A quantum map on a $d$-dimensional space has the general representation: $$ \mathcal{S}(\rho)=\sum_{\alpha,\beta}^{d^2}\chi_{\alpha\beta}\Gamma_{\alpha}\rho \Gamma_{\beta}^{\dagger}, $$ where $\chi$ is the $d^2\times d^2$ process matrix, which is positive semidefinite and trace preserving.

On the other hand, the (unnormalized) maximally entangled bipartite state between a quantum system $S$ and an ancilla system $A$ is $|\psi\rangle=\sum_{k=1}^d|k\rangle_A|k\rangle_S$ , where $\{|k\rangle\}_{k=1}^d$ represents an orthonormal basis. For a quantum process $\mathcal{E}$ acting only on the system $S$ of $|\psi\rangle$, the output state is given by $$ \Upsilon_{\mathcal{E}}=(\mathcal{I} \otimes \mathcal{E})(|\psi\rangle\langle\psi|)=\sum_{k, l=1}^d|k\rangle\langle l| \otimes \mathcal{E}(|k\rangle\langle l|), $$ which is called the Choi matrix of the process $\mathcal{E}$.

Since they can all represent quantum process, so why this paper of process tomography uses $\chi$ matrix?


2 Answers 2


The two descriptions are entirely equivalent. It doesn't matter which you use when, it's just a case of using whichever description you personally find to be mathematically the most convenient.

  • $\begingroup$ Is there any reason as to why someone would use the process matrix? Computing the Choi matrix seems much much easier. $\endgroup$
    – FDGod
    Commented Nov 3, 2023 at 13:46
  • 1
    $\begingroup$ Well, let's take a silly extreme: the map is just a unitary. If I want to apply this unitary to a density matrix, I would much rather just calculate $U\rho U^\dagger$ than have the calculate $\text{Tr}_1(\rho^T\otimes U\cdot |\Omega\rangle\langle\Omega|\cdot I\otimes U^\dagger)$ $\endgroup$
    – DaftWullie
    Commented Nov 3, 2023 at 13:55

This is just a comment, but it's too long for a comment, so writing as an answer. As I haven't read the paper you are asking about, I cannot answer as to particularly why that paper is using the process matrix. Will update my answer later if I get time.

A map $\mathcal{E}$ is completely positive if and only if the Choi matrix $\Upsilon_\mathcal{E}$ is a positive matrix.

Equivalently, the super-operator $\mathcal{E}$, acting on any operator $A$ can be written as $$ \mathcal{E}(A) = \sum_{\alpha,\beta}^{d^2}\chi_{\alpha\beta}\Gamma_{\alpha}A \Gamma_{\beta}^{\dagger} \,,$$ where $\chi$ is a positive matrix in some operator basis $\{\Gamma_{\alpha}\}$. However, if your $\mathcal{E}$ is not a CPTP map, then you will not be able to write the above equation with positive $\chi$.

Both the Choi matrix and the process matrix accomplish the same thing, which is to verify the complete positivity of your map/process $\mathcal{E}$, i.e., $\mathcal{E}$ correspond to some valid physical quantum process or not$^1$.

Generally, if you have $\mathcal{E}$ given, it's easier to calculate the Choi matrix to verify the complete positivity of $\mathcal{E}$.

Fun fact: $\Upsilon_\mathcal{E}$ and $\chi$ are related by just a unitary transformation.
1: If you need some intuition behind this statement, my this answer to another question would be helpful.


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.