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.


1 Answer 1


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ If you want more details on it, see here $\endgroup$ Commented May 12, 2021 at 20:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.