For a time-homogeneous diffusion process $\{X_t\}$, its infinitesimal generator is defined by $$\mathcal{A}f(x)=\lim_{t\to0^+}\frac{\mathbb{E}^x[f(X_t)]-f(x)}{t}=\frac{\partial}{\partial t}|_{t=0^+}\mathbb{E}^xf(X_t)$$.
On the other hand, if $\{X_t\}$ is a Markov process which has transition density function $p(t,x,y)$ and stationary distribution $\pi(x)$, we have $$\mathcal{A}\pi(x)=\lim_{t\to0^+}\frac{1}{t}[\int p(t,x,y)\pi(y)\text{d}y-\pi(x)]=0$$, so $\pi(x)$ can be solved from a differential equation(s) $Af\equiv 0$.
Example 1(Brownian motion)
$$\text{d}X_t=\text{d}B_t, p_1(t,x,y)=\frac{1}{\sqrt{2\pi t}}\exp(-\frac{(x-y)^2}{2t})$$ $$\mathcal{A}=\frac12\Delta=\frac12\frac{\text{d}^2}{\text{d}x^2}, \frac{\partial}{\partial t}p_1(t,x,y)=\frac12\frac{\partial^2}{\partial x^2}p_1(t,x,y)$$
Example 2(Ornstein-Uhlenbeck process)
$$\text{d}X_t=-\mu X_t\text{d}t+\sigma\text{d}B_t,p_2(t,x,y)=p_1(\frac{\sigma^2}{2\mu}(1-e^{-2\mu t}),xe^{-\mu t},y)$$ $$\mathcal{A}=-\mu x\frac{\text{d}}{\text{d}x}+\frac{\sigma^2}{2}\frac{\text{d}^2}{\text{d}x^2}, \frac{\partial}{\partial t}p_2(t,x,y)=(-\mu x\frac{\partial}{\partial x}+\frac{\sigma^2}{2}\frac{\partial^2}{\partial x^2})p_2(t,x,y)$$
Noticing these examples, I guess that for 1-dimensional situation, i.e. $x,y\in\mathbb{R}$, it holds that $$\frac{\partial}{\partial t}p(t,x,y)=\mathcal{A}p(t,x,y)$$, where the right hand side is to be interpreted as $\mathcal{A}$ applied to the function $x\mapsto p(t,x,y)$. Conversely, if there exists a diffential operator $$Q=\sum_j q_j(x)\frac{\text{d}^j}{\text{d}x^j}$$ such that $$\frac{\partial}{\partial t}p(t,x,y)=Qp(t,x,y)=\sum_j q_j(x)\frac{\partial^j}{\partial x^j}p(t,x,y)$$, where $Q$ operates on variable $x$, then one must have $$\mathcal{A}f(x)=Qf(x)\quad\forall f\in\mathcal{D_A}$$. Is this proposition correct? Could it be extended to general cases?
My thoughts: $\mathcal{A}f(x)=\int (Qp(t,x,y))f(y)\text{d}y$ while $Qf(x)=\int p(t,x,y)Qf(x)\text{d}y$, so $\mathcal{A}=Q$ may be explained from $pQ=Qp$(as some special operators), which reminds me of the Kolmogorov's equations $P(t)Q=P'(t)=QP(t)$ about Markov processes taking discrete values.
I'm not sure if it is the same as Kolmogorov's forward equation, which could be written as $$\frac{\partial}{\partial t}p(t,x,y)=\mathcal{A}^\ast p(t,x,y)$$, where $\mathcal{A}^\ast$ operates on the variable $y$, not $x$, and is defined as the adjoint of $\mathcal{A}$.
Update: From wiki I found that Kolmogorov's backward equation is nearly what I need. Wikipedia says that for the SDE: $$\text{d}X_t=\mu(X_t,t)\text{d}t+\sigma(X_t,t)\text{d}W_t$$, the Kolmogorov's backward equation is $$-\frac{\partial}{\partial t}\bar p(x,t)=\mu(x,t)\frac{\partial}{\partial x}\bar p(x,t)+\frac12\sigma^2(x,t)\frac{\partial^2}{\partial x^2}\bar p(x,t)=\mathcal{A}_t \bar p(x,t)$$, where $\bar p(\cdot,t)$ gives the probability distribution of the system at time $t$.
So I think the key point is to state the connection between $\bar p(x,t)$ and $p(t,x,y)$.
