
Re: Little Quantum Mechanics Question
Posted:
Sep 16, 2003 2:53 PM


In article <bk75eq$eoi$1@bunyip.cc.uq.edu.au>, D.McAnally@i'm_a_gnu.uq.net.au (David McAnally) writes: >meron@cars3.uchicago.edu writes: > >>>>>But, also, the postion operator, X , is an observable: >>> >>>>>Xx> = xx> >>> >>>>>So one can expand Psi> as: >>> >>>>>Psi> = Integrate[ f(x)x> , {x,0,L} ] >>> >>>In physics, maybe, but in math we tend to be more careful. > >>Yes, true. I admit that physicists tend to be kinda sloppy about >>these things. Usually they can get away with this, but not always. >>So, I plead guilty as charged. > >>>The position operator has a continuous spectrum. As George Jones >>>observed, x> is not a member of the Hilbert space, and X has >>>_no_ eigenvectors, let alone an orthonormal basis of them. >>>Mathematicians (or mathematical physicists) use various devices >>>to deal with this, e.g. the "projectionvalued measure" form of the >>>Spectral Theorem. > >As I noted elsewhere, x> actually DOES exist in the largest space of >a Gelfand triple, the middle space of which is the Hilbert space. > Yes, only it doesn't exist within Hilbert space proper. In many situations this distinction can be ignored. Sometimes, it cannot.
Mati Meron  "When you argue with a fool, meron@cars.uchicago.edu  chances are he is doing just the same"

