The Exercise

A process on two qubits has a $\chi$ matrix, the $16 \times 16$ array of its coefficients in a fixed basis of sixteen operators on the two qubits. Process tomography finds that matrix from the outputs of the process on known inputs. For one qubit, Box 8.5 of the book does this with a fixed $4 \times 4$ matrix $\Lambda$, written out in (7) below.

For two qubits the inputs are the sixteen operators $|a\rangle\langle b|$, with $a$ and $b$ among $00$, $01$, $10$ and $11$, and their outputs fill a $16 \times 16$ block matrix. From that matrix $\chi$ is to be recovered by a permutation of its rows and columns followed by multiplication with $\Lambda \otimes \Lambda$ on both sides.

The Operator Basis

The operator basis on one qubit assigns to each pair of bits $jk$ an operator $\tilde E_{jk}$ on the qubit, $$\tilde E_{00} = I, \qquad \tilde E_{01} = X, \qquad \tilde E_{10} = -iY, \qquad \tilde E_{11} = Z,$$(1) where $X$, $Y$ and $Z$ are the Pauli matrices $$X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \qquad Y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \qquad Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.$$(2)

A two-qubit label is a pair of single-qubit labels, one for each qubit.^✦ The set of them, $$\mathcal{I}_2 = \bigl(\{0,1\} \times \{0,1\}\bigr) \times \bigl(\{0,1\} \times \{0,1\}\bigr),$$(3)^✦ has sixteen elements. The two-qubit operator $\tilde E_M$ with label $M = (m_1, m_2)$ in $\mathcal{I}_2$ is the tensor product of the single-qubit operators of its two halves, $$\tilde E_M = \tilde E_{m_1} \otimes \tilde E_{m_2}.$$(4)^✦

The Process and Its Outputs

Here the process is given by its $\chi$ matrix, and its outputs are computed from it rather than measured. For a complex matrix $\chi$ with rows and columns labelled by $\mathcal{I}_2$, the process $\mathcal{E}_\chi$ is the map on operators of the two qubits given by $$\mathcal{E}_\chi(\rho) = \sum_{M, N \in \mathcal{I}_2} \chi_{MN} \, \tilde E_M \, \rho \, \tilde E_N^\dagger .$$(5)^✦

The block matrix of outputs $M$ of $\mathcal{E}_\chi$ is the $16 \times 16$ matrix whose block in row $a$ and column $b$ is the output $\mathcal{E}_\chi(|a\rangle\langle b|)$.^✦ With $a$, $b$, $i$ and $j$ two-qubit basis labels, so that $(a, i)$ and $(b, j)$ lie in $\mathcal{I}_2$, its entries are $$M_{(a,i),(b,j)} = \langle i | \, \mathcal{E}_\chi\bigl(|a\rangle\langle b|\bigr) \, | j \rangle .$$(6)^✦

Two Fixed Matrices

The matrix $\Lambda$ of Box 8.5 is the $4 \times 4$ matrix with rows and columns labelled by pairs of bits in the order $00$, $01$, $10$, $11$, the first bit numbering the $2 \times 2$ block and the second the place within it: $$\Lambda = \frac{1}{2} \begin{pmatrix} I & X \\ X & -I \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & -1 & 0 \\ 1 & 0 & 0 & -1 \end{pmatrix}.$$(7)^✦ The two-qubit matrix $\Lambda_2$ is the tensor product of $\Lambda$ with itself, a $16 \times 16$ matrix with rows and columns labelled by $\mathcal{I}_2$: $$\Lambda_2 = \Lambda \otimes \Lambda .$$(8)^✦

The perfect shuffle $\sigma$ is the permutation of $\mathcal{I}_2$ that regroups the bits of a label of $M$, input pair first and output pair second, into one pair for each qubit: $$\sigma\bigl((a_1, a_2), (i_1, i_2)\bigr) = \bigl((a_1, i_1), (a_2, i_2)\bigr).$$(9)^✦ The shuffle matrix $P$ is the permutation matrix of $\sigma$, the $16 \times 16$ matrix on $\mathcal{I}_2$ with entries $$P_{UV} = \delta_{\sigma(U), V} .$$(10)^✦

The Claim

What Is Checked

The claim is stated over the process $\mathcal{E}_\chi$ of (5), built on the operators $\tilde E_M$ of (4), over its block matrix of outputs (6), and over the fixed matrices $\Lambda_2$ and $P$ of (8) and (10).