# 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 $$$$\tag{*}\label{*} \tilde{E}_m \rho_j \tilde{E}_n^\dagger = \sum_k \beta^{mn}_{jk} \rho_k.$$$$ In this case, with the choices given, it is simple to compute the elements $$\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.

• 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.
– JSdJ
Sep 4, 2023 at 19:04

I am suggesting a way of indexing the 16 x 16 $$\beta$$ matrix, but I am not sure if it corresponds to your Eq(*)
$$$$\beta = \frac{1}{4} \left[ \begin{matrix} \beta_{11}^{00} & \beta_{11}^{01} & \beta_{11}^{02} & \beta_{11}^{03} & \ldots & \beta_{11}^{32} & \beta_{11}^{33}\\ \\ \beta_{12}^{00} & \beta_{12}^{01} & \beta_{12}^{02} & \beta_{12}^{03} & \ldots & \beta_{12}^{32} & \beta_{12}^{33}\\ \\ \beta_{13}^{00} & \beta_{13}^{01} & \beta_{13}^{02} & \beta_{13}^{03} & \ldots & \beta_{13}^{32} & \beta_{13}^{33}\\ \\ \beta_{14}^{00} & \beta_{14}^{01} & \beta_{14}^{02} & \beta_{14}^{03} & \ldots & \beta_{14}^{32} & \beta_{14}^{33}\\ \\ \vdots & \vdots & \ldots & \ldots & \ldots & \vdots & \vdots \\ \\ \vdots & \vdots & \ldots & \ldots & \ldots & \vdots & \vdots \\ \\ \beta_{43}^{00} & \beta_{43}^{01} & \ldots & \ldots & \ldots & \beta_{43}^{32} & \beta_{43}^{33}\\ \\ \beta_{44}^{00} & \beta_{44}^{01} & \ldots & \ldots & \ldots & \beta_{44}^{32} & \beta_{44}^{33}\\ \end{matrix} \right].$$$$
I am wondering about this too! I cannot provide an answer, but (maybe) I have a clue: $$\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!