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.
Best Answer
You are completely right with the first part of your question: What is meant by $|b_i\rangle_B$ is indeed $\mathbb I_A\otimes |b_i\rangle_B$. (Note that omitting identities is quite common, e.g., when writing Hamiltonians!)
Now to your question how to show \begin{equation} \begin{split} \sum_{i}{\langle b_{i}|\big(U(\rho_{A}\otimes |0\rangle\langle 0|)U^{\dagger}\big)|b_{i}\rangle}\overset{*}{=}\sum_{i}{E_{i}\rho_{A}E^{\dagger}_{i}}\ . \end{split} \end{equation} To this end, note that \begin{align} \rho_{A}\otimes |0\rangle\langle 0| &= (\rho_A\otimes \mathbb I)(\mathbb I\otimes|0\rangle)(\mathbb I\otimes \langle 0|) \\ &= (\mathbb I\otimes|0\rangle) \rho_A (\mathbb I\otimes \langle 0|)\ . \end{align} (I elaborate below why this equality holds.) Inserting this on the LHS, we obtain \begin{equation} \begin{split} \mbox{LHS}=\sum_{i}{\langle b_{i}|U (\mathbb I\otimes|0\rangle) \rho_A (\mathbb I\otimes \langle 0|) U^{\dagger}|b_{i}\rangle} =\sum_{i}{E_{i}\rho_{A}E^{\dagger}_{i}}\ , \end{split} \end{equation} as desired.
Appendix: Why is $(\rho_A\otimes \mathbb I)(\mathbb I\otimes|0\rangle)= (\mathbb I\otimes|0\rangle) \rho_A$?
First, note that $(A\otimes B)(C\otimes D)=(AC)\otimes(BD)$. Also note that we can regard $|0\rangle$ as a ($d\times 1$) matrix and $1$ as a ($1 \times 1$) matrix. We thus have \begin{align} (\rho_A\otimes \mathbb I)(\mathbb I\otimes|0\rangle) &= (\rho_A\mathbb I)\otimes(\mathbb I|0\rangle) \\ &= (\mathbb I\rho_A)\otimes(|0\rangle\cdot 1) \\ &= (\mathbb I\otimes |0\rangle)(\rho_A\otimes 1) \\&= (\mathbb I\otimes|0\rangle) \rho_A\ . \end{align}