in $C^*-$algebras with unit element, there is the definition of a state, as a functional $\omega$ with $\omega(e)=||\omega||=1.$
Now, of course there is also in classical physics and quantum mechanics the definition of a state.
In classical physics this is either a point in phase space or more generally a probability measure on this space.
In quantum mechanics this is either a wavefunction or a density matrix.
Now there are basically two interesting examples of $C^*-$ algebras I would say: $L(H)$ the space of bounded operators on some Hilbert space $H$ (a non-comm. $C^*-$ algebra or $C_0(X)$ the space of $C_0-$ functions on some locally compact Hausdorff-space (a commutative one).
Obviously, if $X$ is some compact subset of $\mathbb{R}^n,$ then $C(X)$ is a $C^*-$ algebra with unit element and dirac measures on $X$ and more generally probability measures are indeed states as we defined them in the functional analysis sense.
Moreover, if we work on some Hilbert space $H$ then the density matrices $\rho$ define functionals $l:L(H) \rightarrow \mathbb{R}, T \mapsto tr(\rho T).$ So these are also states in the sense of functional analysis.
But this made me think whether
1.) Every state in the sense of functional analysis can be interpreted as a physical state?
2.) Where does the interpretation come from that the commutative $C^*-$algebra cooresponds to classical mechanics and the non-commutative one to quantum mechanics? Is there any deep interpretation of this fact? (besides the fact that non-commutativity is known to be an issue for QM?)Cause this seems to be much deeper here as this is the only distinguishing fact between the two in this setup.
Best Answer
1.) Yes. In the commutative case, this is the statement of the Riesz representation theorem (any linear functional on $C(K)$ is an integral against a measure, which has to be positive by the positivity condition on the state).
In the non-commutative case, the answer may be Yes or No, depending on how you phrase the question. As pointed out in the comments, there may be states $\langle - \rangle$ on $L(H)$ that are not of the form $\operatorname{tr}(\rho(-))$ for a trace-class $\rho$. However, the GNS theorem states that any state gives rise to a representation $\pi$ of $L(H)$ by bounded operators on a possibly different Hilbert space $H'$ such that the state is in fact of the form $\operatorname{tr}(\rho \pi(-))$ for some trace-class (in fact rank-1) $\rho$. Of course, the representation $\pi\colon L(H) \to L(H')$ on $H'$ may not be unitarily equivalent to the defining representation $\mathrm{id}\colon L(H) \to L(H)$ on $H$.
2.) This question is rather subjective, so it is impossible to answer with mathematical precision. Both classical and quantum mechanics came first, along with their own motivations, and the $C^*$-algebraic descriptions were later noticed and retrofitted onto the theories. One thing that can be said is the following. Thinking carefully about taking the classical limit of a quantum system, one can recover from a non-commutative algebra a commutative one with a Poisson bracket. Naturally one can also ask whether a non-commutative algebra can be obtained from a commutative one with a Poisson bracket. Such things are studied under the name deformation quantization.