The propagator being the Green's function of the Euler-Lagrange operator corresponding to the Lagrangian of some QFT, should not depend on the interaction term. But shouldn't the probability amplitude depend on the coupling strength of the Lagrangian?
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$\begingroup$ possible duplicates: physics.stackexchange.com/q/215776/84967, physics.stackexchange.com/q/238560 $\endgroup$– AccidentalFourierTransformCommented Mar 25, 2020 at 21:22
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2 Answers
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FWIW, the bare/free propagator $G_0$ only depends on the free part of the action, while the full/exact propagator $G$ also depends on the interaction part of the action.
answered Mar 25, 2020 at 14:49
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The probability amplitudes will of course depend on the interactions, and the object you might need to calculate the probability amplitudes is called the "fully corrected propagator". It's realized as a sum of Feynman diagrams at all orders of perturbation theory. But this isn't a Green's function for the full Euler-Lagrange operator with interactions (which is nonlinear, so it's already unclear how you would derive and use a bona fide Green's function), because EL equations represent classical equations of motion and the object called the "fully corrected propagator" is a quantum mechanical construction.
A particle's propagator, e.g. your favorite $\frac{1}{p^2-m^2+i\varepsilon}$, is $\textit{by definition}$ the 2-point correlator for a free theory describing the particle. In this case, since the E-L equations are linear, this really is a Green's function in the most traditional sense (Fourier transform what I gave you, apply the Klein-Gordon operator on it, and you get a delta function).
