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.


2 Answers 2


The correct definition is

$\rho$ 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.


There are fixed sets that you can use, but the question is how large that set is.

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.


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