- #31
aleazk
Science Advisor
- 83
- 62
rubi said:
Hi, rubi, I agree with everything you said in this thread. I also know that you said what I quoted in order to show how the equivalence is trivially shown and that it has nothing to do with the Stone-von Neumann theorem, etc. But, let me make the small correction that this argument doesn't necessarily show the equivalence of both pictures in the most general sense in which it's possible, or, most importantly, required. And that's because the definition of expectation values as ##<A>_{\psi}=(\psi\mid A\psi)##, with ##\psi\in D(A)##, is, though correct, not the most general one.
Given an observable ##A## characterized by the PVM ##P^{(A)}## and a state ##\rho## (a positive trace class operator of trace 1), one can define the following probability measure associated to the pair (observable, state): ##\mu_{\rho}^{(A)}(E)=tr(\rho P^{(A)}(E))##, for every Borel set ##E## of the real line. By its definition, it's clear that the interpretation of the measure ##\mu_{\rho}^{(A)}## is that it represents the probability that the measuring of ##A## in state ##\rho## falls in ##E##. In this way, we can define the expectation value of the observable ##A## in state ##\rho## by: ##<A>_{\rho}=\int_{\mathbb{R}}\lambda\mathrm{d}\mu_{\rho}^{(A)}(\lambda)##.
The expectation value will exist when ##\mathbb{R}\ni\lambda\longmapsto\lambda## is in ##L^{1}(\mathbb{R},\mu_{\rho}^{(A)})##. This will depend on the state ##\rho##. Now, let's restrict to pure states only; ##\rho_{\psi}## will denote the pure state determined by the normalized vector ##\psi\in\mathcal{H}##. It can be shown that: i) ##<A>_{\rho_{\psi}}## exists iff ##\psi\in D(\mid A\mid^{\frac{1}{2}})##; ii) if ##\psi\in D(A)## then the expectation value clearly exists since ##D(A)\subset D(\mid A\mid^{\frac{1}{2}})##, in this case we also get ##<A>_{\rho_{\psi}}=(\psi\mid A\psi)##.
Thus, the problem with that argument is that it only shows the equivalence of both pictures for only a proper subset from the set of all pure states for which the expectation value can actually be defined. It could turn out that for these other pure states the equivalence is not valid. Just to exemplify how tricky these things can be, there are examples in which two unbounded operators commute in a common domain in which both are essentially self-adjoint but, nevertheles, the associated (to the self-adjoint extensions) PVMs do not commute.
The solution to this problem is easy, one simply defines everything in terms of the PVMs. After doing this, the equivalence can be rigorously proved for all states in which the expectation value exists, as I mentioned in a previous post.