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?

  • $\begingroup$ 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)? $\endgroup$ – Matthew Daws May 26 '19 at 12:17
  • $\begingroup$ Yes, exactly... $\endgroup$ – Dominique Unruh May 26 '19 at 13:33

The answer is negative if we believe [1] (which comes without proofs, unfortunately):

[1] 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).

[1] 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 [1] 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.

[1] 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.

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    $\begingroup$ It's worth noting that while there are not proofs, the counter-example is Theorem 2 which is very concrete and explicit. $\endgroup$ – 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.

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  • $\begingroup$ 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? $\endgroup$ – Dominique Unruh May 27 '19 at 16:21
  • $\begingroup$ 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. $\endgroup$ – Dominique Unruh May 27 '19 at 16:49
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    $\begingroup$ 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.) $\endgroup$ – Matthew Daws May 28 '19 at 21:19
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    $\begingroup$ @Matthew Daws: You are right. $\phi\mapsto\phi_\infty$ is not weak*-continuous. My proof seems to fall apart. $\endgroup$ – Narutaka OZAWA May 29 '19 at 0:37
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    $\begingroup$ @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. $\endgroup$ – 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 $\Omega$.

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