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In Page 365, Operator-sum representation, Chapter 8, Quantum Computation and Quantum Information by Nielsen and Chuang, it is given that


Given a trace-preserving quantum operation expressed in the operator-sum representation, $\mathcal{E}(ρ)=\sum_k E_kρE_k^\dagger$, we can construct a physical model for it in the following way. From $(8.10)$, we want $U$ to satisfy $$ E_k=\langle e_k|U|e_0\rangle $$ where $U$ is some unitary operator, and $|e_k\rangle$ are orthonormal basis vectors for the environment system. Such a $U$ is conveniently represented as the block matrix $$ U=\begin{bmatrix} [E_1] & \cdots & \cdots \\\\ [E_2] & \cdots & \cdots \\\\ \vdots & \ddots & \vdots \\ \end{bmatrix} $$ in the basis $|e_k\rangle$. Note that the operation elements $E_k$ only determine the first block column of this matrix (unlike elsewhere, here it is convenient to have the first label of the states be the environment, and the second, the principal system). Determination of the rest of the matrix is left up to us; we simply choose the entries such that U is unitary. By the results of Chapter 4, $U$ can be implemented by a quantum circuit.


We have a principal system $Q$ and an environment $E$ and $U$ is a unitary operator acting on the combined system $QE$. \begin{align} \mathcal{E}(\rho)&=tr_E\bigg(U(\rho\otimes|e_0\rangle\langle e_0|)U^\dagger\bigg)\\ &=\sum_k(I\otimes\langle e_k|)(U(\rho\otimes|e_0\rangle\langle e_0|)U^\dagger)(I\otimes |e_k\rangle)\\ &=\sum_k\langle e_k|(U(\rho\otimes|e_0\rangle\langle e_0|)U^\dagger)|e_k\rangle\\ &=\sum_k(I\otimes\langle e_k|)U(I\otimes|e_0\rangle)\rho(I\otimes\langle e_0|)U^\dagger(I\otimes|e_k\rangle)\\ &=\sum_k\langle e_k|U|e_0\rangle\rho\langle e_0|U^\dagger|e_k\rangle\\ &=\sum_k E_k \rho E_k^\dagger \end{align} where $E_k=(I\otimes\langle e_k|)U(I\otimes|e_0\rangle)=\langle e_k|U|e_0\rangle$ is an operator on the state space of the principal system $Q$, $|e_k\rangle$ are the orthonormal basis vectors of the environment, $|e_0\rangle$ be the initial state of the environment.

Please check Prove that $tr\Big(\sum_k E_k^\dagger E_k\rho\Big)=1$ for all $\rho$ implies $\sum_k E_k^\dagger E_k=I$ for the derivation.

In the problem, $E_k=(I\otimes\langle e_k|)U(I\otimes|e_0\rangle)=\langle e_k|U|e_0\rangle$ where $|e_k\rangle$ are some orthonormal basis vectors and $|e_0\rangle$ be the initial state of the environment $E$.

How do we say that $U$ is represented as the block matrix $$ U=\begin{bmatrix} [E_1] & \cdots & \cdots\\ [E_2] & \cdots & \cdots\\ [E_3] & \cdots & \cdots\\ \vdots & \vdots & \vdots \end{bmatrix} $$ in the basis $|e_k\rangle$ ?

My observation:

Given an operator $A:V\to V$ $$ A=\begin{bmatrix}a_{00}&a_{01}\\a_{10}&a_{11}\end{bmatrix}=\begin{bmatrix}\langle0|A|0\rangle&\langle0|A|1\rangle\\\langle1|A|0\rangle&\langle1|A|1\rangle\end{bmatrix}=\sum_{i,j}a_{j}|i\rangle\langle j| $$ If we had an operator $A:V\to V$ and $\{|v_i\rangle\}$ constitute a basis of $V$ then

Let $\{|v_i\rangle\}$ be an orthonormal basis then the change of basis matrix is $P_v=V^{-1}=V^\dagger=\begin{bmatrix}\langle v_0|\\\langle v_1|\end{bmatrix}$

The matrix $A$ in the orthonormal basis $\{|v_i\rangle\}$ is, $$ B=[A]_v=P_vAP_v^{-1}=\begin{bmatrix}\langle v_0|\\\langle v_1|\end{bmatrix}A\begin{bmatrix} |v_0\rangle& |v_1\rangle\end{bmatrix}=\begin{bmatrix}\langle v_0|A|v_0\rangle&\langle v_0|A|v_1\rangle\\\langle v_1|A|v_0\rangle&\langle v_1|A|v_1\rangle\end{bmatrix} $$ $$ A=P_v^\dagger BP_v=\begin{bmatrix} |v_0\rangle& |v_1\rangle\end{bmatrix}\begin{bmatrix}\langle v_0|A|v_0\rangle&\langle v_0|A|v_1\rangle\\\langle v_1|A|v_0\rangle&\langle v_1|A|v_1\rangle\end{bmatrix}\begin{bmatrix}\langle v_0|\\\langle v_1|\end{bmatrix}\\ =\sum_{i,j}\langle v_i|A|v_j\rangle|v_i\rangle\langle v_j| $$ That means, $$ A=\sum_{i,j}a_{ij}|i\rangle\langle j|=\sum_{i,j}b_{ij}|v_i\rangle\langle v_j| $$ where $a_{ij}=\langle i|A|j\rangle$ and $b_{ij}=\langle v_i|A|v_j\rangle$

This is clear!

So the $(i,j)^{th}$ term of $A$ in the basis $|v_i\rangle$ is $b_{ij}=\langle v_i|A|v_j\rangle$ and $B=[A]_v=\sum_{i,j}b_{ij}|i\rangle\langle j|$

Unlike $\langle v_i|W|v_j\rangle$ which is a number and $|v_j\rangle$ is a vector, the term $E_k=(I\otimes\langle e_k|)U(I\otimes|e_0\rangle)=\langle e_k|U|e_0\rangle$ is an operator acting on the first system (principal system $Q$) and $I\otimes|e_0\rangle=|e_0\rangle,I\otimes|e_k\rangle=|e_k\rangle$ are matrices.

My Attempt

Thanks @glS for the hint.

\begin{align} E_k&=(I\otimes\langle e_k|)U(I\otimes|e_0\rangle)=\langle e_k|U|e_0\rangle\\ \end{align} If we interchange the order of systems, i.e., the first system is the environment $E$ and second is the principal system $Q$, then

Decomposing $U$ and $(\langle e_k|\otimes I)U(|e_0\rangle\otimes I)$ in terms of the orthonormal basis vectors of the respective systems as,

\begin{align} &(\langle e_k|\otimes I)U(|e_0\rangle\otimes I)=\\ &=(\langle e_k|\otimes I)(\sum_{m,n,m',n'}\mu_{m,n,m',n'}(|e_n\rangle\otimes|p_m\rangle)(\langle e_{n'}|\otimes\langle p_{m'}|))(|e_{0}\rangle\otimes I)\\ &=\langle e_k|\otimes I\bigg[\sum_{m,n,m',n'}\mu_{m,n,m',n'}|e_n\rangle\langle e_{n'}|\otimes|p_m\rangle\langle p_{m'}|\bigg]|e_{0}\rangle\otimes I\\ &=\sum_{m,n,m',n'}\mu_{m,n,m',n'}\langle e_k|e_n\rangle\langle e_{n'}|e_{0}\rangle\otimes|p_m\rangle\langle p_{m'}|\\ &=\sum_{m,n,m',n'}\mu_{m,n,m',n'}\delta_{k,n}\delta_{n',0}\otimes|p_m\rangle\langle p_{m'}|\\ &=\sum_{m,m'}\mu_{m,k,m',0}|p_m\rangle\langle p_{m'}|\\ \end{align}

\begin{align} U&=\sum_{m,n,m',n'}\mu_{m,n,m',n'}(|e_n\rangle\otimes|p_m\rangle)(\langle e_{n'}|\otimes\langle p_{m'}|)\\ &=\sum_{m,n,m',n'}\mu_{m,n,m',n'}|e_n\rangle\langle e_{n'}|\otimes|p_m\rangle\langle p_{m'}|\\ &=\sum_{n,n'}|e_n\rangle\langle e_{n'}|\otimes\sum_{m,m'}\mu_{m,n,m',n'}|p_m\rangle\langle p_{m'}|\\ \end{align}

If I define the change of basis matrix $P=P_E\otimes P_p$ then (extending the ideas in the previous section) the operators in the new basis are, $$ (\langle e_k|\otimes I)U(|e_0\rangle\otimes I)'=\sum_{m,m'}\mu_{m,k,m',0}|m\rangle\langle {m'}|\\ $$ \begin{align} U'&=\sum_{m,n,m',n'}\mu_{m,n,m',n'}|n\rangle\langle {n'}|\otimes|m\rangle\langle {m'}|\\ &=\sum_{n,n'}|n\rangle\langle {n'}|\otimes \sum_{m,m'}\mu_{m,n,m',n'}|m\rangle\langle {m'}|\\ &=\sum_{n,n'}|n\rangle\langle {n'}|\otimes\sigma_{n,n'} \end{align} where $\{|n\rangle\}$ and $\{|m\rangle\}$ are the standard basis such that $|n\rangle\langle {n'}|$ is the matrix with $1$ as its $(n,n')$ entry, and $\sigma_{n,n'}=\sum_{m,m'}\mu_{m,n,m',n'}|m\rangle\langle {m'}|$ is the $(n,n')^{th}$ block entry of $U'$.

Therefore, the $(k,0)^{th}$ block entry of $U'$ is

$\sigma_{k,0}=\sum_{m,m'}\mu_{m,k,m',0}|m\rangle\langle {m'}|=(\langle e_k|\otimes I)U(|e_0\rangle\otimes I)'$

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  • $\begingroup$ There is a wonderful SE post that I think might help answer you question. The question and answer are on physics StackExchange here physics.stackexchange.com/questions/27657/… its a shame it can't easily be found on QCSE. I hope it helps. $\endgroup$
    – Condo
    Sep 26, 2022 at 19:58
  • $\begingroup$ What does $\left|e\right> \otimes I$ even mean? Tensor product of state and operator? $\endgroup$
    – consthatza
    Sep 29, 2022 at 9:48
  • $\begingroup$ @noobier yes, as per my understanding. $\endgroup$
    – Sooraj S
    Sep 29, 2022 at 10:57
  • $\begingroup$ @SoorajS isn't that an abuse of notation? Haven't seen anything like that before $\endgroup$
    – consthatza
    Sep 30, 2022 at 9:45

1 Answer 1

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You don't really need to do any step. The way of writing $U$ as a matrix whose blocks are the Kraus operators is nothing but a way to picture what $E_k=(I\otimes \langle e_k|)U(I\otimes |e_0\rangle)$ means. In fact, this isn't even that precise: following the standard conventions to represent tensor products as matrices, that $U$ should more precisely correspond to $(\langle e_k|\otimes I)U(|e_0\rangle\otimes I)$.

To see it, think about the expression $(\langle e_k|\otimes I)U(|e_0\rangle\otimes I)$. This is a way to denote a submatrix of $U$. More precisely, it is the submatrix with components $$[(\langle e_k|\otimes I)U(|e_0\rangle\otimes I)]_{ij} = U_{ki,0j}.$$ In other words, $(\langle e_k|\otimes I)U(|e_0\rangle\otimes I)$ is telling you what a specific subset of elements of $U$ is equal to.

If you now picture $U$ as a "matrix of matrices", with the first indices (in the expression above, the indices $k$ and $0$) specifying the position of a block, and the second indices ($i$ and $j$ above) representing the position within said block, you end up representing $U$ as some unitary matrix whose first block of matrices amounts to the Kraus operators stacked vertically. Note that $U$ will also have other elements aside from this first block of columns, it's just that these elements are not specified by the given expression with Kraus operators, so they can be whatever (as long as they maintain the unitarity of the overall matrix).

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  • $\begingroup$ Thanks for answering. You are saying the expression $(\langle e_k|\otimes I)U(|e_0\rangle\otimes I)$ is a submatrix of $U$ such that $[(\langle e_k|\otimes I)U(|e_0\rangle\otimes I)]_{ij}=U_{\color{red}{k}i,\color{red}{0}j}$ with $k,0$ implying the position of the block and $i,j$ implying the position within block. But how can I see this is how it is mathematically given $|e_k\rangle$ are orthonormal basis and $|e_0\rangle$ is some arbitrary state which is a linear combination of $|e_k\rangle$ ? $\endgroup$
    – Sooraj S
    Sep 28, 2022 at 13:31
  • $\begingroup$ @SoorajS I'm not sure I understand the question. Yes, $|e_k\rangle$ are by definition an orthonormal basis for the space. The state $|e_0\rangle$ is by definition the first element of this basis. Is this what you're asking? In practice, the state $|e_0\rangle$ could be anything, and needs not actually be related to the basis $\{|e_k\rangle\}$. Then the picture with $U$ being a block matrix doesn't work; however you can always imagine to perform some change of basis so that $|e_0\rangle$ is in fact an element of the basis. This only affects how you represent the matrix, not the physics $\endgroup$
    – glS
    Sep 28, 2022 at 14:15
  • $\begingroup$ Okay, sorry I think I misread the question, if $|e_0\rangle$ is the first element of the orthonormal basis $\{|e_k\rangle\}$, how can I visualize that $[(\langle e_k|\otimes I)U(|e_0\rangle\otimes I)]_{ij}=U_{\color{red}{k}i,\color{red}{0}j}$ with $k,0$ implying the position of the block and $i,j$ implying the position within block mathematically ? $\endgroup$
    – Sooraj S
    Sep 28, 2022 at 14:54
  • $\begingroup$ @SoorajS that's what I tried to explain in the answer. I'm not sure what's missing though. Maybe the fact that an operator acting in a bipartite space (which $U$ is) can be represented as a four-indices object, like $U_{ki,0j}$, Then $ki$ specify a row, and $0j$ a column. If you fix $k$ and change $i$, you explore a specific subset of columns. If you group these together (which you can do by suitably chosing a basis wrt which to represent the operator), then you can picture $k$ as corresponding to a block of elements $\endgroup$
    – glS
    Sep 28, 2022 at 21:58
  • $\begingroup$ Thanks. I edited ''My Attempt'' section in my post to include the math. Could you please have a look, am I thinking it correct? $\endgroup$
    – Sooraj S
    Sep 29, 2022 at 10:14

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