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I am working on a project and I expect to have expressions of a bunch of quantum channels of interest. The quantum channels will be in matrix form. For example for a 2 qubit system, the quantum channel will be given by a 16 by 16 matrix.

Is there any systematic way to use the matrix representations of quantum channels and find the corresponding Kraus operators etc? Find which channels correspond to unitary channels etc?

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  • $\begingroup$ This might be helpful: quantumcomputing.stackexchange.com/a/5816/13968 $\endgroup$
    – narip
    Nov 21, 2022 at 2:10
  • $\begingroup$ There are multiple representations of channels using a single, necessarily square matrix, most notably the Choi and $\chi$ matrix. See e.g. this previous answer of mine. For both these we have: 1) the eigenvectors correspond to the Kraus operators and 2) if the matrix is single rank it corresponds to a unitary. If you let us know what representation you will work with I can write up a more detailed answer. $\endgroup$
    – JSdJ
    Nov 21, 2022 at 8:28
  • $\begingroup$ you can also find a bunch of explicit examples of matrix representations (with a focus on Chois and Stinespring isometries) in quantumcomputing.stackexchange.com/q/24511/55. Does any of these linked posts answer your question? If not, could you further clarify what you are asking and how it differs from the other related discussions? $\endgroup$
    – glS
    Nov 21, 2022 at 9:32
  • $\begingroup$ I have yet to perform the computation but I expect to have a bunch of matrices in the usual sense. I'll have to throw away the matrices that don't correspond to CPTP maps. I'll then like to identify the remaining matrices/CPTP maps with their Kraus and/or Choi decomposition. $\endgroup$
    – user22511
    Nov 21, 2022 at 16:12
  • $\begingroup$ @user22511 that still doesn't clarify the problem. There are multiple possible matrix representations of a channel. You need to specify which one you are talking about. $\endgroup$
    – glS
    Nov 21, 2022 at 18:13

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In your latest comments you talked about matrices (channels) that meet some commutation relations. This is best tackled with the representation matrix (also known as "natural representation") with respect to vectorization---as linked in one of the comments, cf. also this answer---which is the unique matrix $\hat\Phi$ such that ${\rm vec}(\Phi(\rho))=\hat\Phi{\rm vec}(\rho)$ for all $\rho$. Given any $\Phi:\mathbb C^{n\times n}\to\mathbb C^{n\times n}$ the explicit expression for $\hat\Phi$ reads $$ \hat\Phi=\sum_{i,j,k,l=1}^n\;\langle l|\Phi(|i\rangle\langle j|)|k\rangle\;|k\rangle\langle j|\otimes|l\rangle\langle i| $$ as is readily verified. The key property here is that $\widehat{(\cdot)}$ is a (linear) homomorphism, that is, $\widehat{\Phi_1\Phi_2}=\widehat{\Phi_1}\widehat{\Phi_2}$ for all $\Phi_1,\Phi_2$; thus commutators of linear maps are in one-to-one correspondence with commutators of their representation matrices. Then in order to check complete positivity of $\Phi$ you have to check the Choi matrix $\mathsf C(\Phi)$ which is just a re-shuffling of $\hat\Phi$. More precisely, it is easy to see that $$ \mathsf C(\Phi)=\sum_{i,j,k,l}\;\langle l|\Phi(|i\rangle\langle j|)|k\rangle\;\color{red}{|i\rangle}\langle j|\otimes|l\rangle\color{red}{\langle k|} $$ so the linear (and even unitary) map $S:\mathbb C^{n^2\times n^2}\to \mathbb C^{n^2\times n^2}$ uniquely defined via $S(|k\otimes l\rangle\langle j\otimes i|):=|i\otimes l\rangle\langle j\otimes k|$ satisfies $S(\hat\Phi)=\mathsf C(\Phi)$ (and even $S(\mathsf C(\Phi))=\hat\Phi$ because $S^2={\rm id}$). From the Choi matrix complete positivity and Kraus operators are obtained as usual.

Finally, because you explicitly mentioned it let me say that as per this result a unitary representation matrix $\hat\Phi$ corresponds to a channel if and only if $\hat\Phi=\overline{U}\otimes U$ for some $U\in\mathsf U(n)$. An easy way to check this would be to look at the partial traces of $\hat\Phi$ with respect to the first and second tensor factor and to check that they are both unitary and coincide up to complex conjugation $\overline{(\cdot)}$.

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