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?
Best Answer
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.