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Hilbert space dimension contradiction

  1. Aug 4, 2009 #1
    Hi,

    I was wondering how the state vector for a particle in a 1-D box can be expanded as a linear combination of the discrete energy eigenkets as well as a linear combination of the continuous position eigenkets. It seems to me that this is a contradiction because one basis is countable whereas the other is uncountable. Is there any way of reconciling this?
     
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  3. Aug 4, 2009 #2

    Dale

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    That is an interesting point. I dont know the answer, but I wonder if it has something to do with the fact that the energy eigenvalues are unbounded while the position eigenvalues are bounded. It is not immediately apparent to me how a bounded but uncountable set could be mapped to an unbounded but countable set.
     
  4. Aug 5, 2009 #3
    The state space for this system is the space of square-integrable functions on [0,1]. This has a countable base. You might like to look at some books on measure theory, Lebesgue integration and functional analysis, e.g. Rudin or Lang. It's a big subject and a cornerstone of modern pure math.

    There are ways of defining position eigenkets, but they are not functions and you can't construct a Hilbert space of states out of them.

    Cheers

    Dave
     
  5. Aug 5, 2009 #4

    George Jones

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    Take a look at this thread:

    https://www.physicsforums.com/showthread.php?t=258277.
     
  6. Aug 5, 2009 #5
    Thanks very much, that was really help helpful. I think what you're saying is that since the generalized position eigenstates lie outside the Hilbert space they cannot be considered a basis by definition . This means that the size of [tex]\{|X\rangle\}[/tex] says nothing about the dimension of the subspace of [tex]H[/tex] containing physically viable functions. Is this correct?

    If this is correct, it still seems weird to me that any element belonging to the subspace can be expanded as a linear combination of an uncountable orthogonal set as well as a countable orthogonal basis set. Am I thinking about this in the wrong way? Should I just accept that these two expansions are completely separate entities and that only one of them says something about the size of space?
     
    Last edited: Aug 5, 2009
  7. Aug 5, 2009 #6

    George Jones

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    Sorry, but I'm not an expert on this.
    I think that the nuclear spectral theorem applied to the position operator gives this result.
     
  8. Aug 7, 2009 #7

    Dale

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    Shouldn't the eigenkets of any Hermitian operator form an orthonormal basis? I thought that was a basic property of all Hermitian operators.
     
  9. Aug 7, 2009 #8
    No! This is correct only for bounded operators. And [tex]\hat{x}[/tex] is unbounded.
     
  10. Aug 7, 2009 #9

    Dale

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    Shouldn't x be bounded for the particle in a box?
     
  11. Aug 7, 2009 #10
    [tex]\hat{x}[/tex] is multiplicative operator. Take for exaple function [tex]\frac{x}{1+x^2}[/tex]. This function is in [tex]L_2(\mathcal{R})[/tex].

    And if you multiplicate this function with [tex]x[/tex] you will get function which isn't in

    [tex]L_2(\mathcal{R})[/tex] - [tex]\frac{x^2}{1+x^2}[/tex]. So [tex]\hat{x}[/tex] is unbounded.
     
  12. Aug 7, 2009 #11
    Is the domain of every bounded operator on a Hilbert space the entire space? I think I remember reading that so I'm wondering if this is what you're talking about.

    Also, a few small asides:

    1) Many people in my theoretical chemistry department say a bounded operator is one which has bounded eigenvalues. It's easy to prove that a bounded operator has bounded eigenvalues but do you know if the converse true as well?

    2) Do you know how to reconcile the fact that any wavefunction can be expanded in a countable orthonormal basis as well as in the uncountable generalized eigenfunctions of the position operator? Should I just accept that the latter says nothing about the size of the wavefunction space and leave it at that?
     
    Last edited: Aug 7, 2009
  13. Aug 7, 2009 #12

    Dale

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    That makes sense.

    That is what I was thinking. I guess Petar's example shows that the converse is not always true. For the particle in a box the x operator has bounded eigenvalues but is not a bounded operator.
     
  14. Aug 7, 2009 #13
    Operator [tex]\hat{A}[/tex] is bounded if exist [tex]C[/tex] such that

    [tex]||\hat{A}\varphi||<C||\varphi||[/tex]

    [tex]\forall\varphi[/tex]

    No just for selfadjoint operators.

    2. Because there exist isomorphism between [tex]L^2[/tex] and [tex]l^2[/tex].
     
  15. Aug 7, 2009 #14

    strangerep

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    There's an uncountable infinity of functions f(x) that can be defined on [0,1].
    But only a (countable) subset of them also satisfy the Schrodinger equation,
    which is what's important.

    HTH.
     
  16. Aug 8, 2009 #15

    George Jones

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    Did you mean to write "But the subset of them that also satisfy the Schrodinger equation is separable, which is what's important."?
     
  17. Aug 8, 2009 #16

    George Jones

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    It depends what is meant by "eigenkets." This is true for Hermitian operators on finite-dimensional Hilbert spaces, but the jump in subtlety that occurs in the move from finite-dimensional Hilbert spaces to infinite-dimensional Hilbert spaces is substantial. In terms of honest mathematics, even a bounded Hermitian operator on an infinte-dimensional space need not have any eigenvectors.
    Yes, the position operator for a particle in a box is bounded (the momentum operator is unbounded), but this operator doesn't have any eigenvectors.
    You have shown that the position operator is unbounded for state spaces function defined on all of [itex]\mathbb{R}[/itex], but, for a particle in a box, this operator is bounded.

    The mathematics of general relativity has a reputation of being difficult, but, in my opinion, the mathematics, as honest mathematics, of non-relativistic quantum mechanics is much more subtle and difficult.
     
  18. Aug 8, 2009 #17

    Dale

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    Meaning that the functions which solve the eigenvalue equation are not themselves properly considered vectors in the space of states? And that is because the Dirac delta is not really a function?
     
  19. Aug 8, 2009 #18

    George Jones

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    Right. Delta functions are sometimes called eigendistributions or weak eigenvectors of the multiplication operator.
     
  20. Aug 9, 2009 #19

    strangerep

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    Oops! Yes, I was inexcusably sloppy, wasn't I.

    (For the benefit of other readers, a metric space is said to be "separable"
    if it has a countable dense subset. A Hilbert space is separable iff it has
    a countable orthonormal basis.)

    I share your opinion.
     
  21. Aug 9, 2009 #20
    Why you said that momentum operator is unbounded?

    And why you think that for the particle in box I can't have functions which goes to zero in infinity?
     
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