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 $U_{CNOT}$?


1 Answer 1


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.