Non-Maximal Rank Separable States Are A Set Of Measure Zero Within The Set of Non-Maximal Rank States

It is well known that the set of separable pure states is measure 0 in the set of pure states. Herein we extend this fact and show that the set of rank r separable states is measure 0 in the set of rank r states provided r is not maximal rank. Recently quite a few authors have looked at low rank separable and entangled states. (See [1] and the references therein and [2].) Therefore it makes sense to determine the size of the set of rank r separable states within the set of rank r states. For rank 1, it is well known that the separable states are a set of measure zero. This contrasts with the maximal rank case, where the separable states not only are not measure zero, but contain an open set. The purpose of this note is to show that the maximal rank case is the exception. In particular, suppose we have p particles modelled on the Hilbert space C1 ⊗ · · · ⊗ Cp . Then the following is true. Theorem 1 Let Sr be the set of rank r separable matrices on C n1 ⊗ · · · ⊗ Cp and Dr the set of all rank r density matrices. Sr is measure 0 in Dr, for all r < N = n1 · · ·np. The proof will use Sard’s Theorem [3]to show the set of ranges of separable rank r density matrices is measure 0 in the set of ranges of rank r density matrices, i.e. within the set of r-dimensional subspaces of C ; i.e. within G(N, r), the Grassmann manifold of r−planes in C [4]. That this is what we need to consider follows from the fact Dr is G(N, r) × Herm + 1 (r), where Herm+1 (r) is the space of Hermitian r × r matrices that are positive definite and trace 1. For those not familiar with Sard’s Theorem, it is an extension to nonlinear functions of the well known fact that if T is a linear transformation between two finite dimensional vector spaces, V and W , and the rank of T is less than the dimension of W, then T (V ) has measure zero in W , since it is a