# Why can any quantum channel be represented as a matrix?

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?

• $\rho$ 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. Commented Sep 13, 2022 at 8:54
• this is also often referred to as the natural representation of the channel. See eg cs.uwaterloo.ca/~watrous/TQI
– glS
Commented Sep 13, 2022 at 19:34
• Every linear function between finite-dimensional vector spaces can be represented as a matrix. Commented Sep 13, 2022 at 20:37

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$$\mathrm{vec}_{\mathrm{r}}\left( E_j\rho E_{j}^{\dagger} \right) =\mathrm{vec}_c\left( \bar{E}_j\rho ^TE_{j}^{T} \right) =E_j\otimes \bar{E}_j\mathrm{vec}_c\left( \rho ^T \right) =E_j\otimes \bar{E}_j\mathrm{vec}_r\left( \rho \right)$$