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The fig below is the process matrix of the CNOT gate from this paper:

Process matrix of the CNOT gate

where the legend explains that red corresponds to $\frac14$, green to $-\frac14$ and white to zero.

I know the $U_{CNOT} = \frac{1}{2}(I ⊗ I + I ⊗ X + Z ⊗ I − Z ⊗ X)$ with tensor products of Pauli operators $\{I, X, Y, Z\}$, but how can I understand this fig? Why we use $II...ZZ$ and $II...ZZ$ as basis? (For example, what does it mean by showing a number with $(IX,IX)$?) What is the corresponding process matrix of $U_{CNOT}$?

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Process matrix definition

Recall that the process matrix of a channel $\mathcal{E}(\rho)=\sum_kK_k\rho K_k^\dagger$ with respect to an operator basis $A_i$ is obtained by expressing the Kraus operators as a linear combination $K_k=\sum_ia_{ki}A_i$ of the basis elements $A_i$ and collecting the terms $A_i\rho A_j^\dagger$ from all Kraus operators, c.f. equation $(1)$ in the cited paper $$ \mathcal{E}(\rho)=\sum_{ijk}a_{ki}\overline{a_{kj}}A_i\rho A_j^\dagger=\sum_{ij}\chi_{ij}A_i\rho A_j^\dagger\tag{P} $$ where $\chi_{ij}:=\sum_ka_{ki}\overline{a_{kj}}$ is called the process matrix. In practice, $A_i$ is often the Pauli basis, because it is convenient both for theoretical analysis and experiments.

CNOT gate

CNOT gate is unitary so the corresponding channel $\mathcal{E}_{\text{CNOT}}(\rho)=U_{\text{CNOT}}\rho U_{\text{CNOT}}^\dagger$ has a single Kraus operator. Therefore, the index $k$ in $(P)$ ranges over a one-element set and can be dropped. Thus, $\chi_{ij}=a_i\overline{a_j}$.

As stated in the question, we can express $U_{\text{CNOT}}$ in the Pauli basis as $$U_{\text{CNOT}}=\frac12(II+IX+ZI-ZX).$$Using the names of Pauli operators as indices instead of numbers (following the convention in figure 1 in the cited paper), we can write the $16$-element vector $a_i$ as $$ \begin{array}{ |c|c|c|c|c| } \hline II & IX & ZI & ZX & \text{other} \\ \hline \frac12 & \frac12 & \frac12 & -\frac12 & 0 \\ \hline \end{array} $$ Consequently, the process matrix $\chi_{ij}=a_i\overline{a_j}$ becomes $$ \begin{array}{ |c|c|c|c|c|c| } \hline & II & IX & ZI & ZX & \text{other} \\ \hline II & \frac14 & \frac14 & \frac14 & -\frac14 & 0 \\ \hline IX & \frac14 & \frac14 & \frac14 & -\frac14 & 0 \\ \hline ZI & \frac14 & \frac14 & \frac14 & -\frac14 & 0 \\ \hline ZX & -\frac14 & -\frac14 & -\frac12 & \frac14 & 0 \\ \hline \text{other} & 0 & 0 & 0 & 0 & 0 \\ \hline \end{array} $$ in agreement with figure 1.

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  • $\begingroup$ So $k=1, K^{\dagger}=\sum \overline{a_j} A_j^{\dagger}$, here $U_{\text{CNOT}}$ is $K$, we can use the Pauli basis combination of $K, K^{\dagger}$ to know the efficients $a_i,\overline{a_j}$, then we can get the process matrix, right? $\endgroup$
    – karry
    Jan 11 at 8:49

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