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?
Is the failure of $\mathcal{B}(H)\simeq H\otimes H^*$ in infinite dimensions the reason for non-normal states in quantum information
c-star-algebrasfunctional-analysisoperator-algebrasoperator-theoryquantum mechanics
Related Solutions
The characters of $C_0(X)$ are the maps $\{ ev_x\mid x\in X\}$ where $ev_x: C_0(X)\to\Bbb C$ is given by $f\mapsto f(x)$. As such the associated semi-definite inner-product $(,)_{ev_x}$ is given by: $$(f,g)_{ev_x}:= ev_x(f^* g) =\overline{f(x)}\cdot g(x)$$ And clearly $C_0(X) /N_{ev_x}\cong\Bbb C$ where $N_{ev_x}$ is the null space of $(,)_{ev_x}$. The isomorphism $C_0(X)/N_{ev_x}\to \Bbb C$ is given by $[g]\mapsto g(x)$ as you can check explicitly. Now the action of $C_0(X)$ on this Hilbert space is given by:
$$f\cdot [g] = [f\cdot g]$$ which, under the above isomorphism $C_0(X)/N_{ev_x}\cong \Bbb C$, corresponds to: $$f\cdot z = f(x)\cdot z$$
Now if you put all this together what you get is that the GNS space is: $$H= \bigoplus_{x\in X} \Bbb C = \ell^2(X)$$ and the representation is defined by: $$(\pi(f) v)_{x} = (f(x)\cdot v_x)$$ For some $v = (v_x)_{x\in X}\in \ell^2(X)$.
A key part of the story here is that a $C^*$-algebra whose self-adjoint elements all have finite spectrum must be finite-dimensional. Martin Argerami gave a reference which implies this result by way of several more general results in the context of Banach algebras. However, I decided to write up an account specifically for the $C^*$-algebra case in order to take advantage of the simplifications this context has to offer.
Claim 1: Suppose $A$ is a unital $C^*$-algebra all of whose self-adjoint elements have one point spectrum. Then $A$ is one-dimensional.
Sketch: Use functional calculus to show every self-adjoint element is a multiple of $1_A$, then use the fact that every $C^*$-algebra is the span of its self-adjoint part.
Claim 2: Let $A$ be a $C^*$-algebra in which every self-adjoint element has a finite spectrum and let $p$ be a nonzero projection in $A$ such that the "corner algebra" $pAp$ is not one dimensional. Then, one can write $p=p_1+p_2$ where $p_1$ and $p_2$ are nonzero, orthogonal projections.
Sketch: Working inside $pAp$, whose unit is $p$, the previous claim gives a self-adjoint element whose spectrum is a finite set with at least two elements. Now use functional calculus.
Claim 3: Let $A$ be unital $C^*$-algebra all of whose self-adjoint elements have finite spectrum. Then there exist nonzero, pairwise orthogonal projections $p_1,\ldots,p_n$ summing to $1_A$ such that each of the corner algebras $p_iAp_i$ is one-dimensional.
Sketch: If $A$ is not one-dimensional, subdivide $1_A$ into two projections. If either of the new projections does not determine a one-dimensional corner, subdivide again. This process must eventually terminate and give a decomposition of the desired type. Otherwise, we would have an infinite collection of pairwise orthogonal projections in $A$ which could then be used to embed a copy of $c_0(\mathbb{N})$ into $A$ leading to self-adjoint elements of infinite spectrum, contrary to assumption.
Claim 4: Let $p$ and $q$ be nonzero, orthogonal projections in a $C^*$-algebra $A$ satisfying that $pAp$ and $qAq$ are one-dimensional. Then $qAp$ and $pAq$ are either both one-dimensional or both zero.
Proof: Since $(qAp)^* = pAq$, it suffices to look at $qAp$. Fix any nonzero $a \in qAp$. By the $C^*$-identity, $a^*a \in pAp$ and $aa^* \in qAq$ are nonzero. Since $pAp$ is one-dimensional, up to rescaling $a$, we may assume $a^*a=p$. But then $a$ is a partial isometry, so $aa^*$ is also a projection and, being a nonzero projection in $qAq$, is equal to $q$. The identities $a^*a=p$ and $aa^*=q$ show that $x \mapsto ax : pAp \to qAp$ and $x \mapsto a^* x : qAp \to pAp$ are mutually inverse bijections, and the result follows.
Claim 5: Let $A$ be a unital $C^*$-algebra and let $p_1,\ldots,p_n$ be (necessarily orthogonal) projections satisfying $1_A=p_1 + \ldots + p_n$. Then, $A$ is the internal direct sum of the spaces $p_i A p_j$ where $1 \leq i,j \leq n$.
Basically, each element of $A$ can be thought of as a "matrix of operators" with respect to this partition of the identity.
Final Claim: Let $A$ be a $C^*$-algebra all of whose self-adjoint elements have finite spectrum. Then, $A$ is finite-dimensional.
Proof: First unitize $A$ if necessary (which does not affect the spectra of any of its elements). Then combine Claims 3, 4 and 5.
Best Answer
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).