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It is possible to derive Coulomb's potential from QED by considering the scattering process $e^-+e^-\rightarrow e^-+e^-$. The argument can be found in Peskin and Schroeder p.125, and it is reproduced in this stackexchange answer. The authors state that it is necessary to assume that the particles are distinguishable, meaning that only the $t$-channel diagram is included, and not the $u$-channel diagram. Why is this the case? Indistinguishability is of course not unique to QFT (i.e. it is a QM effect also) so I don't understand why it would be ignored in this limit. I expect that it has something to do with the fact that one electron is treated as fixed, but I don't quite follow how.

xzd209
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    Remark: if we do this derivation for different (=distinguishable) particles(e.g., an electron and a proton), the question does not arise. If we were to account for electrons being indistingushable, the Coulomb interaction between distinguishable particles would be different from that for the indistiguishable ones. – Roger V. Jul 01 '21 at 11:37
  • Good point! I suppose this then begs the question: why do distinguishable and indistinguishable particles experience the Coulomb interaction in the same way? – xzd209 Jul 01 '21 at 12:00
  • I would say that they experience Coulomb interaction differently, because they are exchangeable. Mathematically the problem is clear: we put the same Coulomb interaction in the Hamiltonian, but for indistinguishable particles we get extra exchange diagrams. So when we work backwards, we ignore these diagrams to get the correct answer. But I cannot give a deeper physical explanation. – Roger V. Jul 01 '21 at 12:03
  • That is very post-hoc though and is certainly not a prediction of QED. – xzd209 Jul 01 '21 at 12:20
  • My background is in solid state - that is I am well familiar with the non-relativistic version of this problem. This is why I am not sure, whether Coulomb potential should be derivable from the scattering cross-section or the other quantity that one calculates. For me what you are trying to do is deconvolving two unrelated effects that contribute to the final result: Coulomb interaction has to do with charge, while exchange has to do with the indistinguishibility. – Roger V. Jul 01 '21 at 12:25
  • I see what you mean. In that case, surely including both diagrams should lead to a potential that includes the effect of the exchange interaction? – xzd209 Jul 01 '21 at 14:15
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    In principle, yes. However, it is not quite clear how to write it mathematiclaly... I could think of electron-hole interaction in semiconductors: holes are just valence electrons with renamed operators, $h_k^\dagger =v_{-}k,h_k=v_{-k}^\dagger$. It might be that the potential is derived somewhere in the books on the theory of excitons. Knox's book could be a place to look at. – Roger V. Jul 01 '21 at 14:30

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