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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?

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1 Answer 1

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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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 Meerts 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
    
@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. –  Ron Maimon Mar 22 '12 at 2:47
    
See this –  Nogueira Dec 23 at 18:18

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