Timeline for Why is $\Phi\otimes \operatorname{Id}_n$ being positive on maximally entangled states sufficient to know that $\Phi$ is CP?
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17 events
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| May 10, 2022 at 9:02 | vote | accept | glS♦ | ||
| May 10, 2022 at 7:28 | comment | added | glS♦ | @NorbertSchuch so, about the title, I kinda agree. But the reason I didn't word it like that is that it would make a long title even longer, for no significant advantage. I figure someone looking for this kind of information would naturally land on a question with the current title anyway, and probably find what they are looking for. So I see no disadvantage in the current phrasing, even though technically speaking the title is not entirely accurate | |
| May 10, 2022 at 7:20 | comment | added | Norbert Schuch | Regarding "having different ways of proving results": I fully agree. Though in the way currently asked the question probably falls under "opinion-based" by SE interpretation. Probably asking for alternative proofs would be the right thing to do. | |
| May 10, 2022 at 7:17 | comment | added | Norbert Schuch | I still feel you are sweeping quite a bit of difficulty under the rug, e.g. by the relation you quote above (or something else.) (The alterative is that the way I prove Choi in my course is faaar to complicated ;) ). I agree it is all straightforward, but actually working it out is quite tedious. If not, I would be really interested in understanding it, precisely for the reason above. | |
| May 10, 2022 at 7:04 | comment | added | glS♦ | @NorbertSchuch I don't necessarily think it's complicated, no. The equivalence of $J(\Phi)=\sum_a v_a v_a^\dagger$ and $\Phi(X)=\sum_a A_a XA_a^\dagger$ is immedate eg looking at their components expressions and t the relation $J(\Phi)_{12,34}=\langle E_{13},\Phi(E_{24})\rangle$. Still, there might be methods that are more direct and use a different approach. Like the one you showed. But honestly, simplicity aside, I find it generally good to have different ways of proving the same results | |
| May 9, 2022 at 22:59 | comment | added | Norbert Schuch | Regarding "Kraus representation": The conditions for CP and TP are entirely separate. I would consider this clear, and an important point. I would call this a Kraus representation indep. of TP. | |
| May 9, 2022 at 22:58 | comment | added | Norbert Schuch | Of course you end up with the same channel, that's the point of the isomorphism. But to prove it you have to show that this is indeed the inverse map. This is quite a bit more work. Where would the proof be in your argument? Or are you allowed to use the isomorphism? Then I don't see what's complicated about the argument why n-positivity is enough. | |
| May 9, 2022 at 21:54 | comment | added | glS♦ | @NorbertSchuch I'm not sure what you mean regarding 1 and 2 together not proving that "the channel you get is the one you started with". The Kraus decomposition for $\Phi$ is completely equivalent to the eigendecomposition for $J(\Phi)$, when $v_a$ and $A_a$ are related as mentioned in the post, and such decomposition characterises the action of $\Phi$ itself. In what sense could you end up with a different channel with this procedure? | |
| May 9, 2022 at 21:51 | comment | added | glS♦ | @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 | |
| May 8, 2022 at 23:39 | comment | added | Norbert Schuch | ... 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, 2022 at 23:37 | comment | added | Norbert Schuch | 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, 2022 at 23:36 | review | Close votes | |||
| May 8, 2022 at 23:48 | |||||
| May 8, 2022 at 23:33 | answer | added | Norbert Schuch | timeline score: 6 | |
| May 8, 2022 at 23:31 | comment | added | glS♦ | @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 | |
| May 8, 2022 at 23:20 | comment | added | Norbert Schuch | 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, 2022 at 20:23 | answer | added | Danylo Y | timeline score: 1 | |
| May 8, 2022 at 19:20 | history | asked | glS♦ | CC BY-SA 4.0 |