Lagrangian from Path Integral Suppose I somehow know propagator for a given quantum mechanical system but I don't happen to know either the Lagrangian or Hamiltonian. (For simplicity, assume that this is non-relativistic.) Is there a procedure by which I can recover the original Lagrangian?
 A: 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 \, |x'\rangle\langle x'|H|x\rangle\,dx'-\frac{p^2}{2m}|x\rangle\,, $ or taking any state $|\psi\rangle\,,$ 
$\displaystyle V(x)=\frac{\int_{-\infty}^\infty \, \langle\psi|x'\rangle\langle x'|H|x\rangle\,dx'-\frac{\Delta}{2m}\langle\psi|x\rangle}{\langle\psi|x\rangle}\,, $ and then $L=T-V\,.$ 
So it seems that it should be possible in principle (I did however make some assumption about time independence of the Hamiltonian in my derivation though, but it would seem to me in this moment that you could work it out without this assumption).
A: 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 vertex, then the Lagrangian (most likely) has $\phi^4$-term, or if you have boson-fermion-antifermion vertex, then there is probably a term $e \, \bar{\psi} {\not}{A} \psi$...
A: 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.
