Differential equation (Greens function) satisfied by the kernel using path integrals I'm reading Feynman and Hibbs, Quantum Mechanics and Path Integrals. How do I show that the kernel
$$\tag{2-25} K(x_2 ,t_2;x_1, t_1)=\int_{x=x_1}^{x=x_2}\mathcal{D}x~ e^{\frac{i}{\hbar}S[2,1]}$$
satisfies the differential equation
$$\tag{4-29} \frac{\partial K(2,1)}{\partial t_2}+\frac{i}{\hbar}H_2K(2,1)=\delta(x_2-x_1)\delta(t_2-t_1)~?$$
I know that the kernel satisfies the Schrodinger equation for $t_2 > t_1$, but how do I show the delta's on the RHS as $t_2 \to t_1$. Firstly, it is clear $$<x_2, t_1|x_1, t_1>=\delta(x_2-x_1),$$ so that $$\lim_{t_2 \to t_1}K(2,1)=\delta (x_2-x_1).$$ But is there any method to prove this w/o using the transition amplitude interpretation i.e. directly from the definition as the integral over exponent of actions? Can I evaluate this integral in the $t_2 \to t_1$ limit, and show it as the delta function.
Secondly, why does the LHS acting on the delta function of $x$, give the RHS. It seems to me that the derivative wrt. time, introduces the delta function of time, and the Hamiltonian leaves it unchanged? This is clearly wrong IMO, for e.g. the free particle where the Laplacian should give a term $\delta^{\prime \prime}(x_2-x_1)$. What happens to this term?
 A: I'd like to offer an answer that is more handwavy but probably closer to the original argument that Feynman had in mind.
Feynman defines the kernel to be zero for $t_2<t_1$, so in effect he is redefining the kernel to be a retarded Green function. Explicitly, this is expressed as:
$$G(x_2,t_2;x_1,t_1)=\theta(t_2-t_1)K(x_2,t_2;x_1,t_1).\tag{A}$$
The following two results from the reference are used:
$$i\hbar\frac{\partial K(2,1)}{\partial t_2}-H_2K(2,1)=0 \quad\textrm{for} \quad t_2>t_1, \tag{1}$$
which is equation (4-25) and
$$K(2,1)\rightarrow \delta(x_2-x_1)\quad\textrm{ as}\quad t_2\rightarrow t_1+0, \tag{2}$$
which is from Problem 4-6.
The derivative of the retarded Green function is,
$$i\hbar \frac{\partial G(x_2,t_2;x_1,t_1)}{\partial t_2}=i\hbar\delta(t_2-t_1)K(x_2,t_2;x_1,t_1)+\theta(t_2-t_1)i\hbar\frac{\partial K(x_2,t_2;x_1,t_1)}{\partial t_2}.$$
This was obtained by directly taking the derivative of eq (A). The first term on the RHS can be rewritten as $i\hbar\delta(t_2-t_1)\delta(x_2-x_1)$ using eq (1). The second term on the RHS can also be rewritten as $\theta(t_2-t_1)H_2 K(x_2,t_2;x_1,t_1)=H_2G(x_2,t_2;x_1,t_1)$ using eq (2). Therefore, the derivative of the Green function is,
$$i\hbar \frac{\partial G(x_2,t_2;x_1,t_1)}{\partial t_2}=i\hbar\delta(t_2-t_1)\delta(x_2-x_1)+H_2 G(x_2,t_2;x_1,t_1).$$
Reorganizing this equation gives us,
$$-\frac{\hbar}{i}\frac{\partial G(x_2,t_2;x_1,t_1)}{\partial t_2}-H_2 G(x_2,t_2;x_1,t_1)=-\frac{\hbar}{i}\delta(t_2-t_1)\delta(x_2-x_1).$$
n.b.  Eq (2) was used to insert the spatial delta function into the partial derivative of the Green function w.r.t. to $t_2$, but notice this is valid only when $t_2$ approaches $t_1$ from the right. This piece of information was ignored in the derivation. 
