# Propagator as a Green's function in non-relativistic QM

I have a propagator

$$K=\sum_k\langle x|a_k\rangle\langle a_k| y\rangle \exp\left\{ \frac{-iE_k(t-t_0)}{\hbar} \right\} ~~,\tag{1}$$

which I know satisfies the time-dependent Schrodinger equation

\begin{align} i\hbar\dfrac{\partial}{\partial t}K=\hat HK~~. \end{align}

I am only interested in the retarded propagator so I multiply it by the Heaviside function $$\Theta(t-t_0)$$. This enforces the condition that the probability density at time $$t_0$$ will diffuse only into the future. Therefore, I solve

\begin{align} i\hbar\dfrac{\partial}{\partial t}\Theta K&=\hat H\Theta K\\ i\hbar\left(K\dfrac{\partial}{\partial t} \Theta + \Theta\dfrac{\partial}{\partial t} K \right)&=\hat H\Theta K\\ i\hbar\, K\delta(t-t_0) &=\hat H\Theta K- i\hbar\,\Theta\dfrac{\partial}{\partial t} K ~~. \end{align}

Since the $$\delta$$-function picks out $$t=t_0$$, I can simplify $$K$$ on the LHS using (1) to write

\begin{align} \lim\limits_{t\to t_0}K&=\sum_k\langle x|a_k\rangle\langle a_k| y\rangle \exp\left\{ \frac{-iE_k(t_0-t_0)}{\hbar} \right\} \\ &=\sum_k\langle x|a_k\rangle\langle a_k| y\rangle \\ &= \langle x| y\rangle \\ &= \delta^{(3)}(\vec x-\vec y) ~~, \end{align}

which yields

\begin{align} i\hbar \,\delta(t-t_0) \delta^{(3)}(\vec x-\vec y) =\Theta \left(\hat H- i\hbar\dfrac{\partial}{\partial t}\right)K ~~. \end{align}

Now I see that $$K$$ is the Green's function for the given linear operator. However, in Sakurai's Modern QM, 3rd Ed. (p110), he has a minus sign on the $$i\hbar$$ on the LHS. Where does it come from?

This question is similar but does not explain where the minus sign comes from.

The problem should be the equation \begin{align} i\hbar\dfrac{\partial}{\partial t}\Theta K&=\hat H\Theta K \end{align}
Indeed if you start from \begin{align} i\hbar\dfrac{\partial}{\partial t}K&=\hat H K \end{align} and multiply $$\Theta$$ on both side you have
\begin{align} i\hbar\Theta\dfrac{\partial}{\partial t}K &= \Theta\hat{H}K \\ i\hbar\dfrac{\partial}{\partial t}(\Theta K)-i\hbar\dfrac{\partial}{\partial t}(\Theta) K &=\hat{H}(\Theta K) \\ -i\hbar \delta^{(3)}(\vec{x}-\vec{y})\delta(t-t_0)=(\hat{H}-i\hbar \dfrac{\partial}{\partial t})(\Theta K) \end{align}
Below is the result from the book Quantum field theory for the gifted Amateur Ch16 and $$G^+$$ is itself the retarded propagator which should be consistent with my argument. • Excellent. I was wrong to assume that $\Theta K$ would also be a solution to the Schrodinger equation. Thanks, you are a cool guy! Dec 18, 2020 at 3:46
• Actually, let me ask one more question. If the propagator is supposed to be the Green's function, and the Green's function as you have it is $\Theta K$, then that must mean the sign does not matter. Indeed, if the propagator is (like) a wavefunction, then the sign will go away when we square it to obtain some observable. Therefore, I think the equation is correct as I have written it also. Do you agree? Dec 18, 2020 at 4:09
• I agree if we add a minus sign to the propagator then it's like multiplying a trivial phase to the state. $K$ should satisfy the Schrodinger equation so $(H-i\frac{\partial}{\partial t})K=0$. If $\Theta K$ also satisfy then we should also have $(H-i\frac{\partial}{\partial t})(\Theta K)=0$ but instead we have some $\delta$ function which is non-zero only at certain point. I think in your equation the RHS is not operating on $\Theta K$ so it's a bit different and not merely the problem of minus sign. Dec 18, 2020 at 4:38
• So then it's $\Theta K$ that is the Green's function, and not $K$? Dec 18, 2020 at 16:47