Kraus decomposition In the derivation of the Kraus decomposition, it is not clear to me how to get from the LHS to the expression on the RHS:  $$ \text{Tr}_{env}[U(\rho_1\otimes \rho_{env})U^\dagger]=\sum_k\langle k|U|0\rangle \rho \langle 0|U^\dagger|k\rangle, $$ where we substitute $\rho_{env}=|0\rangle\langle 0|$ and using a basis for the environment $\{|k\rangle\}$. Could someone show me a few steps?
 A: As the derivation you are looking for is not that hard, I guess your confusion probably follows from the fact that your notation does not label states as belonging to the Hilbert spaces $\mathcal{H}_A$ and $\mathcal{H}_B$ for the two subsystems $A$ and $B$ forming your joint system. So I am going to denote states out of $\mathcal{H}_A$ or $\mathcal{H}_{B}$ as for example $|\psi\rangle_A$ or $|\phi\rangle_B$ and operators acting on them as, say, $O_A$ or $O_B$. Along the same lines, $|\Psi\rangle_{AB}$ and $O_{AB}$ are a state out of $\mathcal{H}_{AB}=\mathcal{H}_{A}\otimes\mathcal{H}_{B}$ and an operator acting on it, respectively. Now, given that the partial trace operation is defined as $\text{Tr}_{B}[O_{AB}]=\sum_{k} \langle k|_{B} O_{AB} |k\rangle_{B}$:
$$
\begin{align}
\text{Tr}_B[U_{AB}(\rho_A\otimes\rho_B)U_{AB}^{\dagger}] & =\sum_{k} \langle k|_{B} U_{AB}(\rho_A\otimes\rho_B)U_{AB}^{\dagger} |k\rangle_{B} \\ & = \sum_{k} \langle k|_{B} U_{AB}(\rho_A\otimes |0\rangle_{B}\langle0|_{B})U_{AB}^{\dagger} |k\rangle_{B} \\ & = \sum_{k} \langle k|_{B} U_{AB} |0\rangle_{B} \rho_{A} \langle 0|_{B} U_{AB}^{\dagger} |k\rangle_{B}
\end{align} 
$$
where the last step is motivated by $U_{AB}$ being sandwiched between a bra and a ket out of the same Hilbert space $\mathcal{H}_{B}$. To reconcile with your notation, just replace the subscript $B$ with $env$. Is it clearer now?
