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(Notation) Let $\Phi$ be a generic quantum map sending states in $\mathbb{C}^n$ into states in $\mathbb{C}^m$. We say that $\Phi$ is positive when $\Phi(X)\ge0$ for any positive linear operator $X\in\mathrm{Lin}(\mathbb{C}^n)$. We say that $\Phi$ is completely positive (CP) when $\Phi\otimes \operatorname{Id}_k$ is a positive map for all $k\ge0$.

(A standard approach to proving that CP $\iff$ $n$-positive) It is well-known that $\Phi$ is CP iff its Choi representation, $$J(\Phi)\equiv (\Phi\otimes \operatorname{Id}_n)(|m\rangle\!\langle m|)\in\operatorname{Lin}(\mathbb{C}^m\otimes \mathbb{C}^n),$$ is positive semidefinite. Here $|m\rangle\equiv \sum_{i=1}^n |i,i\rangle$ is the (unnormalised) maximally entangled state. A standard way to show this is to observe that

  1. If $J(\Phi)$ is positive semidefinite then it admits an eigendecomposition $J(\Phi)=\sum_a v_a v_a^\dagger$ for some collection of vectors $v_a\in\mathbb{C}^m\otimes \mathbb{C}^n$;
  2. The eigendecomposition for $J(\Phi)$ corresponds to a Kraus-like decomposition for $\Phi$ itself: $\Phi(X)=\sum_a A_a X A_a^\dagger$ with $A_a$ being the linear operators with the same components as the vectors $v_a$.
  3. Such a Kraus-like decomposition can always be rewritten as a Stinespring-like representation $\Phi(X)=\operatorname{Tr}_1(VXV^\dagger)$ with $V\equiv \sum_a |a\rangle\otimes A_a$, and any map with such a representation is CP, because $$(\Phi\otimes \operatorname{Id}_k)(\mathbb{P}(|\Psi\rangle)) = \operatorname{Tr}_1\!\!\big[\mathbb{P}((V\otimes I_k)|\Psi\rangle)\big], \qquad \mathbb{P}(|\psi\rangle)\equiv |\psi\rangle\!\langle\psi|, \qquad \forall|\Psi\rangle\in \mathbb{C}^{n+k},$$ meaning the action of any finite extension of $\Phi$ on unit-rank projections returns the partial trace of a unit-rank projection, which is always a positive semidefinite operator.

(The question) Now, suppose I'm interested in proving the fact that $\Phi\otimes \operatorname{Id}_n$ sending maximally entangled states to valid states is sufficient to know that any extension of $\Phi$ sends physical states into physical states (i.e. that $\Phi$ is CP). The above approach does of course work, but it involves quite a bit of machinery to show something that on the face of it seems a rather simple statement.

Is there a simpler or more direct way to show that $(\Phi\otimes \operatorname{Id}_n)(|m\rangle\!\langle m|)$ being positive is sufficient to know that $\Phi\otimes \operatorname{Id}_k$ is a positive map for all $k$?

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  • $\begingroup$ This is not particularly complicated (and indeed you seem to be done after 2 as you get the Kraus operators, what's the point of 3?) This seems primarily opinion-based. $\endgroup$ May 8 at 23:20
  • $\begingroup$ @NorbertSchuch well, sure, the third point is not crucial, I probably use it because I prefer to think of maps via Stinespring. But the main rational of the question was to avoid having to pass by the eigendecomposition of the Choi altogether. Or at least, a more direct way to relate positivity on general states to the positivity on the maximally entangled state, that doesn't involve having to discuss different representations of the map etc $\endgroup$
    – glS
    May 8 at 23:31
  • $\begingroup$ I agree that the full Choi proof is quite complicated; indeed, 1 and 2 together don't prove at all that the channel you get is the channel you started with (nor does 3 prove it?), so the full proof would likely be longer. (At least, when I prove the full Choi isomorphism this is taking a while.) -- I posted an argument which might well be considered easier - all you need to know is Schmidt decompositions, I believe. $\endgroup$ May 8 at 23:37
  • $\begingroup$ ... I still don't get at all what 3 is about. I mean, it is clear that a map in Kraus form is CP, is it? (What does Kraus-"like" even mean? This is entirely a Kraus form, isn't it?) $\endgroup$ May 8 at 23:39
  • $\begingroup$ @NorbertSchuch yes I agree that 2 and 3 here are essentially equivalent. Both ultimately rely on the fact that $APA$ is positive if $P$ is. The reason I said "Kraus-like" is because I was not technically working with channels, i.e. $\Phi$ could be not trace-preserving, and then the decomposition is not strictly speaking a "Kraus decomposition" I think? i.e. the "Kraus operators" do not need to satisfy the normalisation $\sum_a A_a^\dagger A_a=I$. I'm not sure whether people would talk about a proper Kraus decomposition in this case $\endgroup$
    – glS
    May 9 at 21:51

2 Answers 2

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One alternative argument would be as follows:

  1. If there exists a $\rho\ge0$ such that $\Phi\otimes \mathrm{Id}_k(\rho)\not\ge0$, then there will also exist a pure $\vert\chi\rangle$ such that $\Phi\otimes \mathrm{Id}_k(\vert\chi\rangle\langle\chi\vert)\not\ge0$. (This follows immediately from convexity, e.g. by taking an ensemble decomposition of said $\rho$ -- it will contain one such $\vert\chi\rangle$.)

  2. Take the Schmidt decomposition of $\vert\chi\rangle$, and denote its Schmidt rank by $\ell$. Clearly, $\ell\le n$. Then, when considering $\Phi\otimes \mathrm{Id}_k(\vert\chi\rangle\langle\chi\vert)$, the extending space (the one with the $\mathrm{Id}$) can be compressed to a space with dimension $\ell$ (spanned by the Schmidt vectors). Call the compressed state $\vert\chi'\rangle$.

  3. Clearly, in the compressed space $\Phi\otimes Id_\ell(\vert\chi'\rangle\langle\chi'\vert)\not\ge0$.

This shows that $n$-positive implies $k$-positive for $k>n$. If, in addition, you want to make sure that $n$-positivity when applied only to the maximally entangled state $\vert\Omega\rangle$ is sufficient, then you can do the following:

  1. Assume wlog $\ell = n$. (Otherwise, embed into an $n$-dimensional space.) Write $\vert\chi'\rangle = (\mathrm{I}\otimes M)\vert\Omega\rangle$. Then, $$ (\mathrm{I}\otimes M)\,\big[(\Phi\otimes \mathrm{Id}_n)(\vert\Omega\rangle\langle\Omega\vert)\big]\,(\mathrm{I}\otimes M^\dagger) = (\Phi\otimes \mathrm{Id}_n)(\vert\chi'\rangle\langle\chi'\vert) \not\ge 0\ , $$ which proves that $(\Phi\otimes \mathrm{Id}_n)(\vert\Omega\rangle\langle\Omega\vert)\ge0$ implies $n$-positivity.

(Fun fact on the side: This argument can also be used to show that in order to check CP, it is sufficient to evaluate the action of the channel on any $\vert\chi'\rangle$ with maximal Schmidt rank, since in that case, $M$ is invertible, and the argument works both ways.)

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  • $\begingroup$ ah, nice, this is the kind of thing I was hoping for. One could also make a trivial modification to the argument to make it "positive" I think: $$(\Phi\otimes\operatorname{Id}_k)(|\chi\rangle\!\langle\chi|)=(I\otimes D)[(\Phi\otimes\operatorname{Id}_n)(|\Omega\rangle\!\langle\Omega|)](I\otimes D)$$ with $D$ diagonal matrix with the Schmidt coefficients of $\chi$, so that we get a direct relation between positivity on $|\Omega\rangle$ and positivity on arbitrary (pure) states $\endgroup$
    – glS
    May 9 at 21:47
  • $\begingroup$ If D is singular this is a one-way relation. But isn't this precisely my point 4? $\endgroup$ May 9 at 22:54
  • $\begingroup$ sure. I just like this equation because it gives a direct explicit relation answering the question: "how does the action of $\Phi\otimes\operatorname{Id}_k$ on a generic state $|\chi\rangle$ relate to the action on maximally entangled states?". It's essentially your point 4, yes, observing that the matrix $D$ is just the diagonal matrix whose elements are the Schmidt coefficients $\endgroup$
    – glS
    May 10 at 7:07
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    $\begingroup$ @gls ... or my 4 using the SVD of M :) In fact, this is one of the advanced questions I like asking in oral exams, if everything goes smoothly ;) (Will have to remember to delete this comment! :-o) $\endgroup$ May 10 at 7:17
  • $\begingroup$ @NorbertSchuch Actually something that always bugs me is whether this theorem implies 2-positivity is enough to prove CP, since your proof essentially says that if it's 2-positive then it's true for $k > 2$. I have never seen this connection stated clearly. $\endgroup$
    – Everiana
    Jun 1 at 1:37
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You can skip point 3 since $\Phi(X) = AXA^\dagger$ is a CP map and a sum of CP maps is again CP. The map $\Phi$ is CP because $(\Phi \otimes {\rm Id})(Z)= (A\otimes I)Z(A\otimes I)^\dagger$ – clearly a positive map.

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