Please be kind. I've been working on this for a long time and can't find an answer. Feel free to edit for clarity if you think the question can be better worded.

**Background**

It may help to see a previous question I asked about quantum channels in order to understand the basis for this question.

A quantum channel is a mapping between algebras of bounded linear operators on Hilbert spaces, $\Phi : L(\mathcal{H}_{A}) \to L(\mathcal{H}_{B})$, where $L(\mathcal{H}_{i})$ is the family of operators on $\mathcal{H}_{i}$. In general, we are interested in completely positive, trace-preserving (CPTP) maps. The operator spaces can be interpreted as $C^{*}$-algebras (with the involution being the standard Hilbert space adjoint, denoted by $\dagger$) and thus we can also view the channel as a mapping between $C^{*}$-algebras, $\Phi : \mathcal{A} \to \mathcal{B}$. Since quantum channels can carry classical information as well, we could write such a combination as $\Phi : L(\mathcal{H}_{A}) \otimes C(X) \to L(\mathcal{H}_{B})$ where $C(X)$ is the space of continuous functions on some set $X$ and is also a $C^{*}$-algebra. In other words, whether or not classical information is processed by the channel, it (the channel) is a mapping between $C^{*}$-algebras. Note, however, that these are not necessarily the same $C^{*}$-algebras. Since the channels are represented by square matrices, the input and output $C^{*}$-algebras must have the same dimension, $d$ (i.e. physicists will often "cheat" and refer to the dimension of an $n\times n$ matrix as simply $n$). Thus we can consider them both subsets of some $d$-dimensional $C^{*}$-algebra, $\mathcal{C}$, i.e. $\mathcal{A} \subset \mathcal{C}$ and $\mathcal{B} \subset \mathcal{C}$. Thus a quantum channel is a mapping from $\mathcal{C}$ to itself.

A quantum channel may be written as a Kraus decomposition,

$T(\rho) = \sum_{i}A_{i}\rho A_{i}^{\dagger}$

where the $\left\{A_{i}\right\}$ are the Kraus operators (and square matrices) and where

$\sum_{i}A_{i}^{\dagger}A_{i}=\textbf{1}$

and $T(\textbf{1})=\textbf{1}$.

Suppose we have two quantum channels, $r$ and $t$

$\begin{eqnarray*} r: \rho \to \sigma & \qquad \textrm{where} \qquad & \sigma=\sum_{i}A_{i}\rho A_{i}^{\dagger} \\ t: \sigma \to \tau & \qquad \textrm{where} \qquad & \tau=\sum_{j}B_{j}\sigma B_{j}^{\dagger} \end{eqnarray*}$

where the $\left\{A_{i}\right\}$ and $\left\{B_{i}\right\}$ are the Kraus operators for the channels respectively. We form the composite $t \circ r: \rho \to \tau$ where

$\begin{align} \tau & = \sum_{j}B_{j}\left(\sum_{i}A_{i}\rho A_{i}^{\dagger}\right)B_{j}^{\dagger} \notag \\ & = \sum_{i,j}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger} \\ & = \sum_{k}C_{k}\rho C_{k}^{\dagger} \notag \end{align}$

where $i \cdot j = k$. Since $A$ and $B$ are summed over separate indices the trace-preserving property is maintained, i.e. $\sum_{k} C_{k}^{\dagger}C_{k}=\textbf{1}$.

**Core argument**

From above, we note that

$\tau=\sum_{i,j}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger}$.

Suppose we only have two Kraus operators for each, i.e. $A_{1}, A_{2}, B_{1}, B_{2}$. Then

$\tau=B_{1}A_{1}\rho A_{1}^{\dagger}B_{1}^{\dagger} + B_{2}A_{1}\rho A_{1}^{\dagger}B_{2}^{\dagger} + B_{1}A_{2}\rho A_{2}^{\dagger}B_{1}^{\dagger} + B_{2}A_{2}\rho A_{2}^{\dagger}B_{2}^{\dagger}$.

($\tau$ of course has a matrix representation (in fact it is a square matrix representation). The following has nothing to do with the size of the matrix representation of $\tau$ and only has to do with the terms in the above summation.)

Using the subscripts as a guide, I can make a matrix

$\begin{equation*} \left( \begin{array}{c c} B_{1}A_{1}\rho A_{1}^{\dagger}B_{1}^{\dagger} & B_{1}A_{2}\rho A_{2}^{\dagger}B_{1}^{\dagger} \\[8pt] B_{2}A_{1}\rho A_{1}^{\dagger}B_{2}^{\dagger} & B_{2}A_{2}\rho A_{2}^{\dagger}B_{2}^{\dagger} \\[8pt] \end{array} \right). \end{equation*}$

This just happens to be the same dimension as the matrix representation of $\sigma \otimes \rho$. If I then do repeated composition I get,

$\begin{equation*} \left( \begin{array}{c c c} B_{1}A_{1}\rho A_{1}^{\dagger}B_{1}^{\dagger} & B_{1}A_{2}\rho A_{2}^{\dagger}B_{1}^{\dagger} & \cdots \\[8pt] B_{2}A_{1}\rho A_{1}^{\dagger}B_{2}^{\dagger} & B_{2}A_{2}\rho A_{2}^{\dagger}B_{2}^{\dagger} & \cdots \\[8pt] \vdots & \vdots & \ddots \end{array} \right). \end{equation*}$

The next step is simply to clarify the purpose. Suppose now that I take the output of a quantum channel and feed it back in on itself. In this case, $\left\{A_{i}\right\}=\left\{B_{i}\right\}$. Thus if we repeatedly apply the same channel $n$ times,

$\begin{equation} T(\rho) \circ T(\rho) \circ \cdots \circ T(\rho) = \sum_{i^{n}}(A_{i})^{n}\rho (A_{i}^{\dagger})^{n} \end{equation}$

we can take the terms of this expansion, form a matrix out of it, and that matrix (which may or may not have any physical significance) turns out to have the same dimension as the matrix representation of,

$T(\rho) \otimes T(\rho) \otimes \cdots \otimes T(\rho) = \bigotimes^{n}_{i=1} T(\rho)$.

Physically, the last equation is like applying $n$ copies of a channel simultaneously. In other words, there may be some kind of strange physical link between applying $n$ copies of a channel *simultaneously* and applying them in succession.

Questions in brief summary:Basically (and you can read the specific questions below) I need to know if a) the math for the core argument is right and b) what the immediate algebraic implications of it are (if there are any).

Question 1:The obvious question is, is this right, i.e. does the composition of quantum channels really have a representation that is morphic to a tensor product of quantum channels of a certain dimension (or is this obvious to a pure mathematician)? Is this generally true at the level of category theory?

Question 2:If it is right, the question that naturally follows is, what are the immediate algebraic implications (if any)?

Sub-question 2a:What kind of morphism is this on the level of monoids and/or categories, i.e. is it an isomorphism, epimorphism, etc.?

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