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Recall what a partially ordered set is.

Let $E(\rho)$ be an entanglement measure. Let $B(\rho) \leq 0$ be a Bell inequality. Define the Bell violation measure with respect to $B$ as $$\tilde{B}(\rho) = 0 \text{ if } B(\rho) \leq 0$$ and $$\tilde{B}(\rho) = B(\rho) \text{ if } B(\rho) > 0.$$ Fix an $E$ and $\tilde{B}$. Do $E$ and $\tilde{B}$ impose the same partial order, denoted $<$, on the set of all density matrices? That is, is it true that $E(\rho_1) < E(\rho_2)$ iff $\tilde{B}(\rho_1) < \tilde{B}(\rho_2)$?

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    $\begingroup$ I think you have a typo in your definition of $\tilde B$. $\endgroup$
    – Tobias Fünke
    Commented Jul 1 at 7:31
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    $\begingroup$ It isn't even generally true that two entanglement measures will give you the same partial orders. Are you using some specific definitions or something? $\endgroup$
    – ors
    Commented Jul 1 at 8:05
  • $\begingroup$ I added clarification. In particular, to fix an entanglement measure and Bell measure. In this case, then, I do not see it as particularly relevant that two distinct entanglement measures may induce different partial orders. @ors $\endgroup$
    – Silly Goose
    Commented Jul 1 at 8:22
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    $\begingroup$ If different entanglement measures give different partial orders they can't both behave the same w.r.t. a given Bell violation. So the answer to the question is "no". $\endgroup$
    – Norbert Schuch
    Commented Jul 1 at 8:27
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    $\begingroup$ @SillyGoose It seems relevant to me, if this argument works for one fixed $E_1$ its going to work for a second $E_2$, and then you've proved that the two entanglement measures give the same partial order which is (generally) false. $\endgroup$
    – ors
    Commented Jul 1 at 8:32

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No.

A possible counterexample is given by a trivial Bell inequality $B(\rho)\equiv 0$ (thus $\tilde B(\rho)\equiv0$) and a non-trivial entanglement measure, e.g. entanglement of formation. Then, for a separable state $\rho_s$ and an entangled state $\rho_e$, $0=E(\rho_s)<E(\rho_e)$, but $\tilde B(\rho_s)=\tilde B(\rho_e)=0$, in contradiction to your claim.

A more interesting counterexample is given by Werner states: Werner (R.F. Werner, Quantum states with Einstein-Podolsky-Rosen correlations admitting a hidden-variable model, Phys. Rev. A 40, 4277 (1989)) shows that there are states which are not separable (so in particular, they have non-zero entanglement of formation), but do not violate any Bell inequality. Thus, your conjecture is wrong for entanglement of formation (or any other entanglement measure which is non-zero on all entangled states) and any Bell inequality.

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