Action-angle coordinates

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. $H=H_1\otimes \cdots \otimes H_n$.

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 $M=\sum _i{v}_i{v}_i^{\ast }$, 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 $M=v_1{v}_1^{\ast }+T$, where T is Hermitian and positive semidefinite. There are two possibilities:

1) span$\{v_1\}\subset$Ker(T). Clearly, in this case, $v_1\in$ Ran(M).

2) Notice 1) is true if and only if Ker(T)${}^{\perp }\subset$ span$\{v_1{\}}^{\perp }$, where $\perp$ denotes orthogonal compliment. By Hermiticity of T, this is the same as Ran(T)$\subset$ span$\{v_1{\}}^{\perp }$. So if 1) does not hold, the intersection Ran(T) $\cap$ span$\{v_1\}$ is nonempty, i.e. there exists some complex number α such that $Tw=\alpha v_1$. So

$Mw=⟨w,v_1⟩v_1+Tw=(⟨w,v_1⟩+\alpha )v_1.$

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

$\rho =\sum _i{\psi }_{1,i}{\psi }_{1,i}^{\ast }\otimes \cdots \otimes {\psi }_{n,i}{\psi }_{n,i}^{\ast }$

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

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

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

References

• P. Horodecki, "Separability Criterion and Inseparable Mixed States with Positive Partial Transposition", Physics Letters A 232, (1997).