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Energy spectrums in QM

  1. Sep 9, 2008 #1
    This is about quantum mechanics, but it is sufficiently difficult existence question dealing with DE and operator theory, that I think it fits the DE subforum the best:

    Let [tex]E:\mathbb{N}\to\mathbb{R}[/tex] be an arbitrary map, but so that [tex]\textrm{Im}(E)[/tex] is bounded from below. Does there exist a measurable function [tex]V:\mathbb{R}\to\mathbb{R}[/tex] such that the eigenvalues of the operator

    [tex]
    H = -\frac{1}{2}\partial_x^2 + M_V
    [/tex]

    are the given [tex]E(n)[/tex]?

    Here [tex]M_V:\mathbb{C}^{\mathbb{R}}\to \mathbb{C}^{\mathbb{R}}[/tex] is the multiplication operator [tex](M_V\psi)(x) = V(x)\psi(x)[/tex].

    For the sake of rigor we can give the following definition for the domain of H,

    [tex]
    D(H) = \{\psi\in L^2(\mathbb{R},\mathbb{C})\;|\; \psi\;\textrm{is piece wisely}\;C^2\quad\textrm{and}\quad \int dx\;\Big|-\frac{1}{2}\partial_x^2\psi(x) + V(x)\psi(x)\Big|^2 < \infty\}
    [/tex]

    So H is a mapping [tex]H:D(H)\to L^2(\mathbb{R},\mathbb{C})[/tex].
     
    Last edited: Sep 9, 2008
  2. jcsd
  3. Sep 9, 2008 #2

    atyy

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    I'm not a math person, so just a wild question:is your question equivalent to asking whether a Hamiltonian exists whose eigenvalues are the zeros of the Riemann zeta function?
     
  4. Sep 9, 2008 #3
    I don't know enough of Riemann zeta function to answer that. I'll say that I don't know how precisely my question would be related to it now.

    If [tex]T:\mathcal{H}\to\mathcal{H}[/tex] is some linear operator on some inner product space, we can define a new non-linear mapping

    [tex]
    f:\mathcal{H}\to\mathcal{H},\quad f(\psi) = T\psi - \frac{(\psi|T\psi)}{\|\psi\|^2}\psi
    [/tex]

    which has the property

    [tex]
    f(\psi)=0\quad\Leftrightarrow\quad T\psi\propto\psi,
    [/tex]

    so in this sense every eigenvalue problem is related to a problem of finding zeros of some function. So... who knows?
     
  5. Sep 9, 2008 #4
    Now I read your post more carefully. You were speaking about eigenvalues being zeros of some function, and I was about eigenvectors being zeros of some function. So this seems to be different thing.
     
  6. Sep 9, 2008 #5

    atyy

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    Let's see, I understood your question to be:
    Given a countable set of real numbers, does a Hamiltonian exist such that its eigenvalues are the elements of the given set?

    (A little more strictly, you asked if a V(x) exists such that the eigenvalues of the Hamiltonian are the elements of the given set.)
     
  7. Sep 10, 2008 #6

    atyy

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  8. Sep 10, 2008 #7
    I see. So the problem I described, seems to be similar to the Berry Conjecture.
     
  9. Sep 10, 2008 #8

    atyy

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    Hmm, reading the Berry conjecture, it seems they need a more general form of the Hamiltonian than in your problem. I wonder whether they've already ruled out "nice" Hamiltonians, and if so, how they did that.
     
  10. Sep 12, 2008 #9

    atyy

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