Propagator in Path Integral Quantum Mechanism as Green Function of Schrodinger Equation I'm studying in Ryder's book of QFT. I'm dealing with QM in the path integral approach and he is trying to prove that the propagator $K(x_f t_f;x_i t_i)$ is the Green function of the Schrodinger (S.) equation:
\begin{equation}
\frac{\hbar^2}{2m} \frac{d^2}{dx_f^2}\psi(x_f t_f) + i \hbar \frac{\partial}{\partial t_f} \psi(x_f t_f) =V(x_ft_f) \psi(x_ft_f) \quad (1)
\label{one}
\end{equation}
We have a generic wave function that satisfies S. equation $\psi$ and we can rewrite it as (eq. 5.27)
\begin{equation}
\psi(x_f t_f)= \phi(x_f t_f) - \frac{i}{\hbar}\int dx dt \; K_0(x_f t_f;x t)V(x,t)\psi(xt) \quad (2)
\label{eq:2}
\end{equation}
where $\phi(x_f t_f)$ is a free plane wave that therefore satisfies the free S. eq:
\begin{equation}
\frac{\hbar^2}{2m} \frac{d^2}{dx_f^2}\phi(x_f t_f) + i \hbar \frac{\partial}{\partial t_f} \phi(x_f t_f) =0 \quad (3)
\end{equation}
and $K_0$ is the free propagator. So plugging eq(1) into eq(2) and using eq (3) Ryder derives this 
\begin{equation}
\frac{\hbar^2}{2m} \frac{d^2}{dx_f^2}K_0(x_f t_f;x t)+ i \hbar \frac{\partial}{\partial t_f} K_0(x_f t_f;x t) = \frac{-i}{\hbar} \delta(x_f-x)\delta(t_f-t) 
\end{equation}
My problem is to get to this equation. Plugging (2) into (1) I can get as far as
\begin{equation}
\frac{-i}{\hbar}\int dx dt \; \left[\frac{\hbar^2}{2m} \frac{d^2}{dx_f^2} + i \hbar \frac{\partial}{\partial t_f}\right] ( K_0(x_f t_f;x t)) \cdot V(xt) \psi(xt) = V(x_ft_f) \phi(x_ft_f) - \frac{i}{\hbar} V(x_ft_f) \int dx dt K_0(x_f t_f;x t)V(x,t)\psi(xt) 
\end{equation}
where I used (3) in the left hand side. 
 A: I am sorry I do not have access to Ryder and would be reluctant to read it with you, anyway. I'll just point out that it is aggressively confusing to be calling $K_0$ the propagator, since it solves the inhomogeneous free TDSE, so it is a Green function, normally called G, that the three linked answers take pains to contrast to the propagator, the fundamental solution of the homogeneous equation. 
It might, or might not, help you to note that, if, instead, you assumed it is a Green function,
\begin{equation}
\frac{\hbar^2}{2m} \frac{d^2}{dx_f^2}K_0(x_f t_f;x t)+ i \hbar \frac{\partial}{\partial t_f} K_0(x_f t_f;x t) = i\hbar \delta(x_f-x)\delta(t_f-t) \tag{5}
\end{equation}
then inserting (2) on the left-hand-side of (1), only, by virtue of (3), you obtain directly
\begin{equation}
\left(\frac{\hbar^2}{2m} \frac{d^2}{dx_f^2}  + i \hbar \frac{\partial}{\partial t_f} \right ) \psi(x_f t_f)\\  =  - \frac{i}{\hbar}\int dx dt \;      \left(\frac{\hbar^2}{2m} \frac{d^2}{dx_f^2}  + i \hbar \frac{\partial}{\partial t_f} \right )   K_0(x_f t_f;x t)V(x,t)\psi(xt)
\\ =  \int dx dt \;  \delta(x_f-x)\delta(t_f-t) V(x,t)\psi(xt)=V(x_ft_f) \psi(x_ft_f),  
\end{equation}
so you obtain the right hand side of (1), that is you derive (1).
You might, or might not, choose to work backwards.
