Sakurai says (in various editions) that the propagator is simply the Green's function for the time-dependent wave equation satisfying

$$\begin{align}&\left [ -\frac{\hbar^2}{2m} \triangledown ''^2+V(\mathbf{x''})-ih\frac{\partial }{\partial t}\right ]K(\mathbf{x''},t;\mathbf{x'},t_0)\cr =&-i\hbar\delta ^3(\mathbf{x''}-\mathbf{x'})\delta (t-t_0)\end{align}\tag{2.5.12/2.6.12}$$

with the boundary condition


for $t<t_0$

I don't have any idea about where the $-i\hbar\delta ^3(\mathbf{x''}-\mathbf{x'})\delta (t-t_0)$ term comes from, and the propagator must be equal to zero when $t<t_0$.

  • 1
    $\begingroup$ ...and your question is? Just compute the propagator and plug it in there, you'll see it does indeed fulfill the equation (when you take the derivatives that are supposed to yield the distributions in the correct fashion). $\endgroup$
    – ACuriousMind
    Mar 10, 2016 at 18:40
  • $\begingroup$ I have computed and plug the propagator in the equation, but the result I got was 0, not $-i\hbar\delta ^3(\mathbf{x''}-\mathbf{x'})\delta (t-t_0)$. $\endgroup$ Mar 10, 2016 at 18:47
  • $\begingroup$ Yeah, that's a common confusion. Do you know how the $\delta$ comes out for other Green's functions? How the fundamental solution/Green's function gives the delta and not zero is really more a math than a physics question $\endgroup$
    – ACuriousMind
    Mar 10, 2016 at 18:50
  • $\begingroup$ I know given a linear differential operator $\mathfrak{L}$, a Green's function $G(x,s)$ is any solution of $\mathfrak{L}G(x,s)=\delta (x-s)$. But I still don't know the relation between the propagator and the Green's function. $\endgroup$ Mar 10, 2016 at 18:57
  • 2
    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/20797/2451 , physics.stackexchange.com/q/22639/2451 , physics.stackexchange.com/q/65489/2451 and links therein. $\endgroup$
    – Qmechanic
    Mar 10, 2016 at 19:13

1 Answer 1


Hint : Check if this "modified" Schrodinger equation is satisfied by the "modified" propagator \begin{equation} \widetilde{K}(\mathbf{x''},t \; \boldsymbol{;} \;\mathbf{x'},t_{0})=\theta(t-t_{0})\;K(\mathbf{x''},t;\mathbf{x'},t_0) \tag{01} \end{equation} where $\;\theta(t-t_{0})\;$ the unit step function with property \begin{equation} \dfrac{\partial \theta (t-t_{0}) }{\partial t}=\dfrac{d \theta (t-t_{0}) }{d t}=\delta (t-t_{0}) \tag{02} \end{equation}

Note that
\begin{equation} \dfrac{\partial \widetilde{K}}{\partial t}=\dfrac{\partial (\theta K) }{\partial t}=\theta\;\dfrac{\partial K}{\partial t}+K\;\dfrac{\partial \theta }{\partial t} \tag{03} \end{equation} and
\begin{equation} K\;\dfrac{\partial \theta }{\partial t}=K(\mathbf{x''},t \; \boldsymbol{;} \;\mathbf{x'},t_{0}) \delta (t-t_{0})=K(\mathbf{x''},t_{0} \; \boldsymbol{;} \;\mathbf{x'},t_{0}) \delta (t-t_{0})=\delta^{3}(\mathbf{x''}-\mathbf{x'})\delta (t-t_{0}) \tag{04} \end{equation}

  • $\begingroup$ But where does $\delta ^3(\mathbf{x''}-\mathbf{x'})$ come from? $\endgroup$ Mar 10, 2016 at 20:19
  • $\begingroup$ @William Huang $\dfrac{\partial \widetilde{K}}{\partial t}=\dfrac{\partial (\theta K) }{\partial t}=\theta\;\dfrac{\partial K}{\partial t}+K\;\dfrac{\partial \theta }{\partial t}$ $K\;\dfrac{\partial \theta }{\partial t}=K(\mathbf{x''},t \; \boldsymbol{;} \;\mathbf{x'},t_{0}) \delta (t-t_{0})=K(\mathbf{x''},t_{0} \; \boldsymbol{;} \;\mathbf{x'},t_{0}) \delta (t-t_{0})=\delta^{3}(\mathbf{x''}-\mathbf{x'})\delta (t-t_{0})$ $\endgroup$
    – Frobenius
    Mar 10, 2016 at 20:32
  • $\begingroup$ So the "unmodified" propagator satisfied $\left [ -\frac{\hbar^2}{2m} \triangledown ''^2+V(\mathbf{x''})-ih\frac{\partial }{\partial t}\right ]K(\mathbf{x''},t;\mathbf{x'},t_0)=0$ ? And the "unmodified" propagator may be not equal to 0 when $t<t_0$ ? $\endgroup$ Mar 10, 2016 at 20:36
  • $\begingroup$ Yes. See related, and check the elementary explicit examples. $\endgroup$ Jan 15, 2017 at 16:06