I was recently made aware (thanks to the answers on [Why does Riesz's Representation Theorem apply in quantum mechanics?](https://mathoverflow.net/questions/382809/why-does-rieszs-representation-theorem-apply-in-quantum-mechanics)) that the $C^\*$ algebra approach and the Hilbert space approach to quantum mechanics don't perfectly align, and the relationship is somewhat subtle.


### Question


If I start with a the Borel algebra on a locally compact Hausdorff space $X$, how do I get a correspondence between probability measures and states (that potentially satisfy a certain property) in the Hilbert space approach?


### Details


In the $C^\*$ algebra approach, this is easy: the states of the algebra $C\_c(X)$ of compactly supported complex-valued continuous functions on $X$ are exactly the ones given by Radon probability measures. (The story is also wrapped nicely in a bow by saying that commutative $C^\*$ algebras always arise in this way.)


But in the Hilbert space approach, I only know how to do this for the case that $X$ is finite and discrete... In that case, let $H=\mathbb{R}^{|X|}$, and then the states (in the Hilbert approach sense, meaning positive semi-definite trace 1 operators) whose eigenbasis is the standard basis are in correspondence with probability measures on $X$.


The story seems more difficult in the case that $X$ is non-finite and non-discrete. Is it true that there is a procedure for this $X$ to induce a Hilbert space in such a way that states that satisfy a certain property correspond to probability measures on $X$? How should one define this $H$? What should its inner product be, and how would it reflect the topology of $X$? What property should we demand of the states (corresponding to the demand that their eigenbasis is the standard basis in the case that $X$ is finite) so that this correspondence holds?


### Clarification (added later)


I am aware that you can do the following construction: Embed $C\_c(X)$ into the algebra of bounded operators on some Hilbert space $H$ via the GNS construction. This is not what I am looking for in this question: it doesn't generalize the discrete finite $X$ case as I described it above. I described a correspondence between probability measures on $X$ and trace 1 diagonal matrices with non-negative entries in $End(\mathbb{R}^{|X|})$. But with the GNS construction, $C\_c(X)$ is itself $\mathbb{R}^{|X|}$, but then the construction embeds it into the algebra of bounded operators on a much bigger Hilbert space (and the states of $C\_c(X)$ are all pure states in this algebra of bounded operators). So I would not consider this as being naturally a Hilbert approach state, but rather degenerating to the $C^\*$ algebra approach.


In order to be a little more concrete, let me try to ask about the following example: Let $X=\mathbb{R}$. Under whatever framework you suggest, even if it goes through the GNS construction, can you describe the standard normal distribution explicitly as some specific operator?



Is there a Hilbert space approach to commutative probability theory on locally compact spaces?
I am under the impression that part of the problem here is the relevant distinction between *states* and *normal states*. So let me briefly recall this distinction, before getting to the actual question.


**States vs. normal states**


Note that there are actually two *$C^\*$-algebra like* settings to describe a dual pair of observables and states:


1. We may consider a general $C^\*$-algebra $A$ and its dual space $A^\*$. In this case, we consider the self-adjoint elements of $A$ as *observables*, and the normalised positive elements of $A^\*$ as *states*.
2. We may consider a von Neumann algebra $A$; this space does not only have a dual space $A^\*$, but also a pre-dual space $A\_\*$ - and we may identify $A\_\*$ with a subspace of $A^\*$ by means of evaluation.  

Again, we have the observables and the states from 1) - but the existence of the pre-dual space $A\_\*$ allows us to define the concept of a *normal state*: a state is called a *normal state* if it is, in addition, an element of $A\_\*$.


Setting 2) includes what you refer to (if I understand you correctly) as the "Hilbert space approach": the space $B(H)$ is not only a $C^\*$-algebra, but even a von Neumann algebra, and the space of trace class operators can be indentified with the pre-dual $B(H)\_\*$. Hence, what is sometimes called the "states" on $B(H)$, namely the positive trace class operators of trace $1$, is - in the terminology above - actually the *normal* states on $B(H)$.


**Classical probabilities (i.e., the commutative case)**


If $X$ is a locally compact Hausdorff space then, as noted in the question, we can consider the $C^\*$-algebra $C\_0(X)$. The observables are then the real-valued functions in $C\_0(X)$, and the states as defined above are simply the Radon probability measures on $X$.


If you would like to consider probability distributions on $X$ in a more "Hilbert space manner" then, as suggested above, it is worthwile to rather start with a von Neumman algebra (note that $C\_0(X)$ is not a von Neumann algebra in most cases).


To this end fix a, say $\sigma$-finite, positive Borel measure $\lambda$ on $X$. The von Neumann algebra we now look at is $L^\infty(X, \lambda)$. The space $L^1(X, \lambda)$ is pre-dual to this von Neumann algebra. Hence, the observables are now the real-valued functions in $L^\infty(X, \lambda)$, and the normal (!) states are the functions $0 \le f \in L^1(X, \lambda)$ with integral $1$. (The general states, however, are the positive normalized functionals on $L^\infty(X, \lambda)$, which constitute an extremely large space.)


(Warning: Note that this approach allows us only to consider probability measures on $X$ that have a density with respect to $\lambda$.)


**Considering classical probabilies as states on $B(H)$**


The von Neumann algebra $L^\infty(X, \lambda)$ can easily be seen as a subspace of an operator space $B(H)$: namely, consider the Hilbert space $H = L^2(X, \lambda)$; then each function $m \in L^\infty(X, \lambda)$ acts as a multiplication operator on $H$, which yields a $C^\*$-embedding $L^\infty(X, \lambda) \hookrightarrow B(L^2(X, \lambda))$. So each observable from $L^\infty(X, \lambda)$ can be seen as self-adjoint multiplication operator in $B(L^2(X, \lambda))$.


If it comes to states, the situation is as follows:


* Every state on $L^\infty(X, \lambda)$ be be extended to a state on $B(L^2(X, \lambda))$; this is a consequence of the Hahn-Banach theorem.
* More relevantly to the question though, every *normal* state on $L^\infty(X, \lambda)$ - i.e., every function from $L^1(X, \lambda)$ - can be extended to a *normal* state on $B(L^2(X, \lambda))$ - i.e., to a trace class operator on $L^2(X, \lambda)$.


Indeed, let $0 \le d \in L^1(X, \lambda)$. Then $\sqrt{d}$ is a function in $L^2(X, \lambda)$, and the rank-$1$ operator $\sqrt{d} \otimes \sqrt{d}$ is a trace class operator on $L^2(X, \mu)$. It is easy to see that, if $m \in L^\infty(X, \lambda)$ and if $M \in B(L^2(X, \lambda))$ denotes the multiplication operator corresponding to $m$, then $\operatorname{tr}\big( (\sqrt{d} \otimes \sqrt{d}) M\big) = \int\_X d m \, \operatorname{d}\lambda$.


**The standard normal distribution on $\mathbb{R}$**


At the end of the question, the normal distribution on $X = \mathbb{R}$ is mentioned. For this example, it is probably easiest to choose $\lambda$ as the Lebesgue measure on $\mathbb{R}$. Let $d \in L^1(\mathbb{R}, \lambda)$ denote the density of the standard normal distribution, i.e.,
$$
 d(x) = \frac{1}{\sqrt{2\pi}} \exp(-\frac{x^2}{2}) \quad \text{for } x \in \mathbb{R}.
$$
Then $d$ is a normal state on the von Neumann algebra $L^\infty(\mathbb{R}, \lambda)$. As explained above, we can extend it to a normal state on the von Neumann Algebra $B(L^2(\mathbb{R}, \lambda))$, namely to the state given by the trace class operator $\sqrt{d} \otimes \sqrt{d}$; the kernel $k$ of this trace class operator is given by
$$
 k(x,y) = \sqrt{d(x)} \sqrt{d(y)} = \frac{1}{\sqrt{2\pi}} \exp(- \frac{x^2 + y^2}{4}) \quad \text{for } x,y \in \mathbb{R}.
$$


