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When you compute the path integral of the relativistic free particle action, it's turns out to be the same as the Green's function of a classical field. This co-incidence is huge because it derives, using free-particle physics, things like the Coulomb potential, which is interaction physics.

Is there a deep reason behind this co-incidence?

Ryder Rude
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1 Answers1

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  1. That the path integral kernel is a Green's function of its associated equation of motion is a general fact, see this question and its answers and/or the Feynman/Hibbs book on path integrals.

  2. There is no "interaction physics" in the free propagator. For instance, the fact that the Fourier transformation of a massless bosonic field produces something that looks like a Coulomb potential doesn't mean anything - the Fourier transform of the propagator does not magically produce interactions where there was none.

    It is the actual interaction terms in an action that produce the connection between the Fourier transform of the propagator and the interaction potential at tree-level (see also this answer of mine). Without the interaction terms you don't have the tree-level diagram that tells you that the interaction potential to first order is the Fourier transform of the free propagator of the carrier particle.

  3. Of course, the fact remains: To first order, the classical interaction potential associated with a force mediated by a certain field is essentially the free propagator of that field. It's a consequence of the tree-level diagram of a minimally coupled interaction being just the propagator of the carrier particle with the two interacting particles attached as external legs.

ACuriousMind
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  • I think I understand how the non-relativistic path integral gives a green's function. It's because it gives the propagator satisfying the Schrodinger equation. But the relativistic path integral is a very different beast, because there are no "upper time" and "lower time" limits on the paths. So I'm not sure if the same arguments from non relativistic QM apply – Ryder Rude Aug 31 '22 at 11:48
  • About your third point, I think it's addressing the co-incidence in my other question, the co-incidence between classical and quantum fields. Is this point related to the correspondence principle between classical and quantum theories? – Ryder Rude Aug 31 '22 at 11:51
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    @RyderRude I don't know what you mean. My third point just explains why the propagator/Green's function is related to the interaction potential. If you mean something else by the propagator "deriving the Coloumb potential" in this question, you'll have to be more specific. – ACuriousMind Aug 31 '22 at 12:06
  • In non relativistic QM, the path integral is calculated for all paths constrained between two time points. This is because the QM propagator/Green's function has the interpretation of "propagating wavefunctions forward from one time to another". But the Klein Gordon Green's function has no such interpretation, nor does the relativistic path integral only consider paths constrained between two time points. This is why I think the arguments from non-relativistic QM don't apply here. – Ryder Rude Aug 31 '22 at 12:27