# How exactly is the propagator a Green's function for the Schrodinger equation

Sakurai mentions that the propagator is a Green's function for the Schrodinger equation because it solves $$\left(H-i\hbar\frac{\partial}{\partial t}\right)K(x,t,x_0,t_0) = -i\hbar\delta^3(x-x_0)\delta(t-t_0)$$

I don't see that. First of all, I don't understand where the $-i\hbar$ 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?

• 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. – KF Gauss May 12 '17 at 13:40
• 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. – Kasper May 13 '17 at 17:31
• Related question – Noix07 May 23 '19 at 14:28

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.
• 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. – Kasper Mar 20 '12 at 18:07
• 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. – josh Mar 20 '12 at 18:52
An quick answer is that the usual Green's function is written like $\left(\frac{\partial}{\partial t}-\frac{H}{i\hbar}\right)K(x,t,x_0,t_0)=\delta^3(x-x_0)\delta(t-t_0)$, as you can check it here: https://en.wikipedia.org/wiki/Green%27s_function.