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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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1  
I don't understand what's a space and what's a linear transformation in your notation. Could you clean it up a little? –  Qiaochu Yuan Feb 22 '10 at 19:37
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Since it is hard for me to parse this question, I will arrogantly assume that it is hard for many other MOers as well. I am going to ask some questions in the interest of clarification. I am not an expert on quantum anything. It is fairly well-established here on MO that if you think your question is clear enough to an expert, you need not clarify it to the point where it makes sense to a general mathematical audience. But if you're looking to get more answers than you have, such clarification will increase the number of people who might be able to help you. –  Pete L. Clark Feb 24 '10 at 7:56
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Regarding the statement in question 6, I was lost completely at "such a combination". As far as I could tell, you haven't specified what the combination was supposed to be. Perhaps I don't know enough physics to get anything from the clause "Since quantum channels can carry classical information as well". From my reading, you had one map \Phi between C^* algebras A, B and then this new map \Phi between L(H_A) \otimes C(X), L(H_B) which you define in that statement, and like Pete L. Clark I couldn't connect the two maps. ... –  j.c. Feb 24 '10 at 22:34
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@ID: None of my comments were snide. All of them were meant to be helpful. I can't even try to teach/help you until I understand what you're talking about. Do you perhaps think that we're affecting a lack of understanding? We're not: we're really trying as hard as we can. Since you seem to get offended by everything I say, I don't think I can be of any help to you. I'm sorry for that. –  Pete L. Clark Feb 25 '10 at 1:27
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One problem with tossing around the word "matrix" in the infinite-dimensional context is that it really only makes sense for a linear transformation between separable Hilbert spaces (in the sense that one can write down "components" in the usual way); in the non-separable case or in the case of arbitrary infinite-dimensional vector spaces it seems to me to be better not to think componentwise since, for example, the product of two "infinite matrices" is not necessarily well-defined. –  Qiaochu Yuan Feb 25 '10 at 5:15

2 Answers 2

I think the answer is "sort of". A mathematical way to think about the "Kraus" operator is as follows. Set C = B(H) (with H finite-dimensional if you wish). Then, assuming the sum in the Kraus operator is finite (again, it could be infinite if you wish) then we can define a map $A:H \rightarrow H^n$ by a "column" operator $$A(x) = ( A_i^\dagger(x) )_{i=1}^n\qquad (x\in H).$$ Then your Kraus operator is $$ T(\rho) = \sum_i A_i \rho A_i^\dagger = A^\dagger (\rho\otimes 1) A. $$ Here $\rho\otimes 1$ is the operator on $H^n$ given by applying $\rho$ to each coordinate: the notation is explained by observing that $H^n$ is the Hilbert space tensor product $H\otimes \ell^2_n$.

So, if you have another operator $S$ given by $(B_j)_{j=1}^m$ we can form $B(x) = (B_j^\dagger(x))$ and then $$ S(T(\rho)) = B^\dagger(T(\rho)\otimes 1)B = B^\dagger (A^\dagger\otimes 1)(\rho\otimes 1\otimes 1)(A\otimes 1)B = C^\dagger (\rho\otimes 1) C,$$ where $C = (A\otimes 1)B : H\rightarrow H^{nm} = H \otimes \ell^2_n \otimes \ell^2_m$.

So, you see why tensors appear. But this is a bit different to what you had...

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OK, dumb question. In your answer, what does $l^{2}_{n}$ represent? –  Ian Durham Feb 23 '10 at 13:04
    
Sorry: this is just an n-dimensional Hilbert space. So $\mathbb C^n$ with the usual $\ell^2$ norm on it. –  Matthew Daws Feb 23 '10 at 14:20
    
@Matt: Seeing as you seem to have done the best out of anyone here at getting what ID was getting at - your calculation is in response to Ian's Q1 right? Is it the case that the composition and the t.p., while not isomorphic, are related by some kind of "intertwine & ampliate"? –  Yemon Choi Feb 25 '10 at 4:33
    
Yemon, yes, vaguely I guess. The pure "tensor" approach would be to look at the map $(A\otimes B)$ I think, but the composition seems to lead to looking at $(A\otimes 1)B$, which is perfectly natural, but not quite "tensoring". If you look at what Ian had originally, in his 2nd big matrix, to get the final solution, you need to sum all the entries: so the fact it's arranged in a matrix at all is a bit misleading. –  Matthew Daws Feb 25 '10 at 8:33

I do not really follow your notation and I know close to nothing about quantum channels, but it is impossible (assuming only that $T(\rho)$ is a linear map and the symbols $\circ$ and $\otimes$ mean what they usually mean) that $$T(\rho) \circ T(\rho) \circ \cdots \circ T(\rho)$$ be the same as $$T(\rho) \otimes T(\rho) \otimes \cdots \otimes T(\rho)$$ because the domains of the two maps are of different dimension (except in the case where the domain of $T(\rho)$ is of dimension $1$...)

LATER: Another shot, upgraded from a comment below. If $A=\sum_{i=1}^nA_i$ and $B=\sum_{j=1}^mB_i$ are linear maps $V\to V$, with $V$ a vector space, then $$A\circ B=\sum_{\substack{1\leq i\leq n\\1\leq j\leq m}}A_i\circ B_j$$ and $$A\otimes B=\sum_{\substack{1\leq i\leq n\\1\leq j\leq m}}A_i\otimes B_j$$ as a direct consequence of the facts that both $\circ$ and $\otimes$ are bilinear operations. From what little I understand in the question and comments here, I think this is the 'similarity' observed...

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Yes, I realize that the size of the resulting matrix in each case is completely different. What I'm saying above is that you can take the terms from the expansion of the first and can make a matrix that just happens to have the same dimension as the tensor expansion. And I want to know what that implies. –  Ian Durham Feb 23 '10 at 12:25
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I have no idea what that means. –  Mariano Suárez-Alvarez Feb 23 '10 at 13:39
    
I'm not yet sure what ID means by $\circ$, but as you say it must not be the ordinary composition of linear maps. (It seems to correspond to some physical juxtaposition of the same quantum channel with itself $n$ times, but I don't understand what that means.) –  Pete L. Clark Feb 24 '10 at 8:24
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@ID, Pete is talking about matrices. –  Mariano Suárez-Alvarez Feb 25 '10 at 1:02
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@Ian: if $A=\sum_{i=1}^nA_i$ and $B=\sum_{j=1}^mB_j$ are linear operators, one the sum of $n$ and the other the sum of $m$ operators, then $A\circ B=\sum_{\substack{1\leq i\leq n\\1\leq j\leq m}}A_i\circ B_j$, and $A\otimes B=\sum_{\substack{1\leq i\leq n\\1\leq j\leq m}}A_i\otimes B_j$. This is just because both composition of linear maps and tensor product of linear maps are bilinear operations. –  Mariano Suárez-Alvarez Feb 25 '10 at 4:22

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