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.

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    $\begingroup$ in that same book, they also discuss that any state can be written as such a convex decomposition of up to $\operatorname{rank}(\rho)^2$ elements. This is also mentioned in quantumcomputing.stackexchange.com/q/13031/55. Is this what you're asking? $\endgroup$
    – glS
    Nov 17, 2023 at 12:36
  • $\begingroup$ @glS yes but they say it by using a definition of separable state that has a finite amount of element in the convex combination. So the result, at leat how it is proved there, is true if we use the definition of separable states as a finite convex combination of product states. Here, I don't see how to use their argument if the separable state has a continuous sum of product states. $\endgroup$ Nov 17, 2023 at 14:49
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    $\begingroup$ See Carathéodory's theorem $\endgroup$
    – Rammus
    Nov 17, 2023 at 18:11
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    $\begingroup$ @Rammus yes, but my question I think still apply, since in the proof of the theorem in the Wikipedia page, they assume that a point in the convex hull is written as a finite sum. So again: is the convex hull just the sum of a finite number of elements in the set? Can I prove a Carathéodory's theorem if my convex hull also consist of continuous sum? I think the question can be also rephrased: there are elements in the convex hull that must be written only as continuous sum, i.e. there is no finite sum that approximate them with arbitrary precision (in some norm)? I hope you see the problem $\endgroup$ Nov 18, 2023 at 10:16
  • $\begingroup$ Possibly related: Holevo (arxiv.org/abs/quant-ph/0504204) defines countably decomposable states as those having an integral representation with "purely atomic" measure. He then provides an example of a separable state that is not countably decomposable in Sec. 4. I don't yet have enough grasp of measure theory to tell if that's what you're looking for. $\endgroup$
    – forky40
    Nov 19, 2023 at 16:23

1 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.

  • $\begingroup$ Thanks! That's exactly what I was looking for. May I add an extra question? In the case I have a convex subset of the separable states this is in general is bounded, but it might not be closed and thus might not be compact. But what if this convex subset is defined in terms of some projector $S$? Is that convex subset closed then? $\endgroup$ Nov 20, 2023 at 11:16

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