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Consider a bipartite state $\rho$, and denote with $\Pi^A\equiv \{\Pi^A_a\}_a$ and $\Pi^A\equiv\{\Pi^B_b\}_b$ local projective measurements. Let the associated joint probability distribution be $p_\Pi$, and denote with $p_{\Pi,A},p_{\Pi,B}$ the marginal distributions: $$p_\Pi(a,b)\equiv \operatorname{Tr}[(\Pi^A_a\otimes \Pi^B_b)\rho], \quad p_{\Pi,A}(a) \equiv \operatorname{Tr}[(\Pi^A_a\otimes I)\rho], \quad p_{\Pi,B}(b) \equiv \operatorname{Tr}[(I\otimes \Pi^B_b)\rho]. \tag A$$ We say that the measurement outcomes are uncorrelated if $p_\Pi(a,b)=p_{\Pi,A}(a)p_{\Pi,B}(b)$.

If $\rho$ is pure and separable (i.e. it is a product state), then regardless of the choice of $\Pi_A,\Pi_B$ we always have uncorrelated outcomes. If $\rho$ is separable but not necessarily pure, this is no longer the case, intuitively because one can have "classical correlations".

A pure, entangled state also always gives rise to correlations, if measured in the basis of its Schmidt decomposition.

What about a generic entangled state? Given an arbitrary entangled $\rho$, is there always a measurement basis with respect to which measurement outcomes are correlated? While this sounds like it must be the case, I'm failing to see a good way to show it at the moment.

More formally, this is akin to asking whether there is an entangled state $\rho$ such that, for any pair of projective measurements $\Pi^A,\Pi^B$ (though I think one can more generally take any local POVM here), we have $$\operatorname{Tr}[(\Pi^A_a\otimes\Pi^B_b)\rho] = \operatorname{Tr}(\Pi^A_a\rho_A) \operatorname{Tr}(\Pi^B_b\rho_B).$$

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    $\begingroup$ I am not sure to entirely understand "What about a generic entangled state? Given an arbitrary entangled ρ, is there always a measurement basis with respect to which measurement outcomes are correlated?". Isn't the Schmidt decomposition that you also talk about precisely a proof that such measurement always exist ? $\endgroup$
    – Marco Fellous-Asiani
    May 16, 2021 at 11:07
  • $\begingroup$ @StarBucK for pure states, yes, but how does that work for mixed ones? I'm vaguely aware that there are generalisations of the Schmidt decomposition to mixed states but I'm not very familiar with those I must say $\endgroup$
    – glS ♦
    May 16, 2021 at 11:46
  • $\begingroup$ Does this lead you towards an answer: arxiv.org/abs/1610.05078 they say "entangled states are a subset of quantum correlated states", making it essentially a matter of definition. $\endgroup$
    – DaftWullie
    May 17, 2021 at 6:57
  • $\begingroup$ @DaftWullie thanks for the pointer. I don't think that works though. As far as I understand, they define "classically correlated states" as a subclass of separable states where you can do a separable decomposition in terms of orthogonal local states. So yes, using those definitions the statement is trivial, but is also different than what I'm asking about. The main difference I believe is that I focus directly on measurement outcomes, not the form of the state itself. I think my question can be reframed as: can you have entanglement that is invisible via any local measurement? $\endgroup$
    – glS ♦
    May 17, 2021 at 7:44

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Yes, they do. If it were true that $$\operatorname{Tr}[(\Pi^A_a\otimes\Pi^B_b)\rho] = \operatorname{Tr}(\Pi^A_a\rho_A) \operatorname{Tr}(\Pi^B_b\rho_B)$$ for all POVMs (or projective measurements, it doesn't make a difference), then we would just have $$\rho = \rho_A \otimes \rho_B,$$ because you'd be in particular doing tomography of your quantum state. Now such a state cannot be even classically correlated, much less entangled.

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  • $\begingroup$ I agree that this sounds obvious, but this doesn't seem like an argument to me. Performing arbitrary local measurements one fully reconstruct the state, I agree, and such a state does not produce any sort of (classical or quantum) correlation, basically by definition. How does it follow from this that the state is entangled? How do you conclude that $\rho$ is a product state from the first condition? $\endgroup$
    – glS ♦
    May 17, 2021 at 9:49
  • $\begingroup$ What I'm saying is that if a state displays no correlations whatsoever, then it is a product state, which is in particular not entangled. $\endgroup$
    – Mateus Araújo
    May 17, 2021 at 10:23
  • $\begingroup$ yes, the question is how you show more formally that you get a product state. Maybe I can reformulate your argument as follows: in general, local measurements only allow to reconstruct $\rho_A\otimes \rho_B$. If the data from the local measurements is put together, one can more generally reconstruct $\rho$. If all local measurements are uncorrelated, the two processes are identical, as there is no need to compare measurement results; it then follows that $\rho=\rho_A\otimes \rho_B$. Ok this might work I think $\endgroup$
    – glS ♦
    May 17, 2021 at 10:44
  • $\begingroup$ @glS Since $\rho = \rho_A \otimes \rho_B$ fits your requirements $\operatorname{Tr}[(\Pi^A_a\otimes\Pi^B_b)\rho] = \operatorname{Tr}(\Pi^A_a\rho_A) \operatorname{Tr}(\Pi^B_b\rho_B)$, and you are fully reconstructing the state, then $\rho = \rho_A \otimes \rho_B$ is the only solution. $\endgroup$
    – Danylo Y
    May 17, 2021 at 13:39
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    $\begingroup$ Another way to look at it: $\operatorname{tr}[M\rho] = \operatorname{tr}[M \rho^A \otimes \rho^B]$ is equivalent to $\operatorname{tr}[M(\rho-\rho^A \otimes \rho^B)] = 0$. If this is true for all $M$ then $\rho-\rho^A\otimes\rho^B = 0$. Now if your local measurements span the Hilbert space, i.e., are tomographically complete, this is equivalent to demanding it to be true for all $M$. $\endgroup$
    – Mateus Araújo
    May 17, 2021 at 15:33

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