# Process matrix of CNOT gate

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

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 $$U_{CNOT}$$?

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 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.
• 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? Jan 11 at 8:49