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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?

Qmechanic
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Kasper
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2 Answers2

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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.

Community
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josh
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A homogeneous equation with an initial condition $\psi(r,t_0)=\psi(r)$ can be thought of as an inhomogeneous equation with Dirichlet boundary conditions $\psi(\infty)=0$ and a source term $\delta(t-t_0+\epsilon)\psi(r)$, where $\epsilon\to 0$.

We get from the usual property of Green's function \begin{align} \psi(r,t)=& \int dr' dt' K(r,t,r',t') \delta(t'-t_0+\epsilon)\psi(r)\\ =& \int dr' K(r,t,r',t_0) \psi(r,t_0), \end{align} where in the line $K(r,t,r',t_0)$ plays the role of propagator in quantum mechanics.

pathintegral
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