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Wavefunctions space in Quantum Physics

  1. Dec 13, 2004 #1
    Well, a week ago my Professor said the space of quantum physical states was a Hilbert space. Thing is, he just said it, and moved on.
    So I have a vector space with a scalar product. Is it, indeed, Hilbert? That is, is it complete?
    I guess I'll see the answer next semester, in real functions analysis or sth., but I'm kinda curious, so thanks in advance for your answers.
     
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  3. Dec 13, 2004 #2

    dextercioby

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    A finite/infinite dimensional vector space on the vectors of which u define an application from the direct product of the space with itself into the body of scalars on which you build the vector space is called a preHilbert space.
    The scalar product is a sesquilinear application which can allow to structure the vector space as a topological vector space,on which u can define a metric and a norm Completion wrt to the norm of a preHilbert space defines a Hilbert space.
     
  4. Dec 13, 2004 #3

    Gokul43201

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    This is correct, but more specifically, it is an infinite dimensional Hilbert space.

    It is a Hilbert Space, only if it is complete with respect to the norm. So, to be a Hilbert Space, it must first be a Banach Space.

    A common misconception is that all Hilbert Spaces are infinite dimensional. This is not true. For instance, any n-dimensional Euclidean Space with the usual dot product is a Hilbert Space.
     
  5. Dec 13, 2004 #4
    Well, you toss out solutions which don't belong to the Hilbert space, so it'll be complete. I guess the world only allows solutions in the Hilbert space. Also normalization may not be to unity, but the dirac-delta function.
     
  6. Dec 13, 2004 #5
    First of all, thanks to you all.
    Now, another question: I was told today that it is also separable. Is L_2 indeed separable? Can you show me the densed group in it?
     
  7. Dec 13, 2004 #6

    dextercioby

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    This is tricky.Requires that bunch of functional analysis which i find horrible.I myself,as a physicist,will never attempt to give demonstrations on delicate matters of topology.[itex] L^{2} (R^{n}) [/itex] is separable,and the proof is to be found in a serious book on functional analysis.I don't know the poof,and i'm not interested in why [itex] L^{2} (R^{n}) [/itex] is separable.It's important for QM that it is separable.A mathematician proved it.I have a hunch it was von Neumann,but i'm not sure.And i won't look for it,as this is just one from the many more other results from mathematics that a physicst uses,but it would not help him a lot to find "why is that??"."Why" is a question for mathematicians.If we physicist would have to come up with the proofs for every mathematical result he uses,then mathematicians would be useless,as we physicists would be making the mathematics.

    Daniel.

    PS.If u're really interested,maybe one of the mathematicians on this forum would pin point to a book where u can find the proof,that,of course,if he does not give you the proof in a post in this thread.I think,among the physics books that i know of,u have a chance of finding it in Prugoveçki's book/bible:"Quantum Mechanics in Hilbert Space".


    PPS:Dense subset,not subgroup.
     
  8. Dec 13, 2004 #7
    Should it be noted that Hilbert space is complex?
     
  9. Dec 13, 2004 #8

    dextercioby

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    It's irrelevant whether it's real or complex.It could be quternions,even. :wink: It's just a body of scalars...

    Daniel.
     
  10. Dec 14, 2004 #9
    They say field in english. We too in France say body. There is no real confusion possible between the "field" algebraic structure and the physicist's field which is a function, or a generalization of it.
     
  11. Dec 14, 2004 #10

    dextercioby

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    Thank you!!!!!!!!!! :wink: This was a really useful post from u.Useful to me,as i had wondered for about 2 months on how to say in English "corp",else than "body". :biggrin:

    Daniel.
     
  12. Dec 14, 2004 #11
    It is sufficient to show that there is a countable basis, that is : that the dimension is countable. For instance, the set [tex]\left\{v_n : n \in \mathbb{Z}\right\}[/tex] with [tex]v_n(x) = \exp^{2\pi \imath nx}[/tex] forms an orthonormal basis of the complex space [tex]L^2([0,1])[/tex].

    From Wikepdia :
     
  13. Dec 14, 2004 #12
    Merci :smile:

    And thanks a lot, dex.
     
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