# How is the $\beta$-matrix interpreted in single qubit QPT?

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.