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Re: Little Quantum Mechanics Question
Posted:
Sep 16, 2003 2:53 PM
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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:
>>>
>>>>>X|x> = x|x>
>>>
>>>>>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 "projection-valued 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"
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