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In Chapter 8 of Quantum Computation & Quantum Information by Nielsen & Chuang, more precisely Box 8.5, there is an example of quantum process tomography for a single qubit. (The same discussion can be found here.) The fixed operators have been chosen as $\tilde{E_0} = I$, $\tilde{E_1} = X$, $\tilde{E_2} = -iY$, and $\tilde{E_3} = Z$. ($\tilde{A_j}$ in the arXiv paper.) The basis elements used are $\rho_1 = | 0 \rangle \langle 0 |$, $\rho_2 = \rho_1 X = | 0 \rangle \langle 1 |$, $\rho_3 = X \rho_1 = | 1 \rangle \langle 0 |$, and $\rho_4 = X \rho_1 X = | 1 \rangle \langle 1 |$. Then $\beta$ is determined from the equation \begin{equation} \tag{*}\label{*} \tilde{E}_m \rho_j \tilde{E}_n^\dagger = \sum_k \beta^{mn}_{jk} \rho_k. \end{equation} In this case, with the choices given, it is simple to compute the elements $\beta^{mn}_{jk}$ explicitly.

Now, Nielsen & Chuang say that $\beta$ can be written as $\beta = \Lambda \otimes \Lambda$, where \begin{equation*} \Lambda = \frac{1}{2} \left[ \begin{matrix} I & X \\ X & -I \end{matrix} \right]. \end{equation*} Then $\beta$ would be the block matrix \begin{equation*} \beta = \frac{1}{4} \left[ \begin{matrix} I & X & & & & & I & X \\ X & -I & & & & & X & -I \\ & & I & X & I & X & & \\ & & X & -I & X & -I & & \\ & & I & X & -I & -X & & \\ & & X & -I & -X & I & & \\ I & X & & & & & -I & -X \\ X & -I & & & & & -X & I \end{matrix} \right]. \end{equation*}

My problem is that I cannot see how the indexing of $\beta$ should be understood so that this would correspond to the elements computed from \eqref{*}. For example, $\beta^{00}_{jk} = \delta_{jk}$, $\beta^{01}_{1k} = \delta_{2k}$, $\beta^{01}_{2k} = \delta_{1k}$, $\beta^{01}_{3k} = \delta_{4k}$, $\beta^{01}_{4k} = \delta_{3k}$, etc. N&C say that the columns of $\beta$ are indexed by $mn$ and the rows by $jk$. I have tried to see if e.g. $m$ would index the blocks of four, and $n$ index for the column within a block, or vice versa, but something like that does not seem to make sense.

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  • $\begingroup$ It's been a while since I looked at Nielsen & Chuang's way of doing QPT (which, btw, is not the 'standard' way of doing it), so I cannot really answer your question (yet). But if $\Lambda$ is the matrix that you describe, then $\beta$ is not the matrix you write down, but only the inner block of the matrix that you wrote. $\endgroup$
    – JSdJ
    Sep 4, 2023 at 19:04

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I am suggesting a way of indexing the 16 x 16 $\beta$ matrix, but I am not sure if it corresponds to your Eq(*)

\begin{equation} \beta = \frac{1}{4} \left[ \begin{matrix} \beta_{11}^{00} & \beta_{11}^{01} & \beta_{11}^{02} & \beta_{11}^{03} & \ldots & \beta_{11}^{32} & \beta_{11}^{33}\\ \\ \beta_{12}^{00} & \beta_{12}^{01} & \beta_{12}^{02} & \beta_{12}^{03} & \ldots & \beta_{12}^{32} & \beta_{12}^{33}\\ \\ \beta_{13}^{00} & \beta_{13}^{01} & \beta_{13}^{02} & \beta_{13}^{03} & \ldots & \beta_{13}^{32} & \beta_{13}^{33}\\ \\ \beta_{14}^{00} & \beta_{14}^{01} & \beta_{14}^{02} & \beta_{14}^{03} & \ldots & \beta_{14}^{32} & \beta_{14}^{33}\\ \\ \vdots & \vdots & \ldots & \ldots & \ldots & \vdots & \vdots \\ \\ \vdots & \vdots & \ldots & \ldots & \ldots & \vdots & \vdots \\ \\ \beta_{43}^{00} & \beta_{43}^{01} & \ldots & \ldots & \ldots & \beta_{43}^{32} & \beta_{43}^{33}\\ \\ \beta_{44}^{00} & \beta_{44}^{01} & \ldots & \ldots & \ldots & \beta_{44}^{32} & \beta_{44}^{33}\\ \end{matrix} \right]. \end{equation}

I have a similar query to related to this(Quantum Process Tomography for 2 qubits). If you @Calle have found any insights then maybe you can share. I understand that this might not qualify as complete answer but I don't have enough reputation to comment.

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  • $\begingroup$ If you have a new question, please ask it by clicking the Ask Question button. Include a link to this question if it helps provide context. - From Review $\endgroup$ Nov 28, 2023 at 7:24
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I am wondering about this too! I cannot provide an answer, but (maybe) I have a clue: $$\begin{pmatrix} \bar{E}_0\\\bar{E}_1\\\bar{E}_2\\\bar{E}_3 \end{pmatrix}=\begin{pmatrix} 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0\\ 0 & 1 & -1 & 0\\ 1 & 0 & 0 & -1\\ \end{pmatrix}\begin{pmatrix} \rho_1\\\rho_2\\\rho_3\\\rho_4 \end{pmatrix}=\Lambda\begin{pmatrix} \rho_1\\\rho_2\\\rho_3\\\rho_4 \end{pmatrix}$$ Could this be where the mysterious $\Lambda$ comes from? How can we proceed? I have no idea!

(Maybe I should not post this as an answer. However, I do not have enough reputation to comment. Sorry for that.)

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