# Is any "measurably independent" bipartite state separable?

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.

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 $$0), then after a measure of the observable $$\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.

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