quantum mechanics - Why is the propagator the Green's function for Schrodinger equation?

Sakurai says that the propagator is simply the Green's function for the time-dependent wave equation satisfying

$$\left [ -\frac{\hbar^2}{2m} \triangledown ''^2+V(\mathbf{x''})-ih\frac{\partial }{\partial t}\right ]K(\mathbf{x''},t;\mathbf{x'},t_0)=-i\hbar\delta ^3(\mathbf{x''}-\mathbf{x'})\delta (t-t_0)$$

with the boundary condition

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

for $t

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

Answer

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