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  1. Is there a use of advanced Green's functions? If yes then when or in which context?

  2. Why in quantum field theory, we always use Feynman's prescription for finding the propagator and not the retarded one?

EDIT: 3. Why the Feynman's prescription of propagator does not make sense in a classical field theory?

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  1. The full Green's function of an equation like the Klein-Gordon equation is the difference of the retarded and advanced Green's functions. It is only when the equation in question is an equation involving time that we often discard the advanced anti-causal part to get physically sensible solutions. In QFT, the full Green's function appears for example as the expectation value of the commutator of a scalar field with itself.

  2. The Feynman propagator is what corresponds to the expectation value of the time ordered product of a field with itself, and it is this time ordered expectation value that appears in the LSZ reduction formula for scattering amplitudes.

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  • $\begingroup$ What does "full" Green's function mean? $\endgroup$ – DanielSank May 19 '15 at 16:04
  • $\begingroup$ @DanielSank: The "full" Green's function is the actual Green's function (i.e. the function which gives the Dirac delta when plugged into the equation). The advanced and retarded Green's functions are just the "positive" and "negative" parts of it (the parts that depends on $t >0$ and $t <0$, in a sense). $\endgroup$ – ACuriousMind May 19 '15 at 16:28
  • $\begingroup$ Yeah, but aren't the different Green's functions really just the differences in the response when you have different boundary conditions? For example, in a lossy system (I'm thinking of circuits but this is true for other systems) you want the retarded function because it's the one where the disturbance emenates out from a source and doesn't come back. $\endgroup$ – DanielSank May 19 '15 at 16:41
  • $\begingroup$ @DanielSank: True. I call the mathematical definition of the Green's function the "full" Green's function because it is an abstract property of the equation. Whether you have to use this full version or the retarded version or even possibly something else is a physical choice not inherent to the equation. $\endgroup$ – ACuriousMind May 19 '15 at 16:45
  • $\begingroup$ I think I'm trying to point out that an e.g. differential equation without boundary conditions is ambiguous. $\endgroup$ – DanielSank May 19 '15 at 16:47
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David Tong notes in his lecture notes that the advanced propagator "is useful if we know the end point of a field configuration and want to figure out where it came from".

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