# Density operators and separable states

I know that a density operator is separable if it can be written in the form:

$$\rho =\sum_k a_k \rho^A_k \otimes \rho^B_k\tag{1}$$ where

$$a_k \ge 0,\quad \sum_k a_k=1\tag{2}$$

My question is will any set of $\rho_k^A \otimes \rho_k^B$ work? I.e. I am asking if the following statement is true (if so how do we prove it and if not - why not):

A density matrix $\rho$ is separable if and only if when written as the sum of "factorized" states $\rho_K^A \otimes \rho_k^B$ (independent of which factorized states are used) the relations (1) and (2) hold.

This is clearly a stronger statement then saying it can be written as (1) and (2) - and if true leads to a quick way to test entanglement.

$\rho$ is separable if and only if there exist $\rho_i^A\ge0$, $\rho_i^B\ge0$, and $p_i\ge0$ such that $$\rho = \sum_i p_i \rho_i^A\otimes \rho_i^B\ .$$
All other properties (that the $\rho_i^\bullet$ have trace 1 and the $p_i$ sum to $1$) follow automatically.
If your possible set of states is infinitely large (every possible pure state of 1 qubit), you can always do it, of course. But, worse, it has to be this large. Imagine you have a finite set $\{\rho^A_i\}$ and you want to describe a pure single-qubit state $|\psi\rangle\langle\psi|$ that is not one of the $\rho^A_i$. Clearly, there is no choice of $\{p_i\}$ such that $\sum_ip_i\rho^A_i=|\psi\rangle\langle\psi|$ because you must sum at least two terms, and hence will have a state with at least rank 2, while the pure state has rank 1.
A good way to visualise this is with the Bloch sphere. Imagine a sphere. Every possible pure state corresponds to a point on the surface. The set $\{\rho^A_i\}$ form a set of points on, or in, the sphere. The set of possible states that you can make via the term $\sum_ip_i\rho^A_i$ is given by the convex hull of the points. You can't reconstruct the surface of the sphere from a convex hull (the shape has flat sides!) unless you have infinitely many points.