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Show that an ONB is complete

  1. Nov 7, 2008 #1
    Is it possible to show that an ONB (orthonormal base is complete). I am quite irritated by the insertions of ones of the form [tex]\sum_{n} \left| \Psi_n \right\rangle \left\langle \Psi_n\right|[/tex] if the vectors are the eigenvectors of some Hamiltonian. So now the more precise questions:
    1) Is there a straight forward approach to show that an ONB is complete in the Schwarz space, C2 or rigged Hilbert space.
    2) If this is a property of Hermitian operators what are the restrictions? (the base formed by eigenfunctions of a finite potential well for example is obviously over complete)
    3) If it is not straight forward: Let's say we have a well known ONB like the eigenfunctions of the harmonic oscillator. Then we take away the eigenfunction for one quantum number like n=5. How is it possible to show that this base is incomplete without the explicit use of that function?
  2. jcsd
  3. Nov 7, 2008 #2
    I imagine if this is from a physics text/course, the result you are using is that the eigenfunctions of a compact, self-adjoint operator [tex]\mathcal{A}[/tex] are dense in [tex]R(\mathcal{A})[/tex].
  4. Nov 7, 2008 #3
    That is already a useful result. Thank you.
    But I am not sure, that I understand the notion of compact, or that I really have an understanding of [tex]R(\mathcal{A})[/tex] (I suppose it's the operator's range) There is a toy problem, where you take the harmonic oscillators potential, and set the potential to infinity for x<0. This way only every second wave function is allowed, because it has to be 0 at the origin. So now the eigenfunctions don't span the same space anymore as the usual harmonic oscillator's eigenfunctions. I suppose then, that the range of the new Hamiltonian is smaller, or is the operator it is not compact anymore?
    If the first is the case, then inserting the "one" actually projects whatever vector to [tex]R(\mathcal{A})[/tex] which is not something a one should do.
    Last edited: Nov 7, 2008
  5. Nov 11, 2008 #4
    You are correct - I mean the range of [tex]\mathcal{A}[/tex] by [tex]R(\mathcal{A})[/tex]. By [tex]\mathcal{A}:X\rightarrow Y[/tex] being compact, I mean that the image under [tex]\mathcal}A[/tex] of any bounded subset of [tex]X[/tex] has compact closure in [tex]Y[/tex], or equivalently for any bounded sequence [tex]\{x_n\}[/tex] in [tex]X[/tex], the sequence [tex]\{\mathcal{A} x_n\}[/tex] in [tex]Y[/tex] contains a convergent subsequence.
  6. Nov 11, 2008 #5


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    the new hamiltonian is only well defined on the space of square integrable functions psi(x) satisfying psi(x)=0 for x<= 0 -- otherwise you would be multiplying a non-zero number by infinite potential. Its range is dense in this space, and so is the space generated by its eigenvalues (so you have an onb).

    no, and it wasn't compact in the first place. It's not even bounded.
  7. Nov 11, 2008 #6


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    for an operator such as position or momentum [itex]p=-i\hbar\partial/\partial x[/itex], the eigenvectors are non-normalizable or delta functions so don't exist in the hilbert space at all. So the sum is just 0. Even generalizing to distributions, they form a continuum and a discrete sum wouldn't make sense. One way to make the expression well defined mathematically is to use the spectral decomposition which replaces the sum as an integral w.r.t. a spectral measure.
  8. Nov 12, 2008 #7
    I learned about that. I think we used projector valued measures, or whatever the English name is. But it is good to know that we mostly don't even know what we are doing, and that we have to understand the domain and range of the operator that we are dealing with really well.
  9. Jun 30, 2009 #8
    Here's something one can think about...

    Most of the books assume completeness (JJ Sakurai, David J Griffiths, R Shankar)

    However, after assuming this completeness, when we have introduced the notion of a wavefunction, we can set up the differential equation for the eigenfunctions of the operator (whose eigen-set we said formed a complete ONB). These differential equations turn out to be Sturm-Liovelle equations. And the proof of the completeness of solutions to the Sturm Liovelle differential equations is well known to Mathematicians. (Search on wiki/ google/ wolfram)

    Of course, one might say, but hadn't we assumed the completeness? Then we're bound to get the same!
    To that, I say, there perhaps was an independent way of getting to the eigen-functions... Which would have then proved the completeness of the ONB!
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