Recent content by Petro z sela

Is something wrong in my assertions below? Suppose we have two quantum systems N and X. Let N is described by discrete observable \hat{n} (bounded s.a. operator with discrete infinite spectrum) with eigenvectors |n\rangle. Let X is described by continuous observable \hat{x} (unbounded s.a...

Sorry for "Gilbert". I am not so good in English ). 1. I agree. Thanks. I must be considerate towards "wikipedia" ). 2 and 3 are still actual.

3. Ok, I am aware that eigenvectors of unbounded (continuous, continuous+discrete spectrum) lie in Ω\times. But I think that only vectors from Ω can be decomposed over the CSCO basis. Thanks, I'm waiting for 1, 2 )

3. So, answer is Ω? 4-5. Operators forming CSCO algebra must have one common invariant domain. You said both Ω and Ω\times. Which of them? I think must be one domain.

3. You say Ω\times. It means that I can decompose any vector from Ω\times over the eigenvectors of CSCO. Is this true for vectors from H\Ω? I don't think so, because nuclear spectral theorem allow us to decompose only vectors from Ω over the eigenvectors of unbounded operator (suppose CSCO...

Suppose that we have rigged Gilbert space Ω\subsetH\subsetΩ\times (H is infinite-dimensional and separable). Is the Ω a separable space? Is the Ω\times a separable space? Consider the complete set of commuting observables (CSCO) which contain both bounded and unbounded operators...