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Is it true that any bipartite state that is "measurably independent" is separable?

I am defining a state $|\psi \rangle \in A \otimes B$ to be "measurably independent" if:

The probabilities of the different outcomes of a measurement of any observable $\hat{O}_B$ of system $B$ is not affected by a preceding measurement of any observable $\hat{O}_A$ of system $A$, and vice versa.

I am asking since entanglement is typically defined in terms of separability, but intuitively I think of entanglement as a violation of measurable independence. It is easy to show that separable states are measurably independent, but might there be measurably independent states that aren't separable?

So far, by setting $\hat{O}_A$ and $\hat{O}_B$ to projection operators $|\phi_A\rangle \langle \phi_A|$ and $|\phi_B\rangle \langle \phi_B|$ for states $|\phi_A\rangle \in A$ and $|\phi_B\rangle \in B$, I can show that: $$ P(|\phi_A\rangle,|\phi_B\rangle)=P(|\phi_A\rangle)P(|\phi_B\rangle) $$ for any measurable state and any states $|\phi_A\rangle$ and $|\phi_B\rangle$ ($P$ stands for probability). If I can show that for measurably independent states we always have some $|\phi_A\rangle$ and $|\phi_B\rangle$ such that $P\left(|\phi_A\rangle\right)=P(|\phi_A\rangle)=1$ then the proof is done, but I'm stuck at this step.

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Let $|\Psi\rangle$ be a measurably independent pure state. The Schmidt decomposition gives us : $$|\Psi\rangle = \sum_i \sqrt{p_i} |x_i\rangle \otimes |y_i\rangle$$

where the $|x_i\rangle$ (resp. $|y_i\rangle$) form an orthonormal basis of $A$ (resp. $B$).

Then, for the observable $\hat O_B =|y_{i_0}\rangle\langle y_{i_0}| $, the expected value is : $$\langle \hat O_B \rangle_\Psi = \sum_i p_i \langle y_i|\hat O_B|y_i\rangle = p_{i_0}$$

If $0<p_{i_0}<1$), then after a measure of the observable $\hat O_A = |x_{i_0}\rangle\langle x_{i_0}|$, the possible states are : $$|\Psi_1\rangle = |x_{i_0}\rangle \otimes |y_{i_0}\rangle \qquad \text{and} \qquad |\Psi_2 \rangle = \sum_{i\neq i_0} \frac{\sqrt{p_i}}{1-p_{i_0}}|x_i\rangle \otimes |y_i\rangle$$

In those states, the expected values of $\hat O_B$ are : $$\langle \hat O_B \rangle_{\Psi_1} = \langle y_{i_0}|\hat O_B|y_{i_0}\rangle = 1 \qquad \text{and}\qquad \langle \hat O_B \rangle_{\Psi_2} = \sum_{i\neq i_0}\frac{p_i}{1-p_{i_0}} \langle y_i|\hat O_B|y_i\rangle = 0$$

This is a contradiction. Therefore we must have $i_0 = 0 \text{ or } 1$ : the state is separable.

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  • $\begingroup$ I did not know about the existence of a Schmidt decomposition! That appears to be the key. $\endgroup$ Commented May 12, 2021 at 20:24
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    $\begingroup$ If you want more details on it, see here $\endgroup$ Commented May 12, 2021 at 20:26

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