I was following Quantum Markov Processes from the book The Theory of Open Quantum Systems by Breuer and Petruccione. In the section The Markovian Quantum Master Equation they proceeds to 'construct the most general form for the generator $\mathcal{L}$ of a quantum dynamical semigroup.' Then I observed that almost the same approach is taken in the book Quantum Dynamical Semigroups and Applications by Alicki and Lendi, so I moved there.
We have the definition of $\mathcal{L}$: $$\frac{d}{dt}\rho_t=\mathcal{L}\rho_t$$ with $\rho_t=\Lambda_t\rho$. And this gives $\Lambda_t=e^{\mathcal{L}t}, t\ge0.$
For unbounded $\mathcal{L}$ they have used the following definition of exponential – $$e^{\mathcal{L}t}\rho=\lim\limits_{n \to \infty}\left(1-\frac{t}{n}\mathcal{L}\right)^{-n}\rho.$$ After that,
We find now a general form of $\mathcal{L}$ in the case of finite-dimensional
Hilbert space $\mathcal{H_S}$ (dim $\mathcal{H_S} = N$). Introducing a linear basis {$F_k$}, $k =$
$0,1,… ,N^2 − 1$ in $\mathcal{B(H_S)}$ such that $F_0 = \mathbb{1}$ we may write a
time-dependent version of $\Lambda\rho=\sum_{\alpha} W_{\alpha}\rho W_{\alpha}^*$ as follows, $$\Lambda_t\rho=\sum_{k,l=0}^{N^2-1}C_{kl}(t)F_k\rho F_l^*,$$ where $C_{kl}(t)$ is a positive-definite matrix.
The part after this is what I'm totally in the dark with. Let me paste the photo as it will be too much effort to type:
How are we getting this expression for $\mathcal{L}\rho$ with all these limits, $\epsilon$'s and all that? What I got is the following: $$\mathcal{L}\rho=\frac{d}{dt}\rho_t=\frac{d}{dt}{[\Lambda_t\rho]}=\frac{d}{dt}\left(\sum_{k,l=0}^{N^2-1}C_{kl}(t)F_k\rho F_l^*\right)$$ but can proceed no further.
Best Answer
You have to be careful with the notation, $\rho = \rho_0$ is the initial state here. The left hand side of your last equation should be $\mathcal L \rho_t$.
We thus have to evaluate the equation at $t=0$, obtaining $$ \mathcal L \rho = \frac{d}{dt} \Bigl( \sum_{kl} C_{kl}(t) F_k \rho F_l^\dagger \Bigr)_{t=0} . $$
The result now follows from $$ \left. \frac{d}{dt} f(t) \right|_{t=0} = \lim_{\epsilon \to 0} \frac{1}{\epsilon} \bigl( f(\epsilon) - f(0) \bigr) , $$ that is, $\mathcal L \rho = \lim_{\epsilon \to 0} \frac{1}{\epsilon} \bigl( \Lambda_\epsilon \rho - \rho \bigr)$.
There is a typo in Alicki / Lendi, the sums in the final result are all supposed to start at $1$ instead of zero. The same calculation can be found for example in the proof of Thm. 5.1 in the Rivas / Huelga book, I think it is a bit easier to understand there.