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In a Feynman diagram, can we consider that the propagator specifying the transition amplitude of a particle (let's say, of a "real" electron, or of a "virtual" photon) between two points or two vertices, is in fact itself the sum of a multiplicity of probability amplitudes, each one corresponding to the contribution, in this sum, of a certain trajectory or world line?

Without going so far as to say that the electron effectively follows "all the world lines at once" -- a phrase which would be at best an interpretation, at worst a gross error of understanding -- it seems to me legitimate to assert that the particle does not follow a well-determined trajectory ; and that, as such, the symbol of a very straight and rigid line for fermions in diagrams could well mislead the beginner in quantum physics (which I am, by the way).

In this respect, the idea of a sum over all paths seems to me to be quite respectful of Heisenberg's principle ("no definite trajectory", which can also be translated as: let us consider all possible trajectories in the initial and final conditions, without privileging any of them); and to draw a well-determined line is not more than a convenience of representation. But on balance, it would perhaps be better to redraw each of these straight lines as one of the "space-time tubes", by means of which we represent the integral of the paths.

What do you think about this? Thanks in advance!

Husserliana
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    To answer this would require an hour long lecture (or two). Please take our [tour] and refer to the [help] for our guidelines. We have a one question per post policy. If there's a way of distilling the question into separate fractions which might be posted in their own threads, that would work. – Jiminy Cricket. Oct 30 '22 at 09:34
  • Also, see https://physics.stackexchange.com/q/230113/50583 and its linked questions for already extant discussion of the nature of virtual particles – ACuriousMind Oct 30 '22 at 11:29
  • My apologies, I'm going to rephrase (condense) my question! – Husserliana Oct 30 '22 at 11:58
  • Is your new question effectively "Would it be better to draw Feynman diagrams using tubes?" I honestly can't tell if there's a question here that has a well-defined answer. – Michael Seifert Oct 30 '22 at 13:19
  • @Michael Seifert Well, the main question would still be this: wouldn't the internal or external lines of the diagrams correspond, each time, to a superposition of different trajectories rather than to well determined trajectories? And this, even if these drawings of rigid lines (for fermions at least) seem to evoke a specific, well determined trajectory-- a potentially misleading image, therefore. – Husserliana Oct 30 '22 at 13:41
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    That's all true. But I also have the impression that it's fairly well-known that Feynman diagrams shouldn't be taken literally. They're really just a device to help us keep track of the various integrals that need to be done in a cross-section calculation at a particular perturbative order. What's more, the lines in a Feynman diagram don't even really correspond to spatial trajectories; they correspond to particular modes in Fourier space. – Michael Seifert Oct 30 '22 at 13:46
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    (You can draw position space Feynman diagrams, in which the vertices correspond to positions that get integrated over. Quite literally, you sum over all paths. There’s nothing wrong with that interpretation, it’s a good one) – QCD_IS_GOOD Oct 30 '22 at 14:34
  • Thank you very much already for your answers!

    @Joshua Lin Yes, that's what I was thinking of! I understand that it is often more convenient in QED to construct your diagrams and calculations in momentum space; but that nothing prevents, in principle, a construction in positions space. I had read a text by E. Witten -- well above my level, but at one point there was some mention of such a construction, and the author suggested that there would then be an analogy with integration over paths.

     – Husserliana Oct 30 '22 at 15:37
  • I just found the excerpt : "Now let us look at a typical Feynman diagram. Instead of evaluating this diagram in momentum space, as is customary, let us think of it in coordinate space (...). According to the usual rules for computing an amplitude from the Feynman diagram, each line corresponds to a propagator. With the representation (1.4.2) for the propagator, each line in the figure represents an integration over the trajectory of a particle that propagated in space-time between the indicated points." – Husserliana Oct 30 '22 at 15:50
  • (1.4.2) being "the well-known path integral formula : ∫_0^∞▒dτ ∫_x^y▒Dx(t)exp{-1/4 ∫_0^τ▒〖dt(〖 x〗^2 ) ̇ 〗} " – Husserliana Oct 30 '22 at 15:52

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