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I would like to know how to answer following questions: Is there classical/quantum correlations in given bipartite pure/mixed state?

I have gathered several definitions. Some of them (it seems) lead to counterintuitive conclusions. I have marked them from 1 to 4. I would be glad if someone could point out mistakes (if there are any) in my logic.

$\underline{\text{Classical correlations}}$ are correlations which can be created by LOCC (local operations and classical communication).

  1. Do local operators acting on bipartite Hilbert space have following form: $U_{A} \otimes I_{B} \text{ and } I_{A} \otimes U_{B} $? If so then locality of these operators have nothing to do with spacial locality. Is that correct?

$\underline{\text{Quantum correlations}}$ are correlations which can not be created by aforementioned way.

  1. Would that mean that quantum correlations will be created by nonlocal(?) operators of form: $U_{A} \otimes U_{B}$? Is there any other way to create quantum correlations? Can classical correlations be created by action of this operator on a given state as a byproduct(at list in case of mixed state)?

I have heard that entanglement is not the only form of quantum correlations. Yet since at the moment I am not interested in exotic states but rather in general concepts I will ignore this knowledge. If this information can't be neglected please let me know.

Here are some statements which seem to be correct yet counterintuitive:

$\underline{\text{Pure states:}}$

  1. No pure (bipartite) state has classical correlations between its subsystems. Proof: If we start from direct product state $|A\rangle\otimes |B\rangle$ no local operator will be able to create any correlations between subsystems. Since classical communication will lead to creation of mixed state we are out of options.

    • Schmidt decomposition tells us whether given pure state is entangled(has quantum correlations) or separable.

$\underline{\text{Mixed states:}}$

  1. Every mixed bipartite state has classical correlations between its subsystems. Proof: Any mixed state can be written as a mixture of pure states. As I understand it this means that any density matrix can be diagonalized. Diagonalized matrix will have eigenvalues on its diagonal. Due to properties of density matrix, eigenvalues $\lambda_{i} \in [0,1]$ and their sum always equals 1. Hence these eigenvalues may be interpreted as probabilities corresponding to some pure states in a mixture. Presence of at least two nonzero eigenvalues would indicate presence of classical correlations between subsystems.

    • Separability tells us whether given mixed state has entanglement(i.e. quantum correlations) or not. For bipartite system one can use so called Peres-Horodecki criterion.
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Let's go through your list.

$\underline{\text{Classical correlations}}$ are correlations which can be created by LOCC (local operations and classical communication).

Yes, that is correct.

  1. Do local operators acting on bipartite Hilbert space have following form: $U_{A} \otimes I_{B} \text{ and } I_{A} \otimes U_{B} $?

Yes, that is correct, though this does not take into account classical communication, so the LOCC set is bigger than that. A full expression is rather clunky (since it needs to account for an arbitrary number of round trips for the classical communication, with a local unitary and projective measurement at each end), but there's good details on Wikipedia if you want them.

If so then locality of these operators have nothing to do with spacial locality. Is that correct?

It doesn't need to have anything to do with spatial locality. You normally enforce this by stipulating that the A and B subsystems are at remote locations and that their measurements occur sufficiently fast that they're within spacelike-separated regions of spacetime. If you don't enforce that, then 'locality' just becomes a marker of the different tensor factors of the state space.

$\underline{\text{Quantum correlations}}$ are correlations which can not be created by aforementioned way.

Generally speaking, yes.

  1. Would that mean that quantum correlations will be created by nonlocal(?) operators of form: $U_{A} \otimes U_{B}$?

Yes.

Is there any other way to create quantum correlations?

It depends on what you count as "ways". There's a bunch of quantum channels that create quantum correlations which are not of the form $U_{A} \otimes U_{B}$; as an example take a unitary channel of the form $U_{A} \otimes U_{B}$ and follow it up with some limited decoherence. Does that count?

Generally speaking, if you have a given quantum channel and you know that it creates quantum correlations, you'll typically be able to decompose it in such a fashion, but that decomposition does not necessarily speak to what is "really going on" inside the system (and you will typically not have any access to that).

I would therefore say that the answer to the question is morally yes, but that it's too ill-defined to say anything concrete.

Can classical correlations be created by action of this operator on a given state as a byproduct (at list in case of mixed state)?

Frankly, this is completely unclear to me.

I have heard that entanglement is not the only form of quantum correlations.

This is correct. You probably want to look at quantum discord as the core example of such correlations.

Yet since at the moment I am not interested in exotic states but rather in general concepts I will ignore this knowledge. If this information can't be neglected please let me know.

That depends on what you mean by "can't be neglected". There are many important conceptual settings in which the information can't be neglected. There are many important conceptual settings in which it can. Some examples of the former are places where quantum contextuality is an important consideration, with the Kochen-Specker theorem taking a role similar to that of Bell's theorem in the study of entanglement. Some examples of the latter are the study of entanglement. Which side you want to listen to is a personal choice.

$\underline{\text{Pure states:}}$

  1. No pure (bipartite) state has classical correlations between its subsystems. Proof: If we start from direct product state $|A\rangle\otimes |B\rangle$ no local operator will be able to create any correlations between subsystems. Since classical communication will lead to creation of mixed state we are out of options.

Yes, that is correct. You may be interested in purity as a quantum resource theory.

$\quad \cdot$ Schmidt decomposition tells us whether given pure state is entangled (has quantum correlations) or separable.

That is correct.

$\underline{\text{Mixed states:}}$

  1. Every mixed bipartite state has classical correlations between its subsystems.

No, this is incorrect. To get a counter-example, simply take any two mixed density matrices $\rho_A$ and $\rho_B$, and set $\rho = \rho_A\otimes \rho_B$ as a mixed separable state with no classical correlations between its subsystems.

Proof: Any mixed state can be written as a mixture of pure states. As I understand it this means that any density matrix can be diagonalized. Diagonalized matrix will have eigenvalues on its diagonal. Due to properties of density matrix, eigenvalues $\lambda_{i} \in [0,1]$ and their sum always equals 1. Hence these eigenvalues may be interpreted as probabilities corresponding to some pure states in a mixture.

So far so true, but

Presence of at least two nonzero eigenvalues would indicate presence of classical correlations between subsystems.

This doesn't follow from anything. Study the counter-example above to see why this fails.

$\quad \cdot$ Separability tells us whether given mixed state has entanglement (i.e. quantum correlations) or not.

Entangled states are, by definition, those states that are not separable. So yes. But it tells you very little.

For bipartite system one can use so called Peres-Horodecki criterion.

That is one possible criterion, but it is not infallible at detecting entanglement. In other words, you get the implication $$ \rho\text{ is separable} \implies \text{its partial transpose is positive semidefinite} $$ and its contraposition $$ \text{the partial transpose of }\rho\text{ is indefinite} \implies \rho\text{ is entangled}, $$ but you don't get the converse, which would be more useful, i.e. it's possible that $\rho$ is entangled but you've just chosen a basis in which the partial transpose is still positive semidefinite.

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  • $\begingroup$ Thank you very much for a great answer! Once I revisited 4th part I do agree with your counterexample. For some reason I used to think that pure state represented as a density matrix could always be rewritten as a tensor product of two density matrices. However now I see that this is clearly not the case. $\endgroup$ – Yaroslav Shustrov Sep 19 '18 at 8:15

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