Timeline for Probabilistic interpretation of von Neumann's approach to quantum mechanics
Current License: CC BY-SA 4.0
7 events
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| May 2, 2023 at 18:25 | comment | added | Christian Remling | @MathMath: You have $\Omega=\mathbb R$, $p=\mu$, and $X(t)=t$. | |
| May 2, 2023 at 16:02 | comment | added | MathMath | @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. | |
| May 2, 2023 at 15:58 | comment | added | Christian Remling | 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. | |
| May 2, 2023 at 14:53 | comment | added | Jochen Glueck | 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 $\psi$. | |
| May 2, 2023 at 14:48 | comment | added | Jochen Glueck | 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. | |
| May 2, 2023 at 14:48 | comment | added | Jochen Glueck | 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. | |
| May 2, 2023 at 14:31 | history | asked | MathMath | CC BY-SA 4.0 |