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Discrete eigenvalues and their eigenfunctions

  1. Oct 9, 2011 #1
    What's the proof that eigenfunctions of discrete eigenvalues are in Hilbert Space?
     
  2. jcsd
  3. Oct 9, 2011 #2

    dextercioby

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    This is included in the definition of an <eigenvalue>. The <eigenfunctions> are elements of a Hilbert space. The set of all eigenvalues is called the <point spectrum> of an operator. What you call <discrete eigenvalue> is probably a point in the <discrete spectrum>, which is a subset of the <point spectrum>, again by definition:

    "The discrete spectrum σd(A) is the
    set of all eigenvalues which are discrete points of the spectrum and whose
    corresponding eigenspace is finite dimensional. The complement of the discrete
    spectrum is called the essential spectrum σess(A) = σ(A)\σd(A)" (*)

    for A a linear operator in a Hilbert space.


    *Quote form G. Teschl's <Mathematical Methods for Quantum Mechanics>, page 145.
     
    Last edited: Oct 9, 2011
  4. Oct 9, 2011 #3
    I don't think it's a matter of definition. If the eigenvalues are continuous, then the eigenfunctions are not necessarily part of the Hilbert Space. e.g. eigenfunctions of the operator [itex]\frac{1}{i} \frac{d}{dx}[/itex] are [itex]A e^{ikx}[/itex] (k = eigenvalue), which isn't in the Hilbert Space.
     
  5. Oct 10, 2011 #4

    dextercioby

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    I don't understand what you're denying. It's true that in the literature there's no common definition for some of the concepts, but apparently everyone agrees that

    <Let A:D(A)[itex]\subset[/itex]H [itex]\rightarrow[/itex] Ran(A)[itex]\subset[/itex]H be a linear operator in a Hilbert space. The complex number [itex]\lambda[/itex] is called eigenvalue of the operator A, iff there's at least one vector [itex]\psi[/itex]≠ 0 in D(A) satisfying

    [tex] Aψ=λψ [/tex]. Such a vector is called eigenvector of the operator A and is assumed by definition to be a member of the Hilbert space> is a definition for eigenvector/eigenvalue.

    The statement <The eigenvalues are continuous> is inaccurate, for eigenvalues are numbers which can't be continuous. I think you meant <If the spectral values are elements of the continuous spectrum of an operator A> which is whole different thing.

    Your second assertion could be taken as a counterexample to a statement such as: For a linear self-adjoint operator in a Hilbert space*, its spectral equation always has solutions in the Hilbert space.

    As for your original question, I believe my previous post above pretty much settles it.

    *The derivative operator (momentum operator) is self-adjoint on a dense everywhere subset of [itex] L^2 (\mathbb{R}, dx) [/itex].
     
  6. Oct 10, 2011 #5
    If the operator is in Hilbert space, then it's a tautology. But an operator like [itex]-i \frac{d}{dx}[/itex] doesn't have to be defined on a Hilbert space alone; it can act on functions outside Hilbert space, like [itex]e^{ikx}[/itex]. But apparently if the spectrum of such an operator is discrete, its eigenfunctions are in Hilbert space. I'm looking for a proof of this proposition.
     
  7. Oct 11, 2011 #6

    dextercioby

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    I would be very interested to learn a way to evade the tautology, namely that you can't prove something which is assumed in a definition (in this case, the definition of an eigenvalue/eigenvector).
     
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