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One of the basic postulates in the mathematical formalism of quantum mechanics is that the probability of a measurement of an observable $A$ in the state $\psi \in \mathscr{H}$ to return a value in a Borel set $E \subset \mathbb{R}$ is given my: $$P_{\psi,A}(E) := \mu_{\psi}(E) \equiv \langle \psi, \chi_{E}(A) \psi\rangle \tag{1}\label{1}$$ where $\mu_{\psi}$ is the spectral measure associated to $\psi$ which, by definition, is given by the inner product $\langle \psi, \chi_{E}(A)\psi\rangle$ with $\chi_{E}(A)$ being the characteristic function $x \mapsto \chi_{E}(x)$ of the Borel set $E$.

If $f$ is a complex-valued measurable function on $\mathbb{R}$, we can consider the expectation: $$\mathbb{E}_{\mu_{\psi}}[f] := \int f(x)d\mu_{\psi}(x) \tag{2}\label{2}$$ whenever this integral is well-defined. In particular, by the Borel functional calculus, we have: $$\int x d\mu_{\psi}(x) = \langle \psi, A\psi \rangle. \tag{3}\label{3}$$

Relations (\ref{2}) and (\ref{3}) are responsible for the probabilistic interpretation of quantum mechanics. The expectation (\ref{2}) tells us that $\mu_{\psi}$ is to be considered as a probability distribution and then relation (\ref{3}) justifies saying that $\langle \psi, A\psi\rangle$ is the average value of $A$.

I understand the point, but something seems to be missing when one defines the average value of $A$ by means of a probability distribution which is in another space instead, the space of Borel measurable functions. I feel I am missing the connection between these two spaces.

So my question is: does the spectral measure (\ref{1}) induces some probability measure on the space of observables, say, as a pushforward measure or something like this? And if so, how can the expectation of an observable $A$ in this measure space be related with (\ref{2}) and (\ref{3}) so to interpret integration with respect to $\mu_{\psi}$ as the proper expectation value of $A$? Maybe an application of the abstract change of variables formula?

Note: I know that, in the $C^{*}$-algebra approach of quantum mechanics, observables are bounded linear operators $\mathcal{L}(\mathscr{H})$ and there is the topology induced by the operator norm on this space. Hence, if we want to define a probability measure on this space, it is quite natural to talk about Borel $\sigma$-algebras and Borel measures. However, in von Neumann's approach, observables are allowed to be unbounded, even though we still have the probabilistic interpretation. This is part of my question: what exactly is the space of observables and how to introduce a $\sigma$-algebra and a probability measure on it so that it becomes $\mathbb{E}_{\mu_{\psi}}$ when averaging $A$?

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    $\begingroup$ I'm not sure if I understand all your points correctly, but there seems to be some misconception in your post: I don't see why you would want a probability measure on the space of observables. When comparing quantum mechanics to classical probability theory, an observable $A$ shoud be interpreted as the quantum mechanical version of a random variable. Computing $\langle \psi, A \psi \rangle$ is similar to integrating a random variable against a probability distribution and thus yields the expected value of the observable/random variable. $\endgroup$
    – Jochen Glueck
    Commented May 2, 2023 at 14:48
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    $\begingroup$ Similarly, in the $C^*$-algebra approach one does not consider probability measures on the $C^*$-algebra. Rather, each element of the $C^*$-algebra is a (non-commutative version of) a random variable and the role of the probability measures is taken by the trace class operators. $\endgroup$
    – Jochen Glueck
    Commented May 2, 2023 at 14:48
  • $\begingroup$ Regarding the question whether observables are bounded or unbounded operators: Think of a real-valued random variable on, say $\mathbb{R}^d$. If it is bounded, then you can integrate it against every probability measure on $\mathbb{R}^d$; hence, the expected value of the random variable exists, no matter which probability measure you consider. For an unbounded random variable though, the expected value might or might not exist, depending on the probability measure. Similarly, for unbounded $A$: whether you can or can't compute $\langle \psi, A \psi \rangle$ depends on $\psi$. $\endgroup$
    – Jochen Glueck
    Commented May 2, 2023 at 14:53
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    $\begingroup$ Putting a probability measure on the space of observables would be completely at odds with the experimental situation the theory is supposed to describe: one decides on an observable to measure (such as position, momentum, energy, etc.), and then what is random is the result of the measurement (= a real number) of this fixed observable. One does not measure an unknown random observable. $\endgroup$
    – Christian Remling
    Commented May 2, 2023 at 15:58
  • $\begingroup$ @JochenGlueck thank you for your comments. Let me address the first one, to clarify my question. In classical probability, one starts with a probability space $(\Omega, \mathcal{F},p)$ and a random variable $X: \Omega \to \mathbb{R}$. Then one defines a measure $\mu$ on $\mathbb{R}$ by the pushforward $\mu(A) = p(X^{-1}(A))$, which is the probability distribution. Of course, if you want to calculate expected values of random variables one has $\int X(\omega)dp(\omega) = \int x d\mu(x)$. My question is exactly how can observables be thought as random variables in this sense. $\endgroup$
    – MathMath
    Commented May 2, 2023 at 16:02

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