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Eigenfunctions (orthogonality & expansion)

  1. Mar 7, 2007 #1
    If you have a particle in 1D bound within range "-a" and "a". You come up with one eigenfunction that is sinusoidal (since it satisfies the problem).
    Now, you get all the necessary constants through the usual way...
    I want to know whether more than one eigenfunction can be produced and how? Because in the end I need to show that these eigenfunctions are orthogonal.

    If your given an eigenfunction say: psi = b(a - |x|)
    what does it mean by "expanding psi in eigenstates of momentum".

    Please note these are Intro questions to QM and I cannot read/understand DIRAC notation or any other type of that nature just yet.
  2. jcsd
  3. Mar 7, 2007 #2


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    You actually come up with all. They are associated to the point spectrum and this is countable, which means that the wavefunctions can be indexed by the set of natural numbers. You need to prove that

    [tex] \langle \psi_{n},\psi{m}\rangle =\delta_{nm} \ ,\forall n,m\in\mathbb{N} [/tex]

    What are the "eigenstates of momentum" ? Have you studied Fourier series/integrals ?
  4. Mar 7, 2007 #3
    1) Ok so if I come up with a wavefunction that is something like:
    Asin(kx), then the associated eigenfunctions lie within 'k' for example?
    Ie. k = 2n(pi)...
    where n represents the first, second, third... eigenfunction?

    2) As for "eigenstates of momentum" I cannot explain further what it is because the question is worded this way.
    Yes I've studied Fourier Series/Integrals and the whole lot. Initially I drew out psi = b(a - |x|) and thought about doing a fourier expansion on it, but I didn't understand why I would or what that would represent.
  5. Mar 7, 2007 #4


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    That's right.

    That would be the solution to your problem, since Fourier expansion and expansion in eigenfunctions of mometum operator are equivalent.
  6. Mar 7, 2007 #5
    Would you be able to explain (or point me to a site/book) why the fourier expansion and expansion in eigenfunctions of momentum operator are equivalent?
    Does the fourier expansion also apply to other operators? From what I know of Fourier its just a way of representing a periodic function.

    Thanks for your help.
  7. Mar 8, 2007 #6


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    The connection i mensioned is rigorously this one:

    The closed self adjoint extension of the momentum operator in [itex] \mathcal{S}\left(\mathbb{R}\right) [/itex] is, by means of the reverse of Stone's theorem, the generator of a uniparametric stongly continuous group of unitary operators representing, in the mathematical framework, the symmetry transformations called "spatial translations". The momentum operator is a selfadjoint operator acting in the rigged Hilbert space [itex] \Phi \subset \mathcal{H}\subset \Phi' [/itex] in which, by the Gelfand-Maurin spectral theorem admits a complete set of generalized eigenvectors, [itex] |p\rangle [/itex].

    It's simple to show that the two known realizations of [itex] \Phi [/itex] as functions spaces, namely [itex] \mathcal{S}(\mathbb{R},dx) [/itex] and [itex] \mathcal{S}(\mathbb{R},dp) [/itex] are connected by the Fourier transformation. It all comes to the simple (looking) assessment (written using bra/ket formalism)

    [tex] \phi(x)=\langle x|\hat{1}|\phi\rangle =\left\langle x\left |\left(\int dp{}|p\rangle\langle p|\right)\right |\phi\right\rangle =\int dp \langle x|p\rangle\langle p|\phi\rangle \simeq \int dp {}e^{ipx}\tilde{\phi}(p) [/tex]
    Last edited: Mar 8, 2007
  8. Mar 8, 2007 #7


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    :rofl: :rofl: Question: was the point of this post to

    a) help the OP?


    b) to show off?
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