# Are states partially ordered in the same way via entanglement and Bell violations?

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 $$\tilde{B}(\rho_1) < \tilde{B}(\rho_2)$$?

• I think you have a typo in your definition of $\tilde B$. Commented Jul 1 at 7:31
• 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?
– ors
Commented Jul 1 at 8:05
• 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 Commented Jul 1 at 8:22
• 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". Commented Jul 1 at 8:27
• @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.
– ors
Commented Jul 1 at 8:32

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 $$\rho_e$$, $$0=E(\rho_s), but $$\tilde B(\rho_s)=\tilde B(\rho_e)=0$$, in contradiction to your claim.