# Is the set of separable quantum states closed?

Let $$\mathcal H,\mathcal H'$$ be Hilbert spaces (not necessarily separable).

A "separable state" is a trace-class operator of the form $$\sum_i \rho_i\otimes\rho_i'$$ where $$\rho_i,\rho_i'$$ are positive trace-class operators over $$\mathcal H,\mathcal H'$$, respectively. (Convergence of the sum is with respect to the trace norm. $$\otimes$$ represents the tensor product.)

Is the set of separable states closed with respect to the trace norm?

• I guess the sum is meant to converge in the space of trace class operators acting on the Hilbert space $\mathcal H \otimes \mathcal H'$ (this being the Hilbert space tensor product)? – Matthew Daws May 26 '19 at 12:17
• Yes, exactly... – Dominique Unruh May 26 '19 at 13:33

The answer is negative if we believe  (which comes without proofs, unfortunately):

 defines the set of separable states as the convex closure of $$\{\sum_{i=1}^n\sigma_i\otimes\tau_i\}$$. I will call that set $$S$$. The set as defined in the original question I call $$T$$. Then $$T\subseteq S$$, and $$T=S$$ iff the answer is yes (i.e., iff $$T$$ is closed).

 shows that there is a $$\rho\in S$$ such that $$\rho$$ cannot be represented as a Bochner integral $$\int\psi\psi^*\otimes\phi\phi^* \pi(d(\psi,\phi))$$ for an atomic measure $$\pi$$. ($$\rho$$ is not "countably decomposable" in their language.)

Any $$\rho\in T$$ can be represented as such an integral with a discrete and hence atomic measure $$\pi$$. Thus $$T\subsetneq S$$.

On the positive side, if we define $$T':=\{\int\psi\psi^*\otimes\phi\phi^* \pi(d(\psi,\phi))\}$$ for probability measures $$\pi$$, then  shows $$S=T'$$. In particular, $$T'$$ is closed. So, in the spirit of the original question, the set of infinite convex combinations of $$\sigma_i\otimes\tau_i$$ is closed, only the notion of "infinite convex combination" must be changed: Not infinite sums, but integrals.

 Werner, R. F.; Kholevo, A. S.; Shirokov, M. E., On the concept of entanglement in Hilbert spaces., Russ. Math. Surv. 60, No. 2, 359-360 (2005); translation from Usp. Mat. Nauk 60, No. 2, 153-154 (2005). ZBL1098.47019.

• It's worth noting that while there are not proofs, the counter-example is Theorem 2 which is very concrete and explicit. – Matthew Daws May 30 '19 at 10:27

[There are difficulties with this proof, see the comments]

Yes, it's closed. Here's a soft proof (which requires a bit of C*-theory).

Let $$A$$ be a (not necessarily unital) C*-algebra and $${\mathcal Q}(A)\subset A^*$$ denotes the quasi-state space (i.e., the space of positive linear functional of norm $$\le1$$), which is compact in the weak*-topology. I claim that if $$\Omega\subset {\mathcal Q}(A)$$ is weak*-closed, then $$\{\sum_{n=1}^\infty \phi_n : \phi_n\in\Omega,\ \sum_{n=1}^\infty \|\phi_n\|\le1\}\subset {\mathcal Q}(A)$$ is also weak*-closed. Specialize this to $$A=K({\mathcal H}\otimes{\mathcal H}')$$ and $$\Omega={\mathcal Q}(K({\mathcal H}))\times{\mathcal Q}(K({\mathcal H}'))$$.

Proof of Claim: Let $$\tilde{A}\subset\ell_\infty({\mathbb N},A)$$ denote the C*-algebra of the norm-convergent sequences in $$A$$. Then $$\tilde{A}$$ is an extension of $$A$$ by $$c_0({\mathbb N},A)$$. Hence any $$\phi\in {\mathcal Q}(\tilde{A})$$ is of the form $$\phi=(\phi_1,\ldots,\phi_\infty)$$, $$\tilde{A}\ni a=(a_n)_n \mapsto (\sum_{n=1}^\infty \phi_n(a_n)) + \phi_\infty(\lim_n a_n) \in{\mathbb C},$$ where $$\phi_n\in {\mathcal Q}(A)$$ satisfy $$\|\phi\|=(\sum_n\|\phi_n\|)+\|\phi_\infty\|\le1$$. It is not difficult to see that $$\tilde{\Omega}:=\{ \phi\in {\mathcal Q}(\tilde{A}) : \phi_n\in\Omega,\ n=1,\ldots,\infty\}$$ is weak*-closed. Thus its restriction to $$A\subset\tilde{A}$$ (diagonal embedding) is also weak*-closed (weak*-compact). This restriction is what we were looking after.

• I have some questions (I am not an expert in C*-algebras). First, just to make sure: The weak*-topology is the topology of pointwise convergence in this context, and compactness of $\cal Q(A)$ is the Banach–Alaoglu theorem? Second, when you say "its restriction to $A\subseteq\tilde A$", you mean the set $e(\tilde\Omega)$ where $e(\phi):=\phi|A$ for $\phi\in \cal Q(\tilde A)$? Then I understand the proof of the claim up to the last step: Why is that restriction weak*-closed? – Dominique Unruh May 27 '19 at 16:21
• And finally: I do not understand the notation $\cal Q(K(\cal H))\times\cal Q(K(\cal H'))$ (I don't know how pairs of functionals are interpreted as functionals), but I assume I should set $\Omega:=\{\phi:\phi(A):=\operatorname{tr}(\sigma\otimes\tau)A, 0\leq\sigma,\tau\leq 1\}$ where $\sigma,\tau$ are trace class and $A$ is compact. I don't know how to show that $\Omega$ is weak*-closed in that case. – Dominique Unruh May 27 '19 at 16:49
• I do not see how to prove that $\tilde\Omega$ is weak$^*$-closed, without already knowing that I can sum absolutely summable sequences in $\Omega$ without leaving $\Omega$. The problem I have is that if $(\phi^{(i)})$ is a net in $\tilde Q$ converging to $\phi$ say, then how can I show that $\phi_\infty \in \Omega$? (That $\phi_n\in\Omega$ for $n<\infty$ is clear.) – Matthew Daws May 28 '19 at 21:19
• @Matthew Daws: You are right. $\phi\mapsto\phi_\infty$ is not weak*-continuous. My proof seems to fall apart. – Narutaka OZAWA May 29 '19 at 0:37
• @DominiqueUnruh The problem is that this argument implicitly swaps the processes of taking the infinite sum over $i$ and taking the limit in $n$. You can't do that. – Matthew Daws May 29 '19 at 10:19

This is not an answer (and I now suspect that the answer is negative), but this maybe still useful and in any case too lengthy for a comment. Assume that $$\mathcal H$$ and $${\mathcal H}'$$ are separable. The norm-closed convex hull of $$\Omega=\{ \phi\otimes\psi : 0\le\phi, \, \|\phi\|\le 1,\,0\le\psi,\, \|\psi\|\le1\}$$ is $$\tilde{\Omega}:=\{ \int_{\mathbb R} f(t) \, dt : f\in L^1({\mathbb R},\Omega),\, \|f\|\le 1\}.$$ Note that the weak*-measurability and norm-measurability for $$f$$ coincide by separability assumption. In particular, $$\tilde{\Omega}$$ is contained in the norm convex hull of $$\Omega$$. I claim $$\tilde{\Omega}$$ is weak*-closed. Let's equip $$\Omega$$ with the weak*-topology coming from $$K({\mathcal H}\otimes{\mathcal H}')$$, which makes $$\Omega$$ compact. Then, the space $$\mathrm{Prob}(\Omega)$$ of Radon probability measures is compact w.r.t. the weak* topology induced by $$C(\Omega)$$. It follows that $$\{\int_\Omega \omega \, d\mu(\omega) : \mu\in \mathrm{Prob}(\Omega)\}.$$ is weak*-compact and so it is the weak*-closed convex hull of $$\Omega$$. As $$L^1(\Omega,\mu)\hookrightarrow L^1({\mathbb R},\mathrm{Leb})$$, we conclude that $$\tilde{\Omega}$$ is the weak*-closed convex hull of $$\Omega$$.