereas in the classical case the particle is treated as a point particle. The delocalized nature of the plane wave is what leads to an integral over all space.

Of course, the quantum treatment with a delocalized plane wave is just a choice of incident wave function that is useful. In particular, because the Schrodinger equation is linear, we can represent some more localized incident particle as a superposition of plane waves and simply take the appropriate linear combination of $f(\theta, \phi)$ to find the corresponding $f$.

Edit out of moral compunction: While I said "simply", the actual calculation will not be so simple, and one will also need to account for time-dependent phase factors if one takes superpositions of plane waves with different energies.

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  • $\begingroup$ When you say take the appropriate linear combination of $ f(\theta, \phi) $ to find the corresponding $ f, $ does this mean that the original scattering amplitude is not the correct one to use for example the interaction cross section of a particle beam (a localized stream) on a target? $\endgroup$ – jesseylin Jul 23 at 18:50
  • $\begingroup$ That's an interesting question about whether the original amplitude is appropriate for a particle beam, and it ties into your original question about similarities and differences in classical and quantum scattering. You may wish to ask another question emphasizing that part of your question (treating a beam of particles as a plane wave), or to edit your question emphasizing that and uncheck-mark me, as I'm less confident about that aspect. $\endgroup$ – user196574 Jul 24 at 23:10
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The answer, given by @user196574 is exshaustive, I only want to add, that integration over all space arises solely due to the choice of basis solution - a plane wave, with a fixed momentum, which exists everywhere in space. If you want to consider a scattering of localized object, for example, a Gaussian wavepacket, then performing first integral over momenta $k$: $$ \int d^3 k \int d^3 r \ e^{-i (k - k_0) r} V(r) e^{-\alpha k^2 / 2} = \ \int d^3 r \ V(r) e^{-(r-r_0)^2 / 2 \alpha} $$ you have an expression, localized in space (decaying fast with the increase of distance).

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