I am reading about the operator-sum representation of quantum operations in Nielsen's and Chuang's 10th Anniversary ed of Quantum Computation and Quantum Information (N&C). I have become quite confused by some of the basic formalism presented therein. I'll get right to it.
N&C gives the definition of a quantum operation $\mathcal{E}$, for an input system $A$ in an initial state $\rho_{A}$ and environment $B$ in an initial pure state $|0_{B}\rangle\langle 0_{B}|$, as \begin{equation} \mathcal{E}(\rho_{A})=\text{Tr}_{B}\big(U(\rho_{A}\otimes |0_{B}\rangle\langle 0_{B}|)U^{\dagger}\big) \end{equation} where $U$ is some transformation on the whole system. N&C then proceeds toward the operator-sum representation by expressing the partial trace $\text{Tr}_{B}$ using a finite orthonormal basis $\{|i_{B}\rangle\}$ for $B$ as \begin{equation} \begin{split} \mathcal{E}(\rho_{A})&=\text{Tr}_{B}\big(U(\rho_{A}\otimes |0_{B}\rangle\langle 0_{B}|)U^{\dagger}\big) &=\sum_{i}{\langle i_{B}|\big(U(\rho_{A}\otimes |0_{B}\rangle\langle 0_{B}|)U^{\dagger}\big)|i_{B}\rangle}\overset{*}{=}\sum_{i}{E_{i}\rho_{A}E^{\dagger}_{i}} \end{split} \end{equation} where the last equality (*) is nothing but the operation-sum representation of the operation $\mathcal{E}$. The elements $\{E_{i}\}$ of this representation are written $E_{k}\equiv \langle k_{B}|U|0_{B}\rangle$ and these are operators.
What I am unable to see is the rationale/reasoning for the equality (*) above. This is a problem to me and is what I would like to solve.
Why I am having trouble is perhaps because of the partial trace. I can't see how it is legitimate to write out the partial trace by using only the basis $\{|i_{B}\rangle\}$ as above, the dimensions do not seem to be in order as to make a meaningful expression. I would rather change to \begin{equation} |i_{B}\rangle\to \mathbb{I}^{A}\otimes |i_{B}\rangle \end{equation} whereby \begin{equation} \mathcal{E}(\rho^{A})=\sum_{i}{\mathbb{I}^{A}\otimes \langle i_{B}|\big(U(\rho_{A}\otimes |0_{B}\rangle\langle 0_{B}|)U^{\dagger}\big)\mathbb{I}^{A}\otimes |i_{B}\rangle} \end{equation} as this is how I understand the partial trace. I would believe that, say, $U^{\dagger}(\mathbb{I}^{A}\otimes |k_{B}\rangle)$ is well-defined as I see dimensions of both operators (matrices) being just right. Then, if my understanding of the partial trace is correct, I would have \begin{equation} E_{k}=\langle k_{B}|U|0_{B}\rangle\to (\mathbb{I}^{A}\otimes \langle k_{B}|)U(\mathbb{I}^{A}\otimes |0_{B}\rangle) \end{equation} which is an expression where one can at least see that an $E_{k}$ is not a scalar, as opposed to how it is written by N&C.
Now, if the transformation $U$ was not present I would have no problem taking the partial trace this way. As this is not the case, I'm stuck. I have no idea as how to rearrange within the terms in the sum so that I get the elements $\{E_{i}\}$.
Any effort to help with this matter is greatly appreciated.
Edit: With the input from Norbert Schuch I have come to understand the following. The trick allowing for suitably rearranging within each term of the partial trace lies in expanding $\rho_{A}\otimes |0_{B}\rangle\langle 0_{B}|$ into products as \begin{equation} \rho_{A}\otimes |0_{B}\rangle\langle 0_{B}| = \underbrace{(\rho_{A}\otimes \mathbb{I}_{B})(\mathbb{I}_{A}\otimes |0_{B}\rangle)}_{=(\mathbb{I}_{A}\otimes |0_{B}\rangle)\rho_{A}}(\mathbb{I}_{A}\otimes \langle 0_{B}|)=(\mathbb{I}_{A}\otimes |0_{B}\rangle)\rho_{A}(\mathbb{I}_{A}\otimes \langle 0_{B}|)\hspace{1mm}, \end{equation} where the last equality can be seen through \begin{equation} \begin{split} (\rho_{A}\otimes \mathbb{I}_{B})(\mathbb{I}_{A}\otimes |0_{B}\rangle)&=(\underbrace{\rho_{A}\mathbb{I}_{A}}_{\text{commutes}})\otimes (\underbrace{\mathbb{I}_{B}|0_{B}\rangle}_{=|0_{B}\rangle=|0_{B}\rangle\cdot 1})\\[2mm] &=(\mathbb{I}_{A}\rho_{A})\otimes (|0_{B}\rangle\cdot 1)\\[2mm] &=(\mathbb{I}_{A}\otimes |0_{B}\rangle)(\underbrace{\rho_{A}\otimes 1}_{=\rho_{A}})\\[2mm] &=(\mathbb{I}_{A}\otimes |0_{B}\rangle)\rho_{A}\hspace{1mm}. \end{split} \end{equation} Note the usage of the so-called mixed product property, i.e., $(A\otimes B)(C\otimes D)=(AC)\otimes(BD)$.
Then, considering the $k$:th term in the partial trace sum, we have that \begin{equation} \begin{split} & \mathbb{I}_{A}\otimes \langle k_{B}|\big(U(\rho_{A}\otimes |0_{B}\rangle\langle 0_{B}|)U^{\dagger}\big)\mathbb{I}_{A}\otimes |k_{B}\rangle\\[2mm] =& \underbrace{(\mathbb{I}_{A}\otimes \langle k_{B}|)U(\mathbb{I}_{A}\otimes |0_{B}\rangle)}_{=E_{k}}\rho_{A}\underbrace{(\mathbb{I}_{A}\otimes \langle 0_{B}|)U^{\dagger}(\mathbb{I}_{A}\otimes |k_{B}\rangle)}_{=E^{\dagger}_{k}} \\[2mm] =& E_{k}\rho_{A} E^{\dagger}_{k} \hspace{1mm}. \end{split} \end{equation}
I am pleased by Norbert's input on this matter and as far as my perspective goes I consider my question to be solved. Thank you very much.