# Trouble with operator-sum representation of a quantum operation

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 $$$$\mathcal{E}(\rho_{A})=\text{Tr}_{B}\big(U(\rho_{A}\otimes |0_{B}\rangle\langle 0_{B}|)U^{\dagger}\big)$$$$ where $$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$$ as $$$$\begin{split} \mathcal{E}(\rho_{A})&=\text{Tr}_{B}\big(U(\rho_{A}\otimes |0_{B}\rangle\langle 0_{B}|)U^{\dagger}\big) &=\sum_{i}{\langle i_{B}|\big(U(\rho_{A}\otimes |0_{B}\rangle\langle 0_{B}|)U^{\dagger}\big)|i_{B}\rangle}\overset{*}{=}\sum_{i}{E_{i}\rho_{A}E^{\dagger}_{i}} \end{split}$$$$ where the last equality (*) is nothing but the operation-sum representation of the operation $$\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 $$$$|i_{B}\rangle\to \mathbb{I}^{A}\otimes |i_{B}\rangle$$$$ whereby $$$$\mathcal{E}(\rho^{A})=\sum_{i}{\mathbb{I}^{A}\otimes \langle i_{B}|\big(U(\rho_{A}\otimes |0_{B}\rangle\langle 0_{B}|)U^{\dagger}\big)\mathbb{I}^{A}\otimes |i_{B}\rangle}$$$$ as this is how I understand the partial trace. I would believe that, say, $$U^{\dagger}(\mathbb{I}^{A}\otimes |k_{B}\rangle)$$ is well-defined as I see dimensions of both operators (matrices) being just right. Then, if my understanding of the partial trace is correct, I would have $$$$E_{k}=\langle k_{B}|U|0_{B}\rangle\to (\mathbb{I}^{A}\otimes \langle k_{B}|)U(\mathbb{I}^{A}\otimes |0_{B}\rangle)$$$$ which is an expression where one can at least see that an $$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 $$$$\rho_{A}\otimes |0_{B}\rangle\langle 0_{B}| = \underbrace{(\rho_{A}\otimes \mathbb{I}_{B})(\mathbb{I}_{A}\otimes |0_{B}\rangle)}_{=(\mathbb{I}_{A}\otimes |0_{B}\rangle)\rho_{A}}(\mathbb{I}_{A}\otimes \langle 0_{B}|)=(\mathbb{I}_{A}\otimes |0_{B}\rangle)\rho_{A}(\mathbb{I}_{A}\otimes \langle 0_{B}|)\hspace{1mm},$$$$ where the last equality can be seen through $$$$\begin{split} (\rho_{A}\otimes \mathbb{I}_{B})(\mathbb{I}_{A}\otimes |0_{B}\rangle)&=(\underbrace{\rho_{A}\mathbb{I}_{A}}_{\text{commutes}})\otimes (\underbrace{\mathbb{I}_{B}|0_{B}\rangle}_{=|0_{B}\rangle=|0_{B}\rangle\cdot 1})\\[2mm] &=(\mathbb{I}_{A}\rho_{A})\otimes (|0_{B}\rangle\cdot 1)\\[2mm] &=(\mathbb{I}_{A}\otimes |0_{B}\rangle)(\underbrace{\rho_{A}\otimes 1}_{=\rho_{A}})\\[2mm] &=(\mathbb{I}_{A}\otimes |0_{B}\rangle)\rho_{A}\hspace{1mm}. \end{split}$$$$ Note the usage of the so-called mixed product property, i.e., $$(A\otimes B)(C\otimes D)=(AC)\otimes(BD)$$.

Then, considering the $$k$$:th term in the partial trace sum, we have that $$$$\begin{split} & \mathbb{I}_{A}\otimes \langle k_{B}|\big(U(\rho_{A}\otimes |0_{B}\rangle\langle 0_{B}|)U^{\dagger}\big)\mathbb{I}_{A}\otimes |k_{B}\rangle\\[2mm] =& \underbrace{(\mathbb{I}_{A}\otimes \langle k_{B}|)U(\mathbb{I}_{A}\otimes |0_{B}\rangle)}_{=E_{k}}\rho_{A}\underbrace{(\mathbb{I}_{A}\otimes \langle 0_{B}|)U^{\dagger}(\mathbb{I}_{A}\otimes |k_{B}\rangle)}_{=E^{\dagger}_{k}} \\[2mm] =& E_{k}\rho_{A} E^{\dagger}_{k} \hspace{1mm}. \end{split}$$$$

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.

• There is no "Nielsen's and Chuang's 10th ed", there is a "10th Anniversary Edition" which is something quite different. – wondering Feb 28 at 13:30

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 $$\begin{split} \sum_{i}{\langle b_{i}|\big(U(\rho_{A}\otimes |0\rangle\langle 0|)U^{\dagger}\big)|b_{i}\rangle}\overset{*}{=}\sum_{i}{E_{i}\rho_{A}E^{\dagger}_{i}}\ . \end{split}$$ 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 $$\begin{split} \mbox{LHS}=\sum_{i}{\langle b_{i}|U (\mathbb I\otimes|0\rangle) \rho_A (\mathbb I\otimes \langle 0|) U^{\dagger}|b_{i}\rangle} =\sum_{i}{E_{i}\rho_{A}E^{\dagger}_{i}}\ , \end{split}$$ 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 $(A\otimes B)(C\otimes D)=(AB)\otimes(CD)$. Also note that we can regard $|0\rangle$ as a ($d\times 1$) 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}

• Thank you for asserting me of my suspicions about the partial trace, and also for pointing out the typo (fixed now). Alas, it does not answer my question as much as it points to that I'm on the right track. Given that I write out the partial trace with one tensor argument as the identity on A, I still cannot grasp the equality marked ( * ). That said, starting from the LHS of equality ( * ) I cannot see how to form the elements E_k and arrive at the RHS of equality ( * ). – Invoker Aug 24 '16 at 11:56
• @Invoker I have added a discussion of what I believe is what you are looking for. Let me know if it answers your question. – Norbert Schuch Aug 24 '16 at 12:14
• I believe your response does answer my question. See my edit to my question where you can tell wether I have caught on or not. Thank you very much. – Invoker Aug 24 '16 at 14:05
• @Invoker Looks got. Though there was a mistake in my argument (which you took over): When moving $\rho_A\otimes \mathbb I$ through $\mathbb I\otimes |0\rangle$, it turns to $\rho_A$ (since the $B$ system is gone). – Norbert Schuch Aug 24 '16 at 14:34
• Ok, I see. This actually changes things. First, on a closer inspection one can see that it's wrong to say that $\rho_{A}\otimes \mathbb{I}_{B}$ and $\mathbb{I}_{A}\otimes |0_{B}\rangle$ commute, as this would imply that $\mathbb{I}_{B}|0_{B}\rangle=|0_{B}\rangle\mathbb{I}_{B}$. As such, it is not commutation which is used here, but something else as to allow for the equality $(\rho_{A}\otimes \mathbb{I}_{B})(\mathbb{I}_{A}\otimes |0_{B}\rangle)=(\mathbb{I}_{A}\otimes |0_{B}\rangle)\rho_{A}$. Still, I can't see how ''$\rho_{A}\otimes\mathbb{I}$ turns to $\rho_{A}$.'' Can you explain? – Invoker Aug 24 '16 at 15:15

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

$$\begin{split} &=(\mathbb{I}_{A}\rho_{A})\otimes (|0_{B}\rangle\cdot 1)\\[2mm] &=(\mathbb{I}_{A}\otimes |0_{B}\rangle)(\underbrace{\rho_{A}\otimes 1}_{=\rho_{A}})\\[2mm] \end{split},$$ 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

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

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

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

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.

• Norbert Schuch's answer is fine. You can't use the mixed product property to justify that step, because the mixed product property (at least in the form your reference has it) keeps all the dimensions fixed, and Schuch's answer uses manipulations where the dimensions change. But they're perfectly valid manipulations. – Peter Shor Jul 10 '18 at 0:17