In what sense is the set of density matrices compact in infinite dimensions?

Question

Consider a complex, infinite-dimensional and separable Hilbert space $H$ and let $\mathcal I(H)$ denote the space of trace-class operators. The set of density operators $$\mathcal S(H):= \{\rho\in \mathcal I(H)\,|\, \mathrm{Tr}\rho=1\, , \, \rho \geq 0 \}$$

enjoys certain nice mathematical and physical features, some explained here. Moreover, it is well-known that $\mathcal S(H)$ is compact for a finite-dimensional Hilbert space, as discussed here.

Question: Is there a notion in which the set of density matrices is compact for an infinite-dimensional $H$? For example, in Ref. 1 it is stated (right after discussing the finite-dimensional case) that

$[\ldots]$ by using a physically natural topology (weak topology for $\rho$) one can still show that $\mathcal S(H)$ is compact.

In which topology is $\mathcal S(H)$ compact and how is it proved? Further, what makes this topology physically natural?

Edit: In Ref. 2, there is a theorem: $\mathcal S(H)$ is a compact convex subset of the real vector space $\mathcal I(H)$, and a proof which goes like this:

Now $\mathcal S(H)\subset \mathcal I(H)=\mathcal B_*(H)$ is a closed unit ball (because $||\rho||_1=\mathrm{Tr}\rho=1$ for all $\rho \in \mathcal S(H)$). Therefore, $\mathcal S(H)$ is weakly compact in the space $\mathcal I(H)$ by the Banach-Alaoglu theorem $[\ldots]$.

But I don't really understand the proof.

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References:

1. Decoherence Suppression in Quantum Systems 2008. World Scientific Publishing. Editors: Mikio Nakahara, Robabeh Rahimi, Akira SaiToh. Page 12

2. Theory of Quantum Information with Memory. M. H. Chang. 2022. De Gruyter. Section 2.4. Page 54, Proposition 2.4.2

Answer 1

The assertion is false, if I understand well it.

First of all some general information on the notion of weak * topology.

One has a (complex) normed space $B$ and its topological dual $B'$ which is, by definition the vector space of continuous linear maps $f: B \to \mathbb{C}$.

This space has a natural structure of normed space as well with the norm $$||f||' := \sup\{|f(x)| \:|\: x \in B \:, ||x||=1\}\:.$$ However we can equip $B'$ with another seminormed topology induced by $B$ called weak * topology, it is characterized by the fact that

$$B' \ni f_n \to f \in B'\quad iff \quad f_n(x)-f(x)\to 0 \quad \forall x\in B\:.$$

The Banach-Alaoglu theorem proves that every norm-closed ball in $B'$ $$\{f \in B' \:|\: ||f|| \leq r\}$$ is compact with respect to the weak * topology provided $B$ is complete (i.e. is a Banach space).

Sometimes it happens that two normed spaces $B_1$ and $B_2$ are in topological duality, that is one is (normed-space) isomorphic to the topological dual of the other. If $B_2$ is isomorphic through $F$ to $B_1' = F(B_2)$, the action of the elements of $B_2$ o the elements of $B_1$ is indicated by means of a pairing: $$\langle b_1, b_2\rangle := (F(b_2))(b_1)\:.$$

Let us pass to the discussed assertion, that the space of the mixed states $S(H)$ is compact with respect to some weak * topology on that. The space $S(H)$ is not linear and it is not a closed ball of the normed (Banach) space of the trace class operators $B_1(H)$ equipped with its natural norm topology induced by the notion of trace. Therefore, one should try to prove that $S(H)$ is a compact subset of $B_1(H)$.

From the sketch of proof, it seems that the authors use the fact that $B_1(H)$ is the topological dual space of some other space. All that in order to apply the Banach-Alaoglu theorem as written above.(*)

As is well known, $B_1(H)$ is in fact (isometrically isomorphic to) the topological dual space of $B_\infty(H)$ (the Banach space of compact operators with the natual norm operator topology).

The duality is represented in terms of the trace operation as $B_1(H)$ is a (both sides *) ideal, as the (left-continuous) pairing

$$\langle A, B \rangle := tr(AB)\:,\:\quad A\in B_1(H)\:, \quad B \in B_\infty(H)$$

The Banach-Alaoglu theorem proves that every closed ball in $B_1(H)$ is therefore compact with respect to the weak * topology induced by the topological duality considered above. In particular, according to this weak* topology, $$B_1(H) \ni \rho_n \to \rho \in B_1(H)\quad iff \quad tr((\rho_n-\rho)A)\to 0 \quad \forall A\in B_1(H)$$ To conclude the proof it would be sufficient to prove that $S(H)$ is closed with respect to that topology, since closed subsets of compact sets are compact as well.

But it is not the case as proved here!

On the other hand, since weak * topology is Hausdorff, compact sets are closed. This proves that the assertion, referred to the said weak * topology, is false as $S(H)$ is not closed with respect to that weak * topology.

Maybe the assertion is true using another weak (natural?) topology.

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($*$) Another possibility, not followed by the authors, is to see $B_1(H) \subset B(H)$ and viewing $B(H)$ as the dual of $B_1(H)$, thus equipping this latter with the weak* topology of $B(H)$ also known as ultraweak topology.

Even another way is to see $B_1(H)$ as subset of $B_2(H)$ the Hilbert-space space of Hilbert Schmidt operators and to use the weak * topology on $B_2(H)$ induced by the Riesz anti isomorphism.

Answer 2

Given two topologies $\mathcal O_1$ and $\mathcal O_2$ on the same set, we say that $\mathcal O_1$ is weaker (or coarser) than $\mathcal O_2$ if $\mathcal O_1\subset \mathcal O_2$ - in other words, every set which is open in $\mathcal O_1$ is open in $\mathcal O_2$. The descriptor "weak" comes from the fact that is in some sense less stringent; a function which is continuous in $\mathcal O_2$ is necessarily continuous in $\mathcal O_1$, and a sequence which converges in $\mathcal O_2$ necessarily converges in $\mathcal O_1$, but the reverse implications do not generally hold.

The weak operator topology $\mathcal O_W$ on the set of bounded operators $B(\mathscr H)$ is defined to be the weakest topology such that, for any $x,y\in \mathscr H$, the map $$f_{xy}:B(\mathscr H)\rightarrow \mathbb C$$ $$T\mapsto \langle x, Ty\rangle$$ is continuous. In terms of open sets, a set $U\subseteq B(\mathscr H)$ is open in the weak operator topology if, for all $x,y\in \mathscr H$, there exists some open set $V\subseteq \mathbb C$ such that $f_{xy}^{-1}(V) = U$.

In contrast, note that the strong operator topology $\mathcal O_S$ - which is the topology on $B(\mathscr H)$ induced by the operator norm - also satisfies the condition that each $f_{xy}$ is continuous, but is stronger (or finer) than $\mathcal O_W$ in the sense that all open sets in $\mathcal O_W$ are also open in $\mathcal O_S$ but the reverse is not true.

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Compactness of $\mathcal S(\mathscr H)$ in $\mathcal O_W$ can be shown by defining the function

$$\pi : \mathscr B_1 \rightarrow \prod_{x,y\in \mathscr H} D_{\Vert x\Vert \cdot \Vert y \Vert}$$ $$ T\mapsto \langle x,Ty\rangle$$

where $$\mathscr B_1 = \bigg\{T\in B(\mathscr H) \ : \ \Vert T\Vert_{op} \leq 1\bigg\}$$ is the unit ball in $B(\mathscr H)$ and $D_{\Vert x\Vert \cdot \Vert y \Vert } \subset \mathbb C$ is the disk (centered at 0) with radius $\Vert x\Vert \cdot \Vert y \Vert$. For whatever scant intuition you can find, this maps each operator in $\mathscr B_1$ to the list of all possible matrix elements.

From here, the logic goes roughly as follows:

1. The image $\pi(\mathscr B_1)$ is closed because it contains all of its limit points, by the definition of the weak operator topology.
2. The product of disks is compact (in the usual product topology), and a closed subset of a compact set is compact, so $\pi(\mathscr B_1)$ is compact.
3. $\pi$ constitutes a homeomorphism between $B_1$ and its image. This is not immediately trivial, but is not so hard to prove. Intuitively, it is invertible because a list of all possible matrix elements uniquely defines an operator, and continuous (with continuous inverse) due to the definition of $\mathcal O_W$. As a result, $\mathscr B_1$ is compact in $\mathcal O_W$.

To Be Finished ...

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To answer your question about the physical relevance of this topology, note that the observable characteristics of an operator can all be framed in terms of its matrix elements. Loosely, a sequence of bounded operators converges in the weak operator topology if and only if all of its matrix elements (and therefore observable characteristics) converge.

As a toy example. consider an orthonormal basis $\phi_i, i\in \mathbb N$ and the sequence of shift operators $T_n: \phi_i \mapsto \phi_{i+n}$. Given any physical state vector $\psi = \sum_i c_i \phi_i$, the sequence $c_i$ must converge to zero. For sufficiently large $n$, the overlap between $\psi$ and $T_n\psi$ becomes smaller and smaller as $n$ increases. It is clear that the sequence $T_n$ does not converge in the strong operator topology, but it does converge in the weak operator topology because for all $x,y\in \mathscr H$, the sequence $\langle x, T_ny\rangle \rightarrow 0$.