Trouble with operator-sum representation of a quantum operation

Question

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.

Answer 1

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.

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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}

Answer 2

I am not convinced with the mathematics of the current answer of Norbert Schuch, that is why I wrote a more rigorous answer (to convince myself). First I want to explain my problem with the current answer. In the proof this step

\begin{equation} \begin{split} &=(\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] \end{split}, \end{equation} is reasoned with the mixed product property. However when we look at this property http://www.math.uwaterloo.ca/~hwolkowi/henry/reports/kronthesisschaecke04.pdf (KRON 7). To apply it, the dimensions of the operator must fit like for the matrix multiplication. Also the last term seems strange

\begin{equation} \begin{split} (\mathbb{I}_{A}\otimes |0_{B}\rangle)\rho_{A}\hspace{1mm}. \end{split} \end{equation}

What should this represent, when not a short form of writing the original term? Otherwise the dimensions do not fit. It seems like there is some black magic around, which I do not understand.

Therefore I propose the following reasoning using the notation in N&C: First the operator must be in the tensor product of the principal system and environment $U\in\mathcal{H}_{\text{princ}}\otimes\mathcal{H}_{\text{env}}$. This means we decompose the operator for some bases of the corresponding Hilbert space U = $\sum_{ij} c_{ij} A_i\otimes B_j$, where $A_i$ are the bases spanning $\mathcal{H}_{\text{princ}}$ and $B_j$ are the bases spanning $\mathcal{H}_{\text{env}}$. Now we insert it

\begin{equation} \begin{split} &\sum_{k}{\langle e_k |\big(U (\rho\otimes |e_0\rangle\langle e_0|)U^{\dagger}\big)|e_k\rangle}\\ &=\sum_{k}{\langle e_k |\big((\sum_{ij} c_{ij} A_i\otimes B_j) (\rho\otimes |e_0\rangle\langle e_0|)(\sum_{ij} c_{ij} A_i\otimes B_j)^{\dagger})\big)|e_k\rangle}\\ &=\sum_{k}{(I_{\mathcal{H}_{\text{princ}}} \otimes \langle e_k |)\big((\sum_{ij} c_{ij} A_i\otimes B_j) (\rho\otimes |e_0\rangle\langle e_0|)(\sum_{ij} c_{ij} A_i^{\dagger}\otimes B_j^{\dagger})\big)(I_{\mathcal{H}_{\text{princ}}} \otimes |e_k\rangle}\\ &=\sum_{k}{(\sum_{ij} c_{ij} A_i\otimes \langle e_k |B_j) (\rho\otimes |e_0\rangle\langle e_0|)(\sum_{ij} c_{ij} A_i^{\dagger}\otimes B_j^{\dagger}|e_k\rangle)}\\ &=\sum_{k}{\sum_{ijkl} c_{ij}c_{kl} A_i \rho A_k^{\dagger} \otimes \langle e_k |B_j|e_0\rangle \langle e_0|B_j^{\dagger}|e_k\rangle}\\ &=\sum_{k}{\sum_{ijkl} c_{ij}c_{kl} A_i \rho A_k^{\dagger} \langle e_k |B_j|e_0\rangle \langle e_0|B_j^{\dagger}|e_k\rangle}\\ &=\sum_{k}{(\sum_{ij}{c_{ij}A_i \langle e_k|B_j|e_0\rangle})\rho(\sum_{ij}{c_{ij}A_i^{\dagger} \langle e_0|B_j^{\dagger}|e_k\rangle})} \end{split} \end{equation} The forelast step comes, because the tensor product with a scalar is the same as multiplication, which we now also apply on the term $E_k$ \begin{equation} E_k = \langle e_k|U|e_0\rangle = (I_{\mathcal{H}_{\text{princ}}} \otimes \langle e_k|)(\sum_{ij}c_{ij}A_i\otimes B_j) (I_{\mathcal{H}_{\text{princ}}} \otimes |e_0\rangle) \\ = \sum_{ij}c_{ij}A_i\otimes \langle e_k|B_j|e_0\rangle = \sum_{ij}c_{ij}A_i \langle e_k|B_j|e_0\rangle \end{equation} We can see that the above term must be \begin{equation}\sum_{k}{(\sum_{ij}{c_{ij}A_i \langle e_k|B_j|e_0\rangle})\rho(\sum_{ij}{c_{ij}A_i^{\dagger} \langle e_k|B_j^{\dagger}|e_0\rangle})} = \sum_{k}E_k\rho E_k^{\dagger}\end{equation}

Now some comments, not important to the answer: I feel like this equivalence and the new notation for a extended dot product should be more explained in the N&C. It took me quite a time to get this.