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Sakurai mentions (in various editions) that the propagator is a Green's function for the Schrodinger equation because it solves $$\begin{align}&\left(H-i\hbar\frac{\partial}{\partial t}\right)K(x,t,x_0,t_0) \cr= &-i\hbar\delta^3(x-x_0)\delta(t-t_0).\end{align}\tag{2.5.12/2.6.12}$$

I don't see that. First of all, I don't understand where the $-i\hbar$ Dirac delta source term comes from.

And if I recall correctly, a Green's function is used to solve inhomogeneous linear equations, yet Schrodinger's equation is homogeneous $$\left(H-i\hbar\frac{\partial}{\partial t}\right)\psi(x,t) = 0,$$ i.e. there is no forcing term. I do understand that the propagator can be used to solve the wave function from initial conditions (and boundary values). Doesn't that make it a kernel? And what does Sakurai's identity mean?

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    $\begingroup$ This is a really old question, but ultimately the answer for why you use the inhomogeneous Green function for the homogeneous case is something known to mathematicians as Duhamel's principle. This is a generic method for Partial Differential Equations, and is the same idea behind Huygen's principle. $\endgroup$
    – KF Gauss
    May 12, 2017 at 13:40
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    $\begingroup$ I'm not sure, Duhamel's principle seems to be about using the solution of an initial value problem to solve the inhomogeneous problem, whilst this is the exact opposite problem, how to solve a homogeneous initial value problem with a fundamental solution for the inhomogeneous problem. It sure is an old question, but looking at some similar questions, I'm now satisfied that the extra terms appear just because of the time ordering, although the given answer here does not explain that very well. $\endgroup$
    – Kasper
    May 13, 2017 at 17:31
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    $\begingroup$ Related question $\endgroup$
    – Noix07
    May 23, 2019 at 14:28

1 Answer 1

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Check out my answer to a related question here. Notice that the next equation Sakurai gives makes the key difference. $$ K({\mathbf x}^{\prime\prime},t; {\mathbf x}^\prime, t_0) = 0 \quad\text{ for } t<t_0\,. $$ This is implicitly the $\Theta(t - t_0)$ function imposing time ordering that I mention. It makes the difference between the Dirac $\delta$ driving terms and not. It also explains the coefficient you ask about. $$ -i\hbar\,\frac{d}{dt}\,\Theta(t - t_0) = -i\hbar\,\delta(t - t_0) $$ The $\delta$-function on the spatial points comes from the answer I linked to.

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  • $\begingroup$ Thank you, I see somewhat clearer now. Yet I still don't understand how and why this procedure solves the Schrodinger equation. If I write $$\psi(x'',t'') = \int d^3x' K(x'',t'';x',t')\psi(x',t')$$ and apply the Schrodinger operator at both sides, I don't get exactly zero as I would assume. I just don't understand why a Green's function, which is used to find a inhomogeneous solution to a linear operator, can be used to solve an initial value problem. $\endgroup$
    – Kasper
    Mar 20, 2012 at 18:07
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    $\begingroup$ You won't solve it this way. You will get the residual term from integrating over the $\delta$ functions. To solve the Schroedinger equation, you would not put the time ordering condition. Then convoling the initial state $\psi(x^\prime,t)$ with $K$ gives the solution at later or earlier times. $\endgroup$
    – josh
    Mar 20, 2012 at 18:52
  • $\begingroup$ @Kasper Meerts: You do get exactly zero: the correct domain of integration is over one time slice only, so that t' is 0 in your integral, and all other integrals are unchanged. This is easiest to make inutitive by solving a SHO by the exact Green's function. $\endgroup$
    – Ron Maimon
    Mar 22, 2012 at 2:47
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    $\begingroup$ See this $\endgroup$
    – Nogueira
    Dec 23, 2014 at 18:18

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