Is there a "zeroth-order" effect in quantum field theory (QED) that two electrons move from two fixed spacetime points to two other fixed spacetime points without the presence of virtual photons? Of course these spacetime points must be "fairly close" because when they are far apart then it seems obvious that the electrons don't feel eachother's presence so no photons are involved but the electrons travel instead independently from their initial points to their final points in all possible ways. But it their initial points and final points are not too far separated in space is there a zeroth order feynman diagram, i.e., two disconnected lines (representing all possible non-interacting paths)?
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3$\begingroup$ How do electrons feel each other's presence without photons? $\endgroup$– Cosmas ZachosApr 18, 2021 at 0:39
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2$\begingroup$ Say you did the experiment that you describe in your question. You prepare two electrons, have them propagate and detect them at their final positions. How would you tell whether they have interacted or not while they were propagating? It is physically impossible to tell, which is why your question isn't well-posed. $\endgroup$– Prof. LegolasovApr 18, 2021 at 1:48
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$\begingroup$ @Prof.Legolasov You could tell they didn't interact when their initial impulses (preparable) were the same as their final impulses (measurable). $\endgroup$– Deschele SchilderApr 18, 2021 at 6:48
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1$\begingroup$ @DescheleSchilder oh but what if they exchanged two virtual photons and the effects of the two exchanges canceled each other such that the final momentum happens to be the same as the initial momentum? $\endgroup$– Prof. LegolasovApr 18, 2021 at 6:54
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1$\begingroup$ @DescheleSchilder over all possible diagrams. Some of them don’t contain the interaction, some do. Your question tries to single out the non interacting diagram, but that’s not how QFT actually works — quantum fields evolve according to the sum of all possible diagrams, a single diagram has no physical meaning $\endgroup$– Prof. LegolasovApr 18, 2021 at 7:12
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1 Answer
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This question appears to be based on the interpretation of internal Feynman propagators as particles. However, the graphs are representations of terms of a perturbation expansion and only the external legs are particles. Therefore the question actually means, are there cases where the Coulomb interaction between two electrons can be ignored. Of course there are, if the electrons are far enough apart at their closest encounter.
In conclusion and to be clear, there is zero probability that two electrons do not interact. However if they are are far enough apart, their interaction may be neglected.
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$\begingroup$ As far as you may take them, they will still interact. It may be very small and not measurable but interact they will. $\endgroup$– OбжорoвApr 18, 2021 at 12:11
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2$\begingroup$ fair enough. The OP on the other hand seems to say ask about no interaction at all. A strange question coming from him, if you see how active he is and on what subjects. $\endgroup$– OбжорoвApr 18, 2021 at 12:15
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$\begingroup$ @Oбжорoв I can remember doing calculations with spinors, propagators, vertex factors, etc. but I can't remeber calculating the zeroth order conteribution. It was already enough to consider mosly a second order calculations. Third order maybe once. $\endgroup$ Apr 18, 2021 at 12:55
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$\begingroup$ Doesn't each diagram involve taking all possible ways (paths) with one (for first order diagram), two (for second order diagrams), or n interactions (for n-th order diagrams) between an initial state and a final state (in the spacetime representation)? You can also work in momentum space (in which I did the calculations) and take only initial momenta and final momenta. So can't you consider just two external lines? Which are actually just two non-interacting straight lines from initial to final momenta (to which the highest contribution is given by particles traveling straight to the endpoint? $\endgroup$ Apr 18, 2021 at 13:11