Let R_{1}, R_{2} be von Neumann algebras on H such that R_{1} ⊆ R′_{2}. Recall that a state ω of R_{12} is called a normal product state just in case ω is normal, and there are states ω_{1} of R_{1} and ω_{2} of R_{2} such that

ω(AB) = ω1(A)ω2(B) —– (1)

∀ A ∈ R_{1}, B ∈ R_{2}. * Werner*, in dealing with the case of B(C

^{n}) ⊗ B(C

^{n}), defined a density operator D to be classically correlated — the term separable is now more commonly used — just in case D can be approximated in trace norm by convex combinations of density operators of form D

_{1}⊗ D

_{2}. Although Werner’s definition of nonseparable states directly generalizes the traditional notion of pure entangled states, he showed that a nonseparable mixed state need not violate a Bell inequality; thus, Bell correlation is in general a sufficient, though not necessary condition for a state’s being non-separable. On the other hand, it has since been shown that

*.*

**nonseparable states often possess more subtle forms of nonlocality, which may be indicated by measurements more general than the single ideal measurements which can indicate Bell correlation**In terms of the linear functional representation of states, Werner’s separable states are those in the norm closed convex hull of the product states of B(C^{n}) ⊗ B(C^{n}). However, in case of the more general setup — i.e., R_{1} ⊆ R′_{2}, where R_{1}, R_{2} are arbitrary von Neumann algebras on H — the choice of topology on the normal state space of R_{12} will yield in general different definitions of separability. Moreover, it has been argued that norm convergence of a sequence of states can never be verified in the laboratory, and as a result, the appropriate notion of physical approximation is given by the (weaker) weak-∗ topology. And the weak-∗ and norm topologies do not generally coincide even on the normal state space.

For the next proposition, then, we will suppose that the separable states of R_{12} are those normal states in the weak-∗ closed convex hull of the normal product states. Note that β(ω) = 1 if ω is a product state, and since β is a convex function on the state space, β(ω) = 1 if ω is a convex combination of product states. Furthermore, since β is lower semicontinuous in the weak-∗ topology, β(ω) = 1 for any separable state. Conversely, any Bell correlated state must be nonseparable.

We now introduce some notation that will aid us in the proof of our result. For a state ω of the von Neumann algebra R and an operator A ∈ R, define the state ω^{A} on R by

ω^{A}(X) ≡ ω(A∗XA)/ω(A∗A) —– (2)

if ω(A∗A) ≠ 0, and let ω^{A} = ω otherwise. Suppose now that ω(A∗A) ≠ 0 and ω is a convex combination of states:

ω = ∑_{i=1}^{n}λ_{i}ω_{i} —– (3)

Then, letting λ^{A}_{i} ≡ ω(A∗A)^{−1}ω_{i}(A∗A)λ_{i}, ω^{A} is again a convex combination

ω^{A} = ∑_{i=1}^{n} λ^{A}_{i}ω^{A}_{i} —– (4)

Moreover, it is not difficult to see that the map ω → ω^{A} is weak-∗ continuous at any point ρ such that ρ(A∗A) ≠ 0. Indeed, let O_{1} = N(ρ^{A} : X_{1},…,X_{n}, δ) be a weak-∗ neighborhood of ρ_{A}. Then, taking O_{2} = N(ρ : A^{∗}A,A^{∗}X_{1}A,…,A^{∗}X_{n}A, δ) and ω ∈ O_{2}, we have

|ρ(A^{∗}X_{i}A) − ω(A^{∗}X_{i}A)| < δ —– (5)

for i = 1,…,n, and

|ρ(A^{∗}A) − ω(A^{∗}A)| < δ —– (6)

By choosing δ < ρ(A^{∗}A) ≠ 0, we also have ω(A^{∗}A) ≠ 0, and thus

|ρ^{A}(X_{i}) − ω^{A}(X_{i})| < O(δ) ≤ δ —– (7)

for an appropriate choice of δ. That is, ω^{A} ∈ O_{1} forall ω ∈ O_{2} and ω → ω^{A} is weak-∗ continuous at ρ.

Specializing to the case where R_{1} ⊆ R′_{2}, and R_{12} = {R_{1} ∪ R_{2}}”, it is clear from the above that for any normal product state ω of R_{12} and for A ∈ R_{1}, ω^{A} is again a normal product state. The same is true if ω is a convex combination of normal product states, or the weak-∗ limit of such combinations. Summarizing the results of this discussion in the following lemma:

Lemma: For any separable state ω of R_{12} and any A ∈ R_{1}, ω^{A} is again separable.

Proposition: Let R_{1},R_{2} be nonabelian von Neumann algebras such that R_{1} ⊆ R′_{2}. If x is cyclic for R_{1}, then ω_{x} is nonseparable across R_{12}.

Proof: There is a normal state ρ of R_{12} such that β(ρ) = √2. But since all normal states are in the (norm) closed convex hull of vector states, and since β is norm continuous and convex, there is a vector v ∈ S such that β(v) > 1. By the continuity of β (on S), there is an open neighborhood O of v in S such that β(y) > 1 ∀ y∈O. Since x is cyclic for R_{1},there is an A ∈ R_{1} such that Ax ∈ O. Thus, β(Ax) > 1 which entails that ω_{Ax} = (ω_{x})^{A} is a nonseparable state for R_{12}. This, by the preceding lemma, entails that ω_{x} is nonseparable.

Note that if R_{1} has at least one cyclic vector x ∈ S, then R_{1} has a dense set of cyclic vectors in S. Since each of the corresponding vector states is nonseparable across R_{12}, Proposition shows that if R_{1} has a cyclic vector, then the (open) set of vectors inducing nonseparable states across R_{12} is dense in S. On the other hand, since the existence of a cyclic vector for R_{1} is not invariant under isomorphisms of R_{12}, Proposition does not entail that if R_{1} has a cyclic vector, then there is a norm dense set of nonseparable states in the entire normal state space of R_{12}. Indeed, if we let R_{1} = B(C^{2}) ⊗ I, R_{2} = I ⊗ B(C^{2}), then any entangled state vector is cyclic for R_{1}; but, the set of nonseparable states of B(C^{2}) ⊗ B(C^{2}) is not norm dense. However, if in addition to R_{1} or R_{2} having a cyclic vector, R_{12} has a separating vector (as is often the case in quantum field theory), then all normal states of R_{12} are vector states, and it follows that the nonseparable states will be norm dense in the entire normal state space of R_{12}.