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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?

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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.

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  • $\begingroup$ Is there any reason as to why someone would use the process matrix? Computing the Choi matrix seems much much easier. $\endgroup$
    – FDGod
    Nov 3, 2023 at 13:46
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    $\begingroup$ 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
    Nov 3, 2023 at 13:55
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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.

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