Quantum channels as categories: question 1.

Question

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. $$\sum_{k} C_{k}^{\dagger}C_{k}=\mathbf{1}.$$ 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). $\square$

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.

Answer 1

Phrasing this in terms of categories is kind of misleading: A category with a single object is just a monoid (associative binary operation with identity). So, per Yemon Choi's correction, you are just trying to demonstrate that the set of quantum channels $L(\mathcal{H}) \to L(\mathcal{H})$ forms a monoid. [Here I'm assuming that "channel" implies CPTP.]

This requires three things:

1. Closure: The product of two channels is a channel.
2. Identity: The identity operator is a channel.
3. Associativity: Multiplication of channels is associative.

1:

There are two different ways to prove closure. You used the characterization of channels as maps given by that Kraus operator form ($\sum_{i} A_{i}\rho A_{i}^{\dagger}$). This works, although I don't feel you made the construction of the $C_k$ from the $A_i$ and $B_j$ clear enough. It also requires that you've already proven this characterization (CPTP linear map $\Leftrightarrow$ Kraus operator form).

You could instead directly use the characterization of a channel as a CPTP linear map. This way is probably easier: It is immediately clear that if r is (a linear map which takes positive matrices to positive matrices and preserves trace) and t is (a linear map which takes positive matrices to positive matrices and preserves trace) then $t\circ r$ will be (a linear map which takes positive matrices to positive matrices and preserves trace). That does it.

2:

I really wish you hadn't said "time evolution". :-) But you basically have the right idea: the identity in this situation is, well, the identity map, which is obviously linear and CPTP.

3:

Practically speaking, you almost never have to prove associativity. This is because as long as your maps are functions deep down on the inside, associativity is an immediate consequence of associativity of function composition. This is one of those cases.

So in response to your question "Is there a clearer or more concise proof?" I would say "Absolutely."

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But again, I think it's unnecessary to put this in the context of categories. UNLESS you plan on generalizing to channels between distinct spaces. Then the concept of a category gains its power.

Good luck!

Answer 2

As I see it, this posted question and some aspects of the answers turn an important but straightforward fact into something needlessly complicated and less general.

Let $\mathcal{A}$ (Alice) and $\mathcal{B}$ (Bob) be $C^*$-algebras of observables, or better yet, von Neumann algebras of observables. Let $\mathcal{A}^\#$ denote the dual space of (finite but not necessarily positive) states on $\mathcal{A}$, and in the von Neumann algebra case let ${}^\#\mathcal{A}$ denote the predual space of normal states. Then a quantum channel, to model a message from Alice to Bob, is a completely positive, unital map $$E:\mathcal{B} \longrightarrow \mathcal{A}.$$ The corresponding CPTP map on states is the transpose: $$E^\#:\mathcal{A}^\# \longrightarrow \mathcal{B}^\#$$ in the von Neumann algebra case, $E$ should be normal and have a pre-transpose: $${}^\#E:{}^\#\mathcal{A} \longrightarrow {}^\#\mathcal{B}$$ Yes, quantum channels should form a category, and yes they do. Yes, you can restrict to the one-object subcategory where the object is $B(\mathcal{H})$. You need to check that quantum channels include the identity (they do) and you need to check that they are closed under composition. It is immediate that preserving 1 (the unital condition) is closed under composition. As for complete positivity, the condition is that $$E \otimes I_\mathcal{C}:\mathcal{B} \otimes \mathcal{C} \longrightarrow \mathcal{A} \otimes \mathcal{C}$$ preserves positive states for all $\mathcal{C}$. Closure of this condition under composition isn't quite immediate, but it's still very easy.

Associativity is immediate because quantum channels are functions.

Answer 3

I'm not sure if it answers your question, but if ${\mathcal C}$ is a finite-dimensional $C^*$-algebra, and if $f$ and $g$ are completely positive, trace-preserving, linear maps from ${\mathcal C}\to {\mathcal C}$, then the composite map $g\circ f$ is going to be completely positive, trace-preserving, and linear. So one can indeed define a certain category to have ${\mathcal C}$ as its sole object, and have as its set of morphisms the collection of all CP-TP linear maps from ${\mathcal C}$ to itself. The associativity rule comes for free just because composition of functions is an associative operation.

If, on the other hand, you define quantum channels to be maps of a certain concrete form (rather than as being maps which preserve certain structure) then probably one needs to do a direct calculation similar to yours.

This might not be quite what you were asking, but I hope it helps.

Answer 4

Plenty of recent work addresses encoding quantum protocols (and hence quantum channels, I guess) using category theory. Check out the work of Bob Coecke and any one related to him. A good starting point is A categorical semantics of quantum protocols by Samson Abramsky and Bob Coecke. It has many of the ingredients you are probably looking for. Or perhaps Kindergarten Quantum Mechanics also by Bob Coecke. There's also Dagger Categories by Peter Selinger, capturing the underlying categorical structures: Dagger compact closed categories and completely positive maps.