# Why is $\Phi\otimes \operatorname{Id}_n$ being positive on maximally entangled states sufficient to know that $\Phi$ is CP?

(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 $$k$$?

• 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. May 8 at 23:20
• @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
– glS
May 8 at 23:31
• 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. May 8 at 23:37
• ... 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?) May 8 at 23:39
• @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
– glS
May 9 at 21:51

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.)

• 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
– glS
May 9 at 21:47
• If D is singular this is a one-way relation. But isn't this precisely my point 4? May 9 at 22:54
• 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
– glS
May 10 at 7:07
• @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) May 10 at 7:17
• @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. Jun 1 at 1:37

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.