Is the failure of $\mathcal{B}(H)\simeq H\otimes H^*$ in infinite dimensions the reason for non-normal states in quantum information?

Question

In the algebraic formulation of quantum physics/information, states $\omega: \mathcal{A}\rightarrow \mathbb{C}$ are defined as linear functionals on a $C^*$-algebra $\mathcal{A}$ (algebra of observables, representable as a subalgebra of $\mathcal{B}(H)$ for some Hilbert space $H$ via the GNS construction) that are positive ($\omega(A^*A)\geq 0\,\forall A\in \mathcal{A}$) and normalized ($\omega(I)=1$ for $\mathcal{A}$ with unit element $I$ or an equivalent condition for non-unital $\mathcal{A}$). These quantum states are then usually represented as density operators defined via $\omega(A)=:\text{Tr}(\omega A)$, but it is well-known that in infinite dimensions there are so-called non-normal states that are not representable this way. Is this due to the fact that for infinite dimensions $\mathcal{B}(\mathcal{H})\simeq H\otimes H^*$ does not hold?

Answer 1

What follows (and all references given) will be based on Chapter 16 of the book "Introduction to Functional Analysis" by Meise & Vogt (1997). There you can also read up on the key spaces in this matter if you are not all too familiar with them yet: the compact operators $\mathcal K(\mathcal H)$ and the trace class $\mathcal B^1(\mathcal H)$ (in the above book denoted by $S_1(\mathcal H)$ for the Schatten-1-class). The inclusions between these spaces in infinite-dimensions read $\mathcal B^1(\mathcal H)\subsetneq\mathcal K(\mathcal H)\subsetneq\mathcal B(\mathcal H)$ (in finite-dimensions they all coincide).

One needs the trace class, obviously, for the trace $\operatorname{tr}(\omega B)$, $B\in\mathcal B(\mathcal H)$ to make sense beyond finite dimensions. One can actually show that every continuous linear functional $\tau:\mathcal B^1(\mathcal H)\to\mathbb C$ (i.e. every dual space element $\tau\in(\mathcal B^1(\mathcal H))'$) is precisely of the form $$ \tau(A)=\operatorname{tr}(AB)\qquad\text{ for some }B\in\mathcal B(\mathcal H)\text{ and all }A\in\mathcal B^1(\mathcal H)\,,\tag{1} $$ cf. Proposition 16.26; for short $(\mathcal B^1(\mathcal H))'\cong\mathcal B(\mathcal H)$. Also the trace class in infinite dimensions is not reflexive (Corollary 16.27) meaning that $\mathcal B^1(\mathcal H)\not\cong (\mathcal B^1(\mathcal H))''\cong (\mathcal B(\mathcal H))'$. In other words, not every dual space element of $\mathcal B(\mathcal H)$ is of this trace form (1).

Instead one can show that every $\varphi\in(\mathcal K(\mathcal H))'$ can be written as $\varphi(K)=\operatorname{tr}(KA)$ for some $A\in\mathcal B^1(\mathcal H)$ (Proposition 16.24). So if one restricts oneself to the compact operators, every functional is of trace form again--but $\mathcal B(\mathcal H)$ is obviously "way larger" than $\mathcal K(\mathcal H)$.

To conclude, all I said can then be nicely summarized in one identity: $$ \boxed{(\mathcal B(\mathcal H))'\supsetneq(\mathcal K(\mathcal H))'\cong \mathcal B^1(\mathcal H)} $$ which--reformulated--says that there exist "non-normal states" in infinite dimensions (those are precisely the states which are not weak-* continuous).