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...
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...