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.

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    $\begingroup$ I might be misunderstanding your terminology: according to the Wikipedia article, a quantum channel is a completely-positive trace-preserving map on an ambient finite-dimensional C*-algebra ${\mathcal C}$. (I don't think calling $d$ the dimension of the $C^*$-algebra is standard, but could be wrong here.) Now it seems that what you are asking is whether the set of all such channels forms the set of arrows for a 1-object category whose object is ${\mathcal C}$. Is that right? $\endgroup$
    – Yemon Choi
    Jan 31, 2010 at 21:21
  • $\begingroup$ Yes. I reread your question a couple of times and I'm pretty sure that the answer is yes. Just to clarify, $\mathcal{C}$ has a specific "dimension" (I use this since it's a relic of the idea of talking about a $d$-dimensional Hilbert space - originally I was using Hilbert spaces as the objects, but then generalized to C$^{*}$-algebra. I felt it better covered any possible classical inputs if that makes sense. $\endgroup$
    – Ian Durham
    Jan 31, 2010 at 21:45

4 Answers 4


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.


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.


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.


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."

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!

  • $\begingroup$ The reason I'm using categories has to do with what I do with these things (which will be the subject of another question here as soon as I'm satisfied I've done this part right). Regarding "time evolution," sorry. I'm a quantum physicist. :-) I'm trying to use category theory to solve a particularly intractable problem in quantum information theory while simultaneously keeping the "physical" nature of the problem in mind. Your comment on the Kraus operators was particularly helpful and is much appreciated. $\endgroup$
    – Ian Durham
    Feb 1, 2010 at 1:45
  • $\begingroup$ Ian's follow-up question appeared at mathoverflow.net/questions/16077/… $\endgroup$ Feb 24, 2010 at 6:48
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    $\begingroup$ Dear Joshua, if you have five minutes for me, I would like to ask you for a favor: Ian has tried to start an nLab entry on quantum channels here: ncatlab.org/nlab/show/quantum+channel , but there is a general feeling that this is quite in need of some improvement. Myself, I don't know the literature on "quantum channels". Just judging from your reply above, I am getting the impression that you know what you are talking. So: could you maybe do us and the world a favor, hit the "edit" button on that nLab entry, move all the discussion you find the below some horizontal line and then... $\endgroup$ Feb 25, 2010 at 15:13
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    $\begingroup$ ... give a clean definition of quantum channels? I gather there is a general abstract definition and then various concrete realizations. I also gather all in all it is not a terribly mysterious idea. If you could just write out a nice clean definition and maybe give a pointer to some good authoragtive reference. That would be much appreciated. $\endgroup$ Feb 25, 2010 at 15:14

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.

  • $\begingroup$ Yeah, the original question was watered down because I was trying to figure out how to ask an appropriate question here. That failed miserably. My original point was I was using category theory to try to solve the quantum Birkhoff theorem problem. I've got alternate approaches now. Much of the discussion of this stuff has been transported to the nLab. $\endgroup$
    – Ian Durham
    Jun 15, 2010 at 12:23

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.

  • $\begingroup$ Well, that's actually an interesting question. In one sense, I am interested in having them preserve some kind of structure. But I'm also interested in having their categorical form preserve (show?) any extremal characteristics they might possess (maybe I should post this as a separate question). $\endgroup$
    – Ian Durham
    Feb 1, 2010 at 0:04
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    $\begingroup$ Since the concrete form of quantum channels used above is equivalent to the definition of quantum channels as completely positive, trace-preserving, linear maps, then all you have to do is prove the equivalence and use Yemon's argument rather than doing the direct calculation. $\endgroup$
    – Peter Shor
    Jun 5, 2010 at 20:19

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.

  • $\begingroup$ Yep. Bob is essentially the person who first introduced me to category theory so I'm aware of all that stuff. Actually, we've basically taken care of much of this over at the nLab. $\endgroup$
    – Ian Durham
    Jun 15, 2010 at 12:21

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