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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

$$K(\mathbf{x''},t;\mathbf{x'},t_0)=0\tag{2.5.13/2.6.13}$$

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 $t<t_0$.

Qmechanic
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William Huang
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    ...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). – ACuriousMind Mar 10 '16 at 18:40
  • I have computed and plug the propagator in the equation, but the result I got was 0, not $-i\hbar\delta ^3(\mathbf{x''}-\mathbf{x'})\delta (t-t_0)$. – William Huang Mar 10 '16 at 18:47
  • 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 – ACuriousMind Mar 10 '16 at 18:50
  • 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. – William Huang Mar 10 '16 at 18:57
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    Possible duplicates: http://physics.stackexchange.com/q/20797/2451 , http://physics.stackexchange.com/q/22639/2451 , http://physics.stackexchange.com/q/65489/2451 and links therein. – Qmechanic Mar 10 '16 at 19:13

1 Answers1

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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}

Frobenius
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  • But where does $\delta ^3(\mathbf{x''}-\mathbf{x'})$ come from? – William Huang Mar 10 '16 at 20:19
  • @William Huang $\dfrac{\partial \widetilde{K}}{\partial t}=\dfrac{\partial (\theta K) }{\partial t}=\theta;\dfrac{\partial K}{\partial t}+K;\dfrac{\partial \theta }{\partial t}$

    $K;\dfrac{\partial \theta }{\partial t}=K(\mathbf{x''},t ; \boldsymbol{;} ;\mathbf{x'},t_{0}) \delta (t-t_{0})=K(\mathbf{x''},t_{0} ; \boldsymbol{;} ;\mathbf{x'},t_{0}) \delta (t-t_{0})=\delta^{3}(\mathbf{x''}-\mathbf{x'})\delta (t-t_{0})$

     – Frobenius Mar 10 '16 at 20:32
  • 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 $t<t_0$ ? – William Huang Mar 10 '16 at 20:36
  • Yes. See related, and check the elementary explicit examples. – Cosmas Zachos Jan 15 '17 at 16:06