Implementing states on a C*-algebra that has a faithful representation on a separable Hilbert space

Question

Let ${\cal A}$ be a C*-algebra that has a faithful representation on a separable Hilbert space over $\mathbb{C}$. This condition is only meant to express an intrinsic property of ${\cal A}$, not to fix any particular representation.

Let $\rho:{\cal A}\to\mathbb{C}$ denote a state, by which I mean a normalized positive linear functional. Under what conditions (on ${\cal A}$ and $\rho$) can $\rho$ be expressed in the form $$ \rho(X) = \sum_n \langle u_n,\,X\hat\rho u_n\rangle \tag{1} $$ for all $X\in{\cal A}$ and some $\hat\rho\in{\cal B}({\cal H})$, where the sum is over an orthonormal basis $u_n\in{\cal H}$ in some representation of ${\cal A}$ on some separable Hilbert space ${\cal H}$ (not necessarily the same representation for all states $\rho$)?

In case asking for necessary-and-sufficient conditions is unrealistic or unnatural, I would be happy with answers to these:

• Are there ${\cal A}$ such that no states $\rho$ have the form (1)?

• Are there ${\cal A}$ such that every state $\rho$ has the form (1), aside from those ${\cal A}$ that can be faithfully represented on a finite-dimensional Hilbert space?

Answer 1

If you allow yourself to change the representation, then on any C$^*$-algebra you can take $\hat \rho$ to be a rank-one projection: this is the GNS construction.

If you don't change representation, then $(1)$, written more easily as $\rho(X)=\operatorname{Tr}(\hat \rho\,X)$ with $\hat\rho$ a trace-class operator, is the precise characterization of the normal states of $B(H)$.

So the question is whether you can have $\mathcal A$ such that no state extends to a normal state of $B(H)$. This is easily seen to be impossible: because as mentioned above among the normal states you have the point states. Thus the only way $\mathcal A$ can fail to have states of the form (1) is if $\langle a\xi,\xi\rangle=0$ for all $\xi\in H$ and for all $a\in \mathcal A$. But having the for all $\xi$ equality implies that $a=0$. In other words, for any $a\in\mathcal A$ there exists $\xi\in H$ with $\langle a\xi,\xi\rangle\ne0$. In conclusion:

• For any $a\in\mathcal A$, there exists a state $\varphi$ of the form (1) such that $\varphi(a)\ne0$.

• If $\mathcal A$ is separable, there exists a faithful state $\varphi$ of the form $(1)$.

If $\mathcal A$ has every state satisfying $(1)$, then every state is normal. Any infinite-dimensional von Neumann algebra has a non-normal state. So $\mathcal A$ has to be finite-dimensional.