Bounded witnesses for every continuous-variable entangled state
- Field
- Topic
Problem
Does every entangled bosonic state admit a bounded entanglement witness?
Let \(m,n\geq1\) be arbitrary integers, with \(\mathcal H_A:=L^2(\mathbb R^m)\) and \(\mathcal H_B:=L^2(\mathbb R^n)\). Let \(\rho\) be an entangled density operator on \(\mathcal H_A\otimes\mathcal H_B\). Separability means membership in the trace-norm closed convex hull of product density operators. States with positive partial transpose \(\rho^{T_B}\) in a fixed product Fock basis are included.
The desired operator \(L\) is bounded and self-adjoint and satisfies
In Eq. (1), \(a\in\mathcal H_A\) and \(b\in\mathcal H_B\). No finite-energy or moment-existence assumption is imposed.
Source
Sperling and Vogel prove this criterion for arbitrary-dimensional Hilbert spaces in Theorem 2 and Eq. (7) [SV09].
Progress
Reports do not certify correctness or automatically change the problem's status. Progress policy.
The answer is affirmative. The existence of a bounded Hermitian witness follows by separating the state from the trace-norm closed convex set of separable states; Sperling and Vogel attribute this existence step to Horodecki, Horodecki, and Horodecki [HHH96]. Sperling–Vogel Theorems 1–2 reformulate the witness criterion as Eq. (1), with Eq. (5) identifying the separable supremum with the pure-product supremum. The criterion also detects positive-partial-transpose entanglement [SV09].
Comment
The resolving result is peer-reviewed. Existence of a bounded witness does not supply an efficient finite-moment algorithm. Arbitrary density operators need not have all polynomial moments.