So I'm having a hard time understanding how the (QM) cross section fits into the general picture, of e.g. collider experiments. So we can calculate cross sections (for one reaction) exactly in QFT with Feynman diagrams (perturbation theory...). When we take the classical limit of these formulas, we find back for example the classical Thomson scattering of electrons on an electromagnetic potential.

To calculate the classical differential scattering (I'll take Thomson scattering as the general example here) we integrate over all possible impact parameters. This means we have a huge incoming UNIFORM flux + our possible targets and we calculate the statistical chance of what the distance is between one of the target particles and the incoming particles ( = impact parameter) so we can classically calculate the trajectory and the weighed possible outcomes, the differential cross section in other words. So this assumes we can't 'aim' our particles at one target particle.

With (relativistic) quantum mechanical calculations we can also get the same Thomson differential scattering, when we take the low velocity limit. So in principle we know we are doing physically relevant things.

Now my question is, when u shoot a (collimated) beam of particles on other target particles, you can calculate the differential cross section from quantum mechanics. These cross sections are calculated for one-on-one particle reactions, like electron scattering on a proton (QFT). But these calculations just 'suppose' a plane wave incoming directly on the center of the proton or the assumed target particles. Then we do something like, "event rate = flux x target particles x cross section". But where did we specify the impact parameter? Where did we say we aimed our incoming particles? In Quantum Mechanics it makes no difference what the 'distance' is between your incoming particle and target particle. The reaction just has 'a chance' of happening. Never ever in perturbation theory (QFT) have I used distances between particles, impact parameters or anything related to that. I know that in QM you can't see this problem classical anymore and we can't speak of exact positions and so on... But there still has to be a lower chance of having a reaction between the "top" of the incoming beam and the target particles at the "bottom". But still these experiments give us the right theoretical predicted results. I just don't see how we include these things into the quantum mechanical calculations of cross sections?

Imagine we just have the scattering of one particle on a target particle? How can we know what impact parameter we have (distance) as this will have an influence on the scattering, or in other words, the reaction chance of happening, or the cross section so to speak? Are we not making huge approximations as to assume we have a plane wave incoming which lands "exactly" in the middle of the (spherical) potential of the target particle? How does this translate to actual experiments? Why is there no 'statistical' treatment of this problem like with the classical variant and how is it that we can do this for 1 particle on 1 particle scattering without keeping in mind the influence of distance-like effects in QFT/QM.

  • $\begingroup$ Interesting question but I don't really think it's about statistics? I suggest maybe rewording the title and if possible making the body of your question a bit more succinct. $\endgroup$ – innisfree Dec 17 '18 at 20:44
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    $\begingroup$ In the derivation of cross section the impact parameter is eliminated in terms of the scattering angle seen in the lab frame. $\endgroup$ – Triatticus Dec 17 '18 at 21:57
  • $\begingroup$ Also in the quantum mechanical derivation? So for example in the cross section for e- p scattering the possibility of an electron passing by 1 km from the proton is also included in the differential cross section? I mean the distance should have any effect on this? I know in the classical part we can eliminate the impact parameter via the scattering angle because the trajectory of the particle is classically determined. $\endgroup$ – CFRedDemon Dec 17 '18 at 22:04
  • $\begingroup$ Possibly related. Huge b corresponds to small θ and QM yields a similar formula. $\endgroup$ – Cosmas Zachos Jun 29 at 20:46

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