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If $L = i \hbar \hat{H} - \dfrac{d}{dt}$, then $ L \psi(x,t) = 0$ is the Schrodinger equation.

It is well known that we can solve the Schrodinger equation with initial condition $\psi(x,0) = f(x)$ with the solution $\psi(x,t) = \int d x_i K(x, t; x_i 0) f(x_i)$, where we the kernel is defined as $ K(x_f, t_f; x_i, t_i) := \langle x_f | U(t_f, t_i) | x_i \rangle$.

It is also almost always in texts that we can relate the kernel to a Green's function of $L$ by stating $G(x_f, t_f; x_i, t_i) = \Theta(t_f - t_i) K(x_f, t_f; x_i, t_i)$. A Green's function has the nice property that $L G(x_f, t_f; x_i, t_i) = i \hbar \delta(x_f- x_i) \delta(t_f - t_i)$. Moreover, it is "fundamental solution" to $L$ in the sense that if we seek a solution of the inhomogeneous $L \psi(x,t) = s(x)$, we can provide a solution by using a Green's function according to $\psi(x,t) = \int G(x,t; x', 0) s(x') dx'$.

My question (finally) is: why does it particularly matter that the propagator is related to the Green's function for the Schrodinger equation? While it is cool that the solution to the homogeneous equation with initial condition looks very similar to the solution to the homogeneous equation with source term equal to the other PDE's initial condition, how is that helpful to us? Duhamel's principle is frequently cited as an answer, but looking through the Wikipedia page seems to suggest the Duhamel's principle lets us solve the inhomogeneous problem in terms of the initial conditions problem, which would seem to be the opposite of useful in terms of solving the Schrodinger equation for physical systems. Is there any explicit way or theorem that would allow our knowledge that the propagator is (almost) the Green's function to give us the solution to the initial condition problem?

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    $\begingroup$ The two are equivalent formulations of the same problem, of utility in different contexts. $\endgroup$ Commented Apr 9, 2019 at 21:32
  • $\begingroup$ Few use the Green's function for the TDSE, but all do in relativistic QFT, where relativistic covariance is manifest. $\endgroup$ Commented Apr 9, 2019 at 21:49
  • $\begingroup$ Virtual duplicate. $\endgroup$ Commented Apr 9, 2019 at 22:03
  • $\begingroup$ Thank you @CosmasZachos. With respect to the equivalence-namely that $L(K) = 0$ while $L(\Theta K) = \delta$ - surely this is a property only of the particular operator $L$ associated with the Schrodinger equation and not one that deals with PDEz in general? $\endgroup$ Commented Apr 9, 2019 at 22:05
  • $\begingroup$ Basically yes, but one could think of exceptions that prove the rule... $\endgroup$ Commented Apr 9, 2019 at 22:25

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