Range criterion

The result

Consider a quantum mechanical system composed of n subsystems. The state space H of such a system is the tensor product of those of the subsystems, i.e. H = H 1 ⊗ ⋯ ⊗ H n .

For simplicity we will assume throughout that all relevant state spaces are finite-dimensional.

The criterion reads as follows: If ρ is a separable mixed state acting on H, then the range of ρ is spanned by a set of product vectors.

Proof

In general, if a matrix M is of the form M = ∑ i v i v i ∗ , it is obvious that the range of M, Ran(M), is contained in the linear span of { v i } . On the other hand, we can also show v i lies in Ran(M), for all i. Assume without loss of generality i = 1. We can write M = v 1 v 1 ∗ + T , where T is Hermitian and positive semidefinite. There are two possibilities:

1) span { v 1 } ⊂ Ker(T). Clearly, in this case, v 1 ∈ Ran(M).

2) Notice 1) is true if and only if Ker(T) ⊥ ⊂ span { v 1 } ⊥ , where ⊥ denotes orthogonal complement. By Hermiticity of T, this is the same as Ran(T) ⊂ span { v 1 } ⊥ . So if 1) does not hold, the intersection Ran(T) ∩ span { v 1 } is nonempty, i.e. there exists some complex number α such that T w = α v 1 . So

Therefore v 1 lies in Ran(M).

Thus Ran(M) coincides with the linear span of { v i } . The range criterion is a special case of this fact.

A density matrix ρ acting on H is separable if and only if it can be written as

where ψ j , i ψ j , i ∗ is a (un-normalized) pure state on the j-th subsystem. This is also

But this is exactly the same form as M from above, with the vectorial product state ψ 1 , i ⊗ ⋯ ⊗ ψ n , i replacing v i . It then immediately follows that the range of ρ is the linear span of these product states. This proves the criterion.