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

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 $$\mathcal{B}(H)$$ 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?

• Just a comment: GNS construction doesn't tell you that $A$ can be represented as $B(H)$. If $\omega$ is faithful, $A$ can be represented as a $C^*$-subalgebra of $B(H)$ via the GNS construction. If $\omega$ is not faithful map $A\rightarrow B(H)$ will have nontrivial kernel. – Mogget Jun 27 '19 at 18:13

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).