arXiv subjects
Publications and source records attributed to Robert B. Lockhart.
This paper is being withdrawn due to an error in the proof. Hao Chen has shown the author in the two qubit case that there is an open set of three dimensional subspaces that are spanned by separable states. This means the author's proof is in error. The problem is that homogeneity cannot be used and so the Jacobian needs to be computed at all points
As separable states are a convex combination of product states, the geometry of the manifold of product states is studied. Prior results by Sanpera, Vidal and Tarrach are extended. Furthermore, it is proven that states in the set tangent to the manifold of product states, at the maximally mixed state are separable; the set normal constains, among others, all maximally entangled states. A canonical decomposition is given. A surprising result is that for the case of two particles, the closest product state to the maximally entangled state is the maximally mixed state. An algorithm is provided to find the closest product state.
The optimal (pure state) ensemble length of a separable state, A, is the minimum number of (pure) product states needed in convex combination to construct A. We study the set of all separable states with optimal (pure state) ensemble length equal to k or fewer. Lower bounds on k are found below which these sets have measure 0 in the set of separable states. In the bipartite case and the multiparticle case where one of the particles has significantly more quantum numbers than the rest, the lower bound for non-pure state ensembles is sharp. A consequence of our results is that for all two particle systems, except possibly those with a qubit or those with a nine dimensional Hilbert space, and for all systems with more than two particles the optimal pure state ensemble length for a randomly picked separable state is with probability 1 greater than the state's rank. In bipartite systems with probability 1 it is greater than 1/4 the rank raised to the 3/2 power and in a system with p qubits with probability 1 it is greater than (2^2p)/(1+2p), which is almost the square of the rank.