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Integral of Exp(I x) and the Dirac Delta

  1. Oct 26, 2009 #1
    I am trying to see why exactly the momentum eignenstates for a free particle are orthogonal. Simply enough, one gets:

    [tex]\int_{-\infty}^{\infty} e^{i (k-k_0) x} dx = \delta(k-k0)[/tex]

    I can see why, if k=k0, this integral goes to zero. But if they differ, I don't see why it goes to zero. You have:

    [tex] \int_{-\infty}^{\infty} e^{i(k-k0)x} dx = \int_{-\infty}^{\infty}( \cos [(k-k0)x] + i \sin [(k-k0)x]) dx [/tex]

    Now the sine vanishes by symmetry, but what about Cos[x]? I would imagine this integral diverges, but it must go to zero for these to be orthogonal...

    I am recalling a bit from complex analysis that might be useful, but for now I am in the dark. Why is this integral the dirac delta function.
  2. jcsd
  3. Oct 26, 2009 #2
    Probably you have to write down what it means. Something like this:
    \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} e^{i(k-k_0)x} g(k)\,dk\,dx = \begin{cases}
    g(k_0)\qquad\text{if }k=k_0
    0\qquad\text{if }k\ne k_0
    for good enough function [itex]g[/itex]
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