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Consider spin population observables, a separable state can be correlated or uncorrelated. For example, if I consider the two particle state $$|+\rangle|-\rangle$$ it is separable and spin measurements will be correlated, because measuring the spin of the first particle will always give +1, while for spin two we will always get -1: here spin populations are perfectly anticorrelated. However $$|+\rangle(|+\rangle+|-\rangle)$$ shows no correlations between spin measurement.

The question is : Are uncorrelated states (for two particles or more, spin half or more) always separable ?

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    $\begingroup$ You can use | and \rangle to create kets, e.g. $|+\rangle$. $\endgroup$ Commented May 14, 2021 at 10:55
  • $\begingroup$ I think there is some confusion here as to which form of entanglement you are dealing with exactly. If I understand the question, you are not considering a two-qubit system, but rather a two-particle/mode system in which each mode carries a two-dimensional degree of freedom. In other words, your state is really $a_{1,+}^\dagger+a^\dagger_{2,-}$. When you say that your $|+,-\rangle$ is "correlated", you are thinking about this state as an entangled state of the form $|1,+\rangle+|2,-\rangle$, where the first dof specified the mode and the second dof the spin state $\endgroup$
    – glS
    Commented May 15, 2021 at 9:57
  • $\begingroup$ in other words: you should clarify what exactly you mean with "(un)correlated states" here. Which kind of correlation are you asking about exactly? $|+\rangle|-\rangle$ is a product state and thus uncorrelated wrt local measurements on first and second mode, but it's fully correlated wrt to "local" mesurements performed on the which-mode and which-spin dofs $\endgroup$
    – glS
    Commented May 15, 2021 at 9:58
  • $\begingroup$ Indeed I was thinking of the correlation wrt to "local" mesurements performed on the which-mode and which-spin dofs, not wrt to local measurements on fir and second mode, choosing a product state was not the best idea to convey my problem. $\endgroup$
    – DarkBulle
    Commented May 15, 2021 at 10:25
  • $\begingroup$ "When you say that your |+,−⟩ is "correlated", you are thinking about this state as an entangled state of the form |1,+⟩+|2,−⟩" I was thinking of a state of the form |1,+⟩x|2,−⟩, I guess it is the same if you extend the states characterizing mode 1 and 2 to the whole hilbert space $\endgroup$
    – DarkBulle
    Commented May 15, 2021 at 10:27

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In the bipartite case (only two particle), correlation in spin measurements implies a state of the form $$ |\psi\rangle = \frac{1}{\sqrt{2}} (|+\rangle|-\rangle+|-\rangle|+\rangle) , $$ for example, because the measurement outcome after the first measurement already tells you what the measurement outcome of the second measurement will be. As such, it represents an entangled state. When you have only one of these terms, you cannot really talk about a correlation, because each measurement only ever produces one specific value.

Bipartite entanglement always implies some form of correlation. By simple logic implication, the lack of correlation in a bipartite system must then imply separability.

As argued in the comments, the situation for more than two particles is more complicated. If you only measure the spin of two particles and don't see any correlations, you may still have a system with some form of multipartite entanglement.

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    $\begingroup$ "Entanglement always implies correlation" - while this seems obvious, one might ask whether this is actually they case. Here's an answer by glS that gives an argument that this is true. $\endgroup$
    – ACuriousMind
    Commented May 14, 2021 at 11:33
  • $\begingroup$ Isn't $|+-+\rangle+|-++\rangle$ a non separable state where spin 3 is not correlated to spin 1 or 2 while spin 1 and 2 are perfectly anticorrelated ? Entanglement always imply correlation can maybe be nuanced with something like a "certain degree of correlation" ? or am I wrong ? $\endgroup$
    – DarkBulle
    Commented May 14, 2021 at 11:35
  • $\begingroup$ I meran you didn't say perfect correlations so I guess you're right $\endgroup$
    – DarkBulle
    Commented May 14, 2021 at 11:41
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    $\begingroup$ This is only true for pure states. $\endgroup$ Commented May 15, 2021 at 11:41
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    $\begingroup$ @glS The first sentence of the answer. (Well, it is incorrect in more than a way, but crucially, it says that correlations imply entanglement. This is incorrect.) $\endgroup$ Commented May 15, 2021 at 22:28

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