I was recently made aware (thanks to the answers on Why does Riesz's 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.


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?


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?

  • 2
    $\begingroup$ You should search for GNS (Gelfand, Naimark, Segal) representation. $\endgroup$ – Ruy Feb 6 at 1:40
  • 4
    $\begingroup$ Small remark: please note that the space $C_c(X)$ that you refer to in your section "Details" is not a $C^*$-algebra since it is not norm complete. The $C^*$-algebra you are looking for is probably $C_0(X)$ (i.e., the space of continuous functions that vanish at infinity). $\endgroup$ – Jochen Glueck Feb 6 at 13:03
  • 2
    $\begingroup$ You mention $\mathbb{R}^{|X|}$ for your $H$ in finite dimensions... perhaps you mean $\mathbb{C}^{|X|}$? One usually considers complex Hilbert spaces in these contexts... I apologize in advance if I am missing something... $\endgroup$ – Ujan Chakraborty Feb 6 at 13:11
  • 3
    $\begingroup$ Following up on @UjanChakraborty's comment, the notation $K^{|X|}$ is rather unusual in this context anyway. If you want a Hilbert space of dimension $|X|$, a standard realization would be $\ell^2(X)$. $\endgroup$ – Christian Remling Feb 6 at 14:43
  • 3
    $\begingroup$ Also, your final remarks, on the GNS Hilbert space of $C(X)$, seem to be a bit off the trodden path. For example, if $X=[0,1]$, then a natural Hilbert space would be not $\ell^2(X)$ (way too big) but $L^2(0,1)$. $\endgroup$ – Christian Remling Feb 6 at 14:48

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 $$ k(x,y) = \sqrt{d(x)} \sqrt{d(y)} = \frac{1}{\sqrt{2\pi}} \exp(- \frac{x^2 + y^2}{4}) \quad \text{for } x,y \in \mathbb{R}. $$

  • $\begingroup$ This seems to be a great answer, and exactly what I need to clear things up in my mind. Give me some time to process, and I'll write back here if I have questions. Thanks! $\endgroup$ – Andrew NC Feb 7 at 3:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.