A Derivation of the differential Chapman-Kolmogorov Equation

vancouver_water

Summary
I am following the book "The Theory of Open Quantum Systems" by Breuer and Petruccione. I can follow the derivation of the integral CK equation but do not understand their derivation of the differential CK equation.
The integral equation is $$T(x_3,t_3|x_1,t_1)=\int \text{d}x_2T(x_3,t_3|x_2,t_2)T(x_2,t_2|x_1,t_1)$$ where $T(x_3,t_3|x_1,t_1)$ is the probability density of a Markov process taking the value $x_3$ at time $t_3$ given that it took the value of $x_1$ at time $t_1$. So far so good. To derive the differential CK equation, they take the time derivative of the integral equation and the result is $$\frac{\partial}{\partial t}T(x,t|x',t')=A(t)T(x,t|x',t')$$ where $A(t)$ is time translation linear operator defined in terms of a density as $$A(t)\rho(x)=\lim_{\Delta t\rightarrow 0}\frac{1}{\Delta t}\int \text{d}x'[T(x,t+\Delta t|x',t)-\delta(x-x')]\rho(x').$$ I don't understand how to derive that form of $A(t)$. What I would think (since they are differentiating w.r.t. the first time parameter) is that
\begin{align*} \frac{\partial}{\partial t_3}T(x_3,t_3|x_1,t_1) &= \lim_{\Delta t\rightarrow 0}\frac{1}{\Delta t}\int \text{d}x_2[T(x_3,t_3+\Delta t|x_2,t_2)T(x_2,t_2|x_1,t_1)-T(x_3,t_3|x_2,t_2)T(x_2,t_2|x_1,t_1)] \\ &= \lim_{\Delta t\rightarrow 0}\frac{1}{\Delta t}\int \text{d}x_2[T(x_3,t_3+\Delta t|x_2,t_2)-T(x_3,t_3|x_2,t_2)]T(x_2,t_2|x_1,t_1)] \end{align*}
but I cannot get this into the same functional form as what they get. They only have one time parameter in their $A(t)$ but I still have all three time parameters.

What am I missing?

Thanks!

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Stephen Tashi

"Derivation of the differential Chapman-Kolmogorov Equation"

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