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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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