Is possible to write a separable state as a finite or countable infinite sum of product states?

Question

Let us consider the tensor product of two finite Hilbert spaces $\mathcal{H}_1\otimes \mathcal{H}_1$.

According to Watrous book, the set of separable states is the convex hull of the set of product states, that is any finite sum of the form $$ \sum_i p_i \varrho^i \otimes \sigma^i $$ Now, assuming I'm given a state of this form $$ \varrho = \int d\psi(\sigma) \sigma \otimes \sigma $$ where $d\psi(\sigma)$ is a certain measure on the set of state. This is separable by construction, even if it is not given as a finite sum.

My question is: can I rewrite this $\varrho$ as a finite sum of product states (even non pure)?

Results on the possible way to write separable states (and on the maximum number of elements in the convex mixture) applies only if one assume that separable state are given as a finite convex combination. I could not find any reference in the literature discussing separability with continuous measure and its possible representation.

Answer 1

Yes, you can always write an integral of separable states as a (finite) convex combination of product states, owing to the fact that the set of separable states is compact.

A quick way to see this is to use the Horodecki criterion (although it is more general than this and doesn't really have much to do with separability or the Horodecki criterion specifically). Let's say we have a state like this:

$$ \rho = \int \mathrm{d}\mu(z) \, \sigma(z) \otimes \xi(z). $$

We don't need to worry too much about the specifics of what we're integrating over — I'm just assuming that $\mu$ is a probability measure and $\sigma(z)$ and $\xi(z)$ are density operators. For every positive map $\Phi$ we have that $\Phi(\sigma(z))$ is positive semidefinite, so

$$ (\Phi\otimes\mathbb{1})(\rho) = \int \mathrm{d}\mu(z) \, \Phi(\sigma(z)) \otimes \xi(z) \geq 0. $$

It follows by the Horedecki criterion that $\rho$ is separable.

I said that this can be made more general and doesn't have much to do with separable states specifically. I'll leave this for a mathematician to explain more eloquently — but the idea is that if you're integrating over a probability measure and the integrand is contained in some compact convex set, then the integral can't fall outside of the convex set. If it did, there would be a separating hyperplane between it and the convex set, and you could then use the fact that integration is linear to derive a contradiction, where we have an integral of a nonnegative function being negative.