$\DeclareMathOperator\tr{tr}$One begins with a quantum mechanical system, i.e. a unital $C^*$-algebra $A$.
It is common to begin the discussion with embedding $A$ into the algebra of bounded operators $\mathcal{B}$ on some Hilbert space $H$.
A state is defined as a positive linear functional $\varphi: A\rightarrow \mathbb{C}$ taking $1$ to $1$. Since neither $A$ nor $\mathcal{B}$ is a Hilbert space, we can't use Riesz's Representation Theorem directly.
In Physics texts that I have encountered, however, a state is often cast as a trace $1$ operator $\alpha$ in $\mathcal{B}$, and its action on an observable $\rho$ is by $\tr(\rho\alpha)$.
Recall that observables are self dual, and so $\tr(\rho\alpha)=\tr(\rho^* \alpha)$, which is highly suggestive of the subalgebra of $\mathcal{B}$ given by the Hilbert–Schmidt operators, which is a Hilbert space with the inner product $\alpha\cdot\beta=\tr(\alpha^*\beta)$.
Indeed if you restrict $\varphi$ to the algebra of Hilbert–Schmidt operators, then $\varphi$ can be associated with a trace $1$ operator via Riesz's Representation Theorem.
Questions
- What is going on here? Are we saying that states (defined as positive linear functionals taking $1$ to $1$) are completely determined by their restriction to the subalgebra of Hilbert–Schmidt operators in $A$? Or are there two competing definitions of "states" here? And if so, what is the merit of having these two different definitions?
- In https://math.stackexchange.com/questions/77820/a-question-about-pure-state they appear to suggest that not all pure states are represented by projections to one dimensional subspaces of $H$. This confuses me, because I thought those are exactly the pure states. Is this somehow an issue with diverging definitions, related perhaps to my first question?
Best Answer
Okay, there is a lot of confusion in this question.
First, I'm not sure why you say ``it is common to begin the discussion with embedding $A$'' into $B(H)$. The point of the C${}^*$-algebra approach to quantum mechanics is doing things in a representation-independent manner, so I would say it's unusual to begin the discussion this way.
You are right that a state on a C${}^*$-algebra is a unital positive linear functional. This isn't really related to the Riesz representation theorem. You seem to want to apply that theorem to the Hilbert-Schmidt operators, which constitute a Hilbert space, but they do not constitute a C*-algebra, and many C${}^*$-algebras contain no Hilbert-Schmidt operators. (So the answer to question 1 is an emphatic ``no''.)
Part of the confusion may have to do with the distinction between pure and mixed states. In the Hilbert space approach to quantum mechanics, pure states are represented by unit vectors in the Hilbert space and mixed states are represented by positive, norm one, trace-class operators. If you're working with C${}^*$-algebras then the mixed states are the states defined above, and the pure states are the extreme points of the set of mixed states. You can use the GNS representation to put a C${}^*$-algebra into a Hilbert space in such a way that the states on $A$ (in the C${}^*$-algebra sense) extend to states on $B(H)$ (in the Hilbert space sense).
Finally, the issue in the question you linked to arises because someone is applying the C${}^*$-algebra definition of states to $B(H)$. This is a little more subtle because we can turn C${}^*$-algebra states into Hilbert space states by embedding the C${}^*$-algebra into some $B(H)$, but if you start with $B(H)$ you might have to embed it in a larger $B(K)$.
(Maybe I should add that looking at states on $B(H)$ in the C${}^*$-algebra sense is something a C${}^*$-algebraist might do, but it's not something mathematical physicists typically do.)
My answer to this question goes into further detail about why one would bother with the C${}^*$-algebra approach to quantum mechanics, etc.