In quantum mechanics, in particular quantum information, the Range criterion is a necessary condition that a state must satisfy in order to be separable. In other words, it is a separability 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. ${\displaystyle H=H_{1}\otimes \cdots \otimes H_{n))$.

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 ${\displaystyle M=\sum _{i}v_{i}v_{i}^{*))$, the range of M, Ran(M), is contained in the linear span of ${\displaystyle \;\{v_{i}\))$. On the other hand, we can also show ${\displaystyle v_{i))$ lies in Ran(M), for all i. Assume without loss of generality i = 1. We can write ${\displaystyle M=v_{1}v_{1}^{*}+T}$, where T is Hermitian and positive semidefinite. There are two possibilities:

1) span${\displaystyle \{v_{1}\}\subset }$Ker(T). Clearly, in this case, ${\displaystyle v_{1}\in }$ Ran(M).

2) Notice 1) is true if and only if Ker(T)${\displaystyle \;^{\perp }\subset }$ span${\displaystyle \{v_{1}\}^{\perp ))$, where ${\displaystyle \perp }$ denotes orthogonal complement. By Hermiticity of T, this is the same as Ran(T)${\displaystyle \subset }$ span${\displaystyle \{v_{1}\}^{\perp ))$. So if 1) does not hold, the intersection Ran(T) ${\displaystyle \cap }$ span${\displaystyle \{v_{1}\))$ is nonempty, i.e. there exists some complex number α such that ${\displaystyle \;Tw=\alpha v_{1))$. So

${\displaystyle Mw=\langle w,v_{1}\rangle v_{1}+Tw=(\langle w,v_{1}\rangle +\alpha )v_{1}.}$

Therefore ${\displaystyle v_{1))$ lies in Ran(M).

Thus Ran(M) coincides with the linear span of ${\displaystyle \;\{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

${\displaystyle \rho =\sum _{i}\psi _{1,i}\psi _{1,i}^{*}\otimes \cdots \otimes \psi _{n,i}\psi _{n,i}^{*))$

where ${\displaystyle \psi _{j,i}\psi _{j,i}^{*))$ is a (un-normalized) pure state on the j-th subsystem. This is also

${\displaystyle \rho =\sum _{i}(\psi _{1,i}\otimes \cdots \otimes \psi _{n,i})(\psi _{1,i}^{*}\otimes \cdots \otimes \psi _{n,i}^{*}).}$

But this is exactly the same form as M from above, with the vectorial product state ${\displaystyle \psi _{1,i}\otimes \cdots \otimes \psi _{n,i))$ replacing ${\displaystyle v_{i))$. It then immediately follows that the range of ρ is the linear span of these product states. This proves the criterion.