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3

I) Let us for clarity use a subscript "$S$" (and "$H$") to denote the Schrödinger (Heisenberg) picture, where bras and kets evolve (are unchanged) and operators are unchanged (evolve), respectively. Moreover, let us assume that the two pictures coincide at the instant $t_0$ (which Ref. 1 assumes is $t_0=0$). II) Recall first of all the possibly confusing ...

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As the OP asks to try to avoid Schrödinger equation, I think that it could be worthwhile to exhibit some variation of my answer from the similar post http://physics.stackexchange.com/a/202298/1335: The great thing of the exponential measure in Feynman path integral is that when we evaluate the probability it transforms to a sort of derivative of the action, ...

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At a time when my hammer was the Dirac's delta distribution, I conjectured that the answer was Feynman Integral is a generalization of a Dirac's delta, the use of this delta being to find the extreme of the action. Given a function $f(x)$, find a Dirac measure $\delta_f$ concentrated in the critical points of $f$. The answer is obviously $\delta(f'(x))$, ...

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If you know the propagator, ie. $\langle x'|e^{itH}|x\rangle\,,$ then you could differentiate with respect to time at $t=0$ to get $\langle x'|H|x\rangle\,.$ From this we have, using the resolution of the identity, $H|x\rangle=\int_{-\infty}^\infty dx'\, |x'\rangle\langle x'|H|x\rangle\,,$ from which we have $V(x)|x\rangle=\int_{-\infty}^\infty \, ... 2 The propagators themselves are not indicative for the form of the Lagrangian. They only provide information regarding the nature of the field - e.g. scalar / fermion / vector boson, etc (gravity metric?). Things that allude what the Lagrangian looks like are vertices / interactions. As a simple example: if you have a theory of field$\phi$with a 4-prong ... 0 I have a hunch that it might not be possible in the general case. Since you integrated over the fields already, it would be similar to trying to find the original integral from a real number. Also the basic path integral$Z[0] =1$no matter the field, for instance. 0 There are several ways to see why the path integral's phase is built in such a way. That phase was first discovered by Dirac, who simply calculated the propagator of a particle and noted that the amplitude was$\propto e^{\frac{i}{\hbar}S_{cl}}\$. You can also show the path integral formalism from the Schrödinger formalism, simply with the identity ...

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