Derivation of a quantum dynamical map on open quantum system Let us consider a initial total quantum system as $\rho(0) = \rho_S(0)\otimes\rho_B$ where $\rho_S(0)$ is initial open system and $\rho_B$ is density matrix for environment. 
We can use partial trace to get open system after some time, $\rho_S(t) = \mathrm{tr}_B \{ U(t, 0) [\rho_S(0) \otimes \rho_B] U^{\dagger}(t, 0) \}$
This will be a dynamical map $V(t): S(\mathcal{H}_S) \mapsto S(\mathcal{H}_S)$ where $\rho_S(t) =V(t)\rho_S(0)$. And given $\rho_B=\sum_\alpha \lambda_\alpha  | \phi_\alpha > <\phi_\alpha|$ .
The question is, I am not sure how in detail get the following 
$V(t)\rho_S = \sum_{\alpha, \beta} W_{\alpha, \beta}(t) \rho_S W_{\alpha, \beta}^{\dagger}(t)^{}$ where $W_{\alpha, \beta}(t) = \sqrt{\lambda_\beta} <\phi_\alpha|U(t, 0)|\phi_\beta>$.
I tried to put $\rho_B$ into the $\rho_S(t)$ as 
$\rho_S(t) = \mathrm{tr}_B \{ U(t, 0) [\rho_S(0) \otimes (\sum_\alpha \lambda_\alpha  | \phi_\alpha > <\phi_\alpha|)] U^{\dagger}(t, 0) \}$
But how to proceed from here ?
 A: You are on the right track. Just use the definition of the partial trace, which is 
$$\mathrm{tr}_B(A) = \sum_{\alpha}\langle\phi_{\alpha}\rvert A \lvert \phi_{\alpha}\rangle$$
for any operator $A$, where the $\lvert \phi_{\alpha}\rangle$ constitute a complete orthonormal basis for the environment Hilbert space. Using this in your final expression leads directly to the required result. 
A: Usually when investigating the time evolution of a subsystem one finds that
a memory term enters the formalism. I do not see such a term in your
approach.
A standard procedure is to use the Zwanzig projector formalism
\begin{eqnarray*}
\rho (0) &=&\rho _{S}(0)\otimes \rho _{B} \\
\rho (t) &=&U(t,0)\rho (0)U^{\dagger }(t,0) \\
\partial _{t}\rho (t) &=&-i[H,\rho (t)]=-iL\rho (t)
\end{eqnarray*}
Projector
\begin{eqnarray*}
Pf &=&\rho _{B}\mathrm{tr}_{B}f\Rightarrow \;P^{2}f=Pf,\;Q=1-P \\
P\rho (0) &=&\rho _{B}\mathrm{tr}_{B}\rho _{S}(0)\otimes \rho _{B}=\rho
(0),\;Q\rho (0)=0
\end{eqnarray*}
Zwanzig (http://en.wikipedia.org/wiki/Zwanzig_projection_operator)
\begin{eqnarray*}
\partial _{t}P\rho (t) &=&-iPLP\rho (t)-iPLQ\rho (t) \\
\partial _{t}Q\rho (t) &=&-iQLQ\rho (t)-iQLP\rho (t) \\
Q\rho (t) &=&\exp [-iQLQt]Q\rho (0)-i\int_{0}^{t}ds\exp [-iQLQ(t-s)]QLP\rho
(s) \\
\partial _{t}P\rho (t) &=&-iPLP\rho (t)-\int_{0}^{t}dsPLQ\exp
[-iQLQ(t-s)]QLP\rho (s)
\end{eqnarray*}
After Laplace transforming
\begin{eqnarray*}
\hat{f}(z) &=&\int_{0}^{\infty }dt\exp [izt]f(t),\;{Im}z>0, \\
\int_{0}^{\infty }dt\exp [izt]\partial _{t}f(t) &=&-f(0)-iz\hat{f}(z)
\end{eqnarray*}
$$
-P\rho (0)-izP\hat{\rho}(z)=-iPLP\hat{\rho}(z)-PLQ[z-QLQ]^{-1}QLP\hat{\rho}%
(z)
$$
Another way to obtain this is first to take the Laplace transform of $\rho
(t)=\exp [-iLt]\rho (0)$ and then use the Feshbach projection formula.
