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Normalizing the Momentum Eigenfunctions

  1. Jun 24, 2008 #1
    I know that the momentum eigenfunctions are of the form [tex]\phi = Ce^{ikx}[/tex], but how would we normalize them? We just get

    [tex]\int_{-\infty}^{\infty} C^2 dx = 1[/tex]

    which means that [tex]C[/tex] is infintesimally small...
  2. jcsd
  3. Jun 24, 2008 #2


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    Indeed, so these wave-functions are non-normalizable. You either choose C = 0 and you have a trivial wave function, or you have a diverging integral.
  4. Jun 24, 2008 #3
    Well, see the thing is that there's this formula in my book which I have absolutely no idea as to how they got... here it is... could someone show a proof of this?

    [tex]\int_{-\infty}^{\infty} e^{ix(k'-k)} dx = 2\pi \cdot \delta(k' - k)[/tex]

    It makes absolutely no sense to me... if we have that [tex]k' \neq k[/tex], then

    [tex]\int_{-\infty}^{\infty} e^{ix(k'-k)} dx = \int_{-\infty}^{\infty} \cos [(k'-k) x] dx[/tex]

    since sin is an odd function. But even this is a large step to claim... it's like saying that 1 - 1 + 1 - 1 + 1 - 1 + ... = 0. But can we just say that cos is a sideways shift of sin, so we can just say that it evaluates to zero as well? Or do the limits mean that we take a value that approaches infinity and it always equals zero? I think I understand why the k'=k case works, however...delta(0) = infinity, which is also the lhs. Lastly, why does the 2 pi have to be on the RHS?
    Last edited: Jun 24, 2008
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