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In this PDF (page 43), it is argued that, given an arbitrary quantum channel with Kraus decomposition:

$$ E(\rho) = \sum_{j} K_j \rho K_j^{\dagger} $$

Such map can be represented with a matrix in $\mathbb{C}^{d²}$:

$$ \hat E = \sum_j K_j \otimes \bar{K_j} $$

I can't figure out a proof, do you have any ideas?

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    $\begingroup$ $\rho$ is an element of a vector space with $n^2$ elements. If you stack the columns of $\rho$ on top of one another you'll get a vector in the standard column vector form. Then the linear transformation $E(\rho)$ should be representable by a matrix. You can try to show that the matrix that represents its action is $\hat{E}$. Hint: try to prove $\mathrm{vec}(AXB) = A\otimes B^T \mathrm{vec}(X)$ where $\mathrm{vec}$ is the column stacking map. $\endgroup$
    – Rammus
    Sep 13, 2022 at 8:54
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    $\begingroup$ this is also often referred to as the natural representation of the channel. See eg cs.uwaterloo.ca/~watrous/TQI $\endgroup$
    – glS
    Sep 13, 2022 at 19:34
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    $\begingroup$ Every linear function between finite-dimensional vector spaces can be represented as a matrix. $\endgroup$ Sep 13, 2022 at 20:37

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Mind that $E\left( \cdot \right) $ is a linear map, and can be written as matrix act on a vector. If we write matrix in vector form as follows: $$\operatorname{vec}_c(\rho)=\left(\begin{array}{c} \rho_{00} \\ \rho_{10} \\ \vdots \\ \rho_{n n} \end{array}\right)$$ With this form, you can check one fact:$K_j\rho $ has vector form $K_j\otimes I\mathrm{vec}\left( \rho \right) $ and ${\rho K_j}$ has vector form $I\otimes {K_j}^T\mathrm{vec}\left( \rho \right) $, combine them together we have $K_j\rho K_{j}^{\dagger}$ has vector form $K_j\otimes \bar{K}_j\mathrm{vec}\left( \rho \right) $. Edit Okay, I made a mistake. In your pdf link, the author state he do the vectorization in style $|k\rangle\langle l|\leftrightarrow| k, l\rangle$, this is actually stack rows of matrix, i.e. $$\text{vec}_r\begin{pmatrix} \alpha & \beta\\ \gamma & \delta \end{pmatrix} = \begin{pmatrix} \alpha\\ \beta\\ \gamma\\ \delta \end{pmatrix}. $$ With vec$_c$, we should have formula:$${\displaystyle \operatorname {vec}_c (ABC)=(C^{\mathrm {T} }\otimes A)\operatorname {vec} (B)}$$ Hence we have$$ \mathrm{vec}_{\mathrm{r}}\left( E_j\rho E_{j}^{\dagger} \right) =\mathrm{vec}_c\left( \bar{E}_j\rho ^TE_{j}^{T} \right) =E_j\otimes \bar{E}_j\mathrm{vec}_c\left( \rho ^T \right) =E_j\otimes \bar{E}_j\mathrm{vec}_r\left( \rho \right) $$

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