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.


1 Answer 1


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. enter image description here

  • $\begingroup$ Excellent. I was wrong to assume that $\Theta K$ would also be a solution to the Schrodinger equation. Thanks, you are a cool guy! $\endgroup$ Commented Dec 18, 2020 at 3:46
  • $\begingroup$ 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? $\endgroup$ Commented Dec 18, 2020 at 4:09
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Dec 18, 2020 at 4:38
  • $\begingroup$ So then it's $\Theta K$ that is the Green's function, and not $K$? $\endgroup$ Commented Dec 18, 2020 at 16:47

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