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A quantum channel is a mapping between 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 CPTP maps. The operator spaces can be interpreted as $C^{*}$-algebras 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, $d$. 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.

Proposition A quantum channel given by $t: L(\mathcal{H}) \to L(\mathcal{H})$, together with the $d$-dimensional $C^{*}$-algebra, $\mathcal{C}$, on which it acts, forms a category we call $\mathrm{\mathbf{Chan}}(d)$ where $\mathcal{C}$ is the sole object and $t$ is the sole arrow.

Proof: Consider the quantum channels

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

where the usual properties of such channels are assumed (e.g. trace preserving, etc.). We form the composite $t \circ r: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\tau})$ 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}

and the $A_{i}$, $B_{i}$, and $C_{i}$ are Kraus operators.

Since $A$ and $B$ are summed over separate indices the trace-preserving property is maintained, i.e. $\sum_{k} C_{k}^{\dagger}C_{k}=1$ (not sure how to do a blackboard 1 on this site - \mathbb doesn't work since it's a package). $\sum_{k} C_{k}^{\dagger}C_{k}=\mathbf{1}.$$For a similar methodology see Nayak and Sen (http://arxiv.org/abs/0605041). We take the identity arrow, $1_{\rho}: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\rho})$, to be the time evolution of the state$\rho$in the absence of any channel. Since this definition is suitably general we have that $t \circ 1_{A}=t=1_{B} \circ t \quad \forall \,\, t: A \to B$. Consider the three unital quantum channels $r: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\sigma})$, $t: L(\mathcal{H}_{\sigma}) \to L(\mathcal{H}_{\tau})$, and $v: L(\mathcal{H}_{\tau}) \to L(\mathcal{H}_{\upsilon})$ where $\sigma=\sum_{i}A_{i}\rho A_{i}^{\dagger}$, $\tau=\sum_{j}B_{j}\sigma B_{j}^{\dagger}$, and $\eta=\sum_{k}C_{k}\tau C_{k}^{\dagger}$. We have $\begin{align} v \circ (t \circ r) & = v \circ \left(\sum_{i,j}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger}\right) = \sum_{k}C_{k} \left(\sum_{i,j}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger}\right) C_{k}^{\dagger} \notag \\ & = \sum_{i,j,k}C_{k}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger}C_{k}^{\dagger} = \sum_{i,j,k}C_{k}B_{j}\left(A_{i}\rho A_{i}^{\dagger}\right)B_{j}^{\dagger}C_{k}^{\dagger} \notag \\ & = \left(\sum_{i,j,k}C_{k}B_{j}\tau B_{j}^{\dagger}C_{k}^{\dagger}\right) \circ r = (v \circ t) \circ r \notag \end{align}$ and thus we have associativity. Note that similar arguments may be made for the inverse process of the channel if it exists (it is not necessary for the channel here to be reversible). End proof (\Box doesn't seem to work here either).$\square$Question 1: Am I doing the last line in the associativity argument correct and/or are there any other problems here? Is there a clearer or more concise proof? I have another question I am going to ask as a separate post about a construction I did with categories and groups that assumes the above is correct but I didn't want to post it until I made sure this is correct. I'm new here, so feel free to change the tags or point out anything that isn't appropriate. 2 Added a note about Kraus operators. A quantum channel is a mapping between 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 CPTP maps. The operator spaces can be interpreted as $C^{*}$-algebras 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,$d$. 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. Proposition A quantum channel given by $t: L(\mathcal{H}) \to L(\mathcal{H})$, together with the$d$-dimensional $C^{*}$-algebra,$\mathcal{C}$, on which it acts, forms a category we call$\mathrm{\mathbf{Chan}}(d)$where$\mathcal{C}$is the sole object and$t$is the sole arrow. Proof: Consider the quantum channels $\begin{eqnarray*} r: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\sigma}) & \qquad \textrm{where} \qquad & \sigma=\sum_{i}A_{i}\rho A_{i}^{\dagger} \\ t: L(\mathcal{H}_{\sigma}) \to L(\mathcal{H}_{\tau}) & \qquad \textrm{where} \qquad & \tau=\sum_{j}B_{j}\sigma B_{j}^{\dagger} \end{eqnarray*}$ where the usual properties of such channels are assumed (e.g. trace preserving, etc.). We form the composite $t \circ r: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\tau})$ where $\begin{align} \tau & = \sum_{j}B_{j}\left(\sum_{i}A_{i}\rho A_{i}^{\dagger}\right)B_{j}^{\dagger} \notag \\ & = \sum_{i,j}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger} \\ & = \sum_{k}C_{k}\rho C_{k}^{\dagger} \notag \end{align}$ and the $A_{i}$, $B_{i}$, and $C_{i}$ are Kraus operators. Since$A$and$B$are summed over separate indices the trace-preserving property is maintained, i.e. $\sum_{k} C_{k}^{\dagger}C_{k}=1$ (not sure how to do a blackboard 1 on this site - \mathbb doesn't work since it's a package). For a similar methodology see Nayak and Sen (http://arxiv.org/abs/0605041). We take the identity arrow, $1_{\rho}: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\rho})$, to be the time evolution of the state$\rho$in the absence of any channel. Since this definition is suitably general we have that $t \circ 1_{A}=t=1_{B} \circ t \quad \forall \,\, t: A \to B$. Consider the three unital quantum channels $r: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\sigma})$, $t: L(\mathcal{H}_{\sigma}) \to L(\mathcal{H}_{\tau})$, and $v: L(\mathcal{H}_{\tau}) \to L(\mathcal{H}_{\upsilon})$ where $\sigma=\sum_{i}A_{i}\rho A_{i}^{\dagger}$, $\tau=\sum_{j}B_{j}\sigma B_{j}^{\dagger}$, and $\eta=\sum_{k}C_{k}\tau C_{k}^{\dagger}$. We have $\begin{align} v \circ (t \circ r) & = v \circ \left(\sum_{i,j}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger}\right) = \sum_{k}C_{k} \left(\sum_{i,j}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger}\right) C_{k}^{\dagger} \notag \\ & = \sum_{i,j,k}C_{k}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger}C_{k}^{\dagger} = \sum_{i,j,k}C_{k}B_{j}\left(A_{i}\rho A_{i}^{\dagger}\right)B_{j}^{\dagger}C_{k}^{\dagger} \notag \\ & = \left(\sum_{i,j,k}C_{k}B_{j}\tau B_{j}^{\dagger}C_{k}^{\dagger}\right) \circ r = (v \circ t) \circ r \notag \end{align}$ and thus we have associativity. Note that similar arguments may be made for the inverse process of the channel if it exists (it is not necessary for the channel here to be reversible). End proof (\Box doesn't seem to work here either). Question 1: Am I doing the last line in the associativity argument correct and/or are there any other problems here? Is there a clearer or more concise proof? I have another question I am going to ask as a separate post about a construction I did with categories and groups that assumes the above is correct but I didn't want to post it until I made sure this is correct. I'm new here, so feel free to change the tags or point out anything that isn't appropriate. 1 Quantum channels as categories: question 1. A quantum channel is a mapping between Hilbert spaces, $\Phi : L(\mathcal{H}_{A}) \to L(\mathcal{H}_{B})$, where $L(\mathcal{H}_{i})$ is the family of operators on $\mathcal{H}_{i}$. The operator spaces can be interpreted as $C^{*}$-algebras 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,$d$. 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. Proposition A quantum channel given by $t: L(\mathcal{H}) \to L(\mathcal{H})$, together with the$d$-dimensional $C^{*}$-algebra,$\mathcal{C}$, on which it acts, forms a category we call$\mathrm{\mathbf{Chan}}(d)$where$\mathcal{C}$is the sole object and$t$is the sole arrow. Proof: Consider the quantum channels $\begin{eqnarray*} r: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\sigma}) & \qquad \textrm{where} \qquad & \sigma=\sum_{i}A_{i}\rho A_{i}^{\dagger} \\ t: L(\mathcal{H}_{\sigma}) \to L(\mathcal{H}_{\tau}) & \qquad \textrm{where} \qquad & \tau=\sum_{j}B_{j}\sigma B_{j}^{\dagger} \end{eqnarray*}$ where the usual properties of such channels are assumed (e.g. trace preserving, etc.). We form the composite $t \circ r: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\tau})$ where $\begin{align} \tau & = \sum_{j}B_{j}\left(\sum_{i}A_{i}\rho A_{i}^{\dagger}\right)B_{j}^{\dagger} \notag \\ & = \sum_{i,j}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger} \\ & = \sum_{k}C_{k}\rho C_{k}^{\dagger} \notag \end{align}$. Since$A$and$B$are summed over separate indices the trace-preserving property is maintained, i.e. $\sum_{k} C_{k}^{\dagger}C_{k}=1$ (not sure how to do a blackboard 1 on this site - \mathbb doesn't work since it's a package). For a similar methodology see Nayak and Sen (http://arxiv.org/abs/0605041). We take the identity arrow, $1_{\rho}: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\rho})$, to be the time evolution of the state$\rho$in the absence of any channel. Since this definition is suitably general we have that $t \circ 1_{A}=t=1_{B} \circ t \quad \forall \,\, t: A \to B$. Consider the three unital quantum channels $r: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\sigma})$, $t: L(\mathcal{H}_{\sigma}) \to L(\mathcal{H}_{\tau})$, and $v: L(\mathcal{H}_{\tau}) \to L(\mathcal{H}_{\upsilon})$ where $\sigma=\sum_{i}A_{i}\rho A_{i}^{\dagger}$, $\tau=\sum_{j}B_{j}\sigma B_{j}^{\dagger}$, and $\eta=\sum_{k}C_{k}\tau C_{k}^{\dagger}$. We have $\begin{align} v \circ (t \circ r) & = v \circ \left(\sum_{i,j}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger}\right) = \sum_{k}C_{k} \left(\sum_{i,j}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger}\right) C_{k}^{\dagger} \notag \\ & = \sum_{i,j,k}C_{k}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger}C_{k}^{\dagger} = \sum_{i,j,k}C_{k}B_{j}\left(A_{i}\rho A_{i}^{\dagger}\right)B_{j}^{\dagger}C_{k}^{\dagger} \notag \\ & = \left(\sum_{i,j,k}C_{k}B_{j}\tau B_{j}^{\dagger}C_{k}^{\dagger}\right) \circ r = (v \circ t) \circ r \notag \end{align}\$

and thus we have associativity. Note that similar arguments may be made for the inverse process of the channel if it exists (it is not necessary for the channel here to be reversible). End proof (\Box doesn't seem to work here either).

Question 1: Am I doing the last line in the associativity argument correct and/or are there any other problems here? Is there a clearer or more concise proof? I have another question I am going to ask as a separate post about a construction I did with categories and groups that assumes the above is correct but I didn't want to post it until I made sure this is correct.

I'm new here, so feel free to change the tags or point out anything that isn't appropriate.