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Consider the complex $n$-by-$n$ matrices $M_n$. Suppose that $A_i$, for $i=1,\ldots,n^2$, satisfy $\mathrm{Tr}(A_i^* A_j)=\delta_{ij}$, so that together they form an orthonormal basis for $M_n$. Define a linear map $T \colon M_n \to M_n \otimes M_n$ by $T(A_i) = A_i \otimes A_i$.

Question: when is $T$ completely positive?

For example, if $A_i$ are the matrices with a single entry one and the rest zeroes in some fixed basis of $\mathbb{C}^n$, then $T$ is completely positive. In fact, I think these might be the only examples. If $T$ is completely positive, then the following are equivalent to $A_i$ being matrix units as in the above example:

  • each $A_i$ has rank one;
  • each positive semidefinite $A_i$ has trace one;
  • the set $\{0,A_1,\ldots,A_{n^2}\}$ is closed under multiplication;
  • $T(1)$ is idempotent;
  • $T^*(1) \leq 1$;
  • $T$ preserves trace.

These are sufficient conditions, but proving they are sufficient doesn't use $\mathrm{Tr}(A_i^* A_j)=\delta_{ij}$ at all. Are they necessary?

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  • $\begingroup$ do you assume $A_i\geq 0$? $\endgroup$ – Kate Juschenko May 9 '11 at 17:55
  • $\begingroup$ We can't assume $A_i\geq 0$ because self-adjoint matrices can never span all of $M_n$ (unless $n=1$). A good example to understand the question with is the following: $A_i$ are the matrices with a single entry one and the rest zeroes in some fixed basis of $\mathbb{C}^n$. Notice that not all $A_i$ in this example are $\geq 0$, indeed, they need not even be self-adjoint. $\endgroup$ – Chris Heunen May 9 '11 at 18:11
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    $\begingroup$ the second condition confused me, now I see what you mean $\endgroup$ – Kate Juschenko May 9 '11 at 18:26
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    $\begingroup$ I'm confused. Why are these conditions equivalent? Shouldn't we be able to find sets of $A_i$ so there are no positive semidefinite $A_i$ (just take the negative of any $A_i$ which is positive semidefinite), so condition (ii) holds, but where none of the other conditions holds. $\endgroup$ – Peter Shor May 10 '11 at 0:58
  • $\begingroup$ The text after the question proper was meant to be helpful extra information halfway to an answer, but I see how it could be confusing. I've edited it, hopefully it is clearer now. $\endgroup$ – Chris Heunen May 10 '11 at 9:41
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This is not a complete answer, but it might help.

The map $T: A_i \mapsto A_i \otimes A_i$ sends, by its very definition, the orthonormal family $(A_i)$ to an orthonormal family. It is therefore an isometry for the Hilbert-Schmidt norms.

But there are not that many completely positive maps $M_n \to M_m$ which are also isometries for the Hilbert-Schmidt-norms. Namely such a map is of the form $T(x)= D \pi(x)$, for some (not necessarily unital) $*$-homomorphism $\pi$ and some positive operator $D=T(1)\in M_m$ commuting with the range of $\pi$. This is an if-and-only-if condition provided that $\|D\|_{HS}=\sqrt n$. This statement is probably known. If you want I can expand the proof I have in mind.

This implies that such a map satisfies $Tr\circ T=c Tr$ for some positive $c=Tr(D)/n$, and more generally that for any $p>0$, $\|Tx\|_p = c_p \|x\|_p$ for $c_p=\|D\|_p/n^{1/p}$.

Coming back to your problem, I do not see how to conclude, you can already find a couple of necessary conditions on the $A_i$'s for the map $T(A_i)=A_i \otimes A_i$ to be completely positive.


Matthew asked for a proof of

A linear map $T:M_n\to M_m$ is completely positive and isometric for the Hilbert-Schmidt norm if and only if $T$ is of the form $T(x)= D \pi(x)$, for some (not necessarily unital) $*$-homomorphism $\pi$ and some positive operator $D=T(1)\in M_m$ commuting with the range of $\pi$, and such that $\|D\|_{HS}=\sqrt n$.

I only prove the "only if" direction. Assume that $T$ is cp and isometric for the Hilbert-Schmidt norm. Using the fact that $T$ is cp, by Stinespring's theorem, there is a (finite dimensional) Hilbert space $H$ and a linear map $V:\mathbb C^m \to H\otimes \mathbb C^n$ such that $T$ can be decomposed as $T(x)=V^* 1_H \otimes x V$. I claim that the assumption that $T$ is isometric implies that $VV^*$ is of the form $A \otimes 1_n$ for some positive $A \in B(H)$ (in particular $V V^*$ commutes with $1\otimes x$ for all $x \in M_n$). This will imply that $T(x) T(y) = T(1) T(xy) = T(xy) T(1)$ for all $x,y \in M_n$, and hence putting $\pi(x) = T(x) T(1)^{-1}$ (with the convention $0/0=0$) we get the proposition.

The claim is not complicated to check. By the trace property, $\langle Tx,Ty \rangle = Tr(VV^* (1\otimes x) V V^* (1\otimes y^* ))$. Writing $VV^* = \sum B_{i,j} \otimes e_{i,j}$, taking $x=e_{i,j}$, $y=e_{s,t}$, and using that $T$ preserves the scalar product, one gets $\langle e_{i,j},e_{s,t}\rangle= Tr(B_{s,i}B_{j,t})$. But $VV^*$ being self-adjoint, this becomes $\delta_{i,s}\delta_{j,t}= \langle B_{s,i},B_{t,j}\rangle$. This implies that $B_{s,i}=0$ if $s\neq i$ and that the matrices $B_{i,i}$ are all of Hilbert-Schmidt norm $1$, and that $\langle B_{i,i},B_{j,j}\rangle=1$. Thus (equality in Cauchy-Schwartz inequality), the $B_{i,i}$'s are all equal, to some matrix $U$. This proves the claim.

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  • $\begingroup$ Thanks Mikael! Isometry is necessary, but alone it is not sufficient: for example, if ($n=2$ and) $A_i$ are the normalized Pauli matrices, then $T$ is isometric but not completely positive (indeed not even positive). As you note, $T(1)$ seems to play an important role in getting a necessary condition. $\endgroup$ – Chris Heunen May 11 '11 at 20:16
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I think I finally have the answer to this question (some 7 years later, and 3 years after first hearing of the problem myself, but never mind). TL;DR: the matrices with a single entry one and the rest zeroes in some fixed basis of $\mathbb{C}^n$ are indeed the only examples. A detailed proof (with diagrams!) will appear shortly on the arXiv, and I will link to it once it's ready, but I will summarise the gist here for posterity.

A linear map $\delta:M_n \rightarrow M_n \otimes M_n$ defined by $\delta(A_{ij}) := A_{ij} \otimes A_{ij}$ for some orthonormal basis $(A_{ij})_{i,j=1,...,n}$ of $M_n$ is necessarily a special commutative $\dagger$-Frobenius algebra in the category $\operatorname{fHilb}$ of finite-dimensional Hilbert spaces and complex linear maps, with co-unit $\epsilon: M_n \rightarrow \mathbb{C}$ defined by setting $\epsilon(A_{ij}) := 1$ (together with their respective adjoints $\delta^\dagger$ and $\epsilon^\dagger$). If $\delta$ is a CP map, then so is $\epsilon$, and $(\delta,\epsilon,\delta^\dagger,\epsilon^\dagger)$ is also a special commutative $\dagger$-Frobenius algebra in the category $\operatorname{CPM}(\operatorname{fHilb})$ of finite-dimensional Hilbert spaces and completely positive maps.

The point is that every isometric comonoid in $\operatorname{CPM}(\operatorname{fHilb})$, such as $(\delta,\epsilon)$ would be when $\delta$ is CP, must necessarily involve pure maps $\delta,\epsilon$. This is because of the purity principle: whenever for pure CP maps $\Psi_i$ and $F$ we have $\sum_i \Psi_i = F$, necessarily $\Psi_i = q_i F$ for all $i$ and some coefficients $q_i \in \mathbb{R}^+$ (not necessarily all non-zero). The broad strokes of the proof are as follows.

  1. The isometry condition $\delta^\dagger \circ \delta = id$ implies that $\delta$ must be $\mathbb{R}^+$-linear combination (not necessarily convex) $\delta = \sum_i q_i V_i$ of isometries $V_i$ with pairwise orthogonal ranges (i.e. $V_j^\dagger \circ V_i$ is the identity $id$ for $i=j$ and the zero map for $i\neq j$).

  2. The unit laws $(id \otimes \epsilon)\circ \delta = id = (\epsilon \otimes id)\circ \delta $ imply that $\epsilon$ must be pure.

  3. By post-composing the associativity law with $\epsilon$ in all three possible ways, an associativity law for the isometries can be derived.

  4. By post-composing the associativity law for isometries with $\epsilon$, all isometries $V_i$ are identified, and orthogonality of ranges proves that $\delta$ must in fact be pure.

  5. Because $\delta$ and $\epsilon$ are pure, i.e. they arise from the comultiplication and counit of a special commutative $\dagger$-Frobenius algebra of $\operatorname{fHilb}$. Those are exactly the examples given by taking matrices $A_{ij} := |i\rangle\langle j|$ with a single entry one and the rest zero in some fixed orthonormal basis $\big(|i\rangle\big)_{i=1,...,n}$ of $\mathbb{C}^n$.

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