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Orthogonal Integral

  1. Jun 29, 2008 #1

    nicksauce

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    On page 102 of Introduction to Quantum Mechanics, Griffiths writes that
    [tex]\int_{-\infty}^{\infty}e^{i\lambda x}e^{-i \mu x}dx = 2\pi\delta(\lambda-\mu)[/tex]

    I don't see how this follows. If you replace lambda with mu, then you get a cancellation, and the integral of 1dx. Am I missing something?
     
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  3. Jun 30, 2008 #2

    Fredrik

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    I think you would have to know a bunch of stuff about distributions and integration theory to understand that. I can't say that I do, so I can only give you a very non-rigorous argument based on some ugly manipulations with Fourier transforms:

    We know that if f is a nice enough function, it has a Fourier transform g that can be defined by

    [tex]g(p)=\frac{1}{\sqrt{2\pi}}\int f(y)e^{-ipy}dy[/tex]

    The theorem that guarantees that also says that f can now be expressed as

    [tex]f(x)=\frac{1}{\sqrt{2\pi}}\int g(p)e^{ipx}dp[/tex].

    If we insert the first expression into the second and recklessly switch the order of the integrations, we get

    [tex]f(x)=\int f(y)\bigg(\frac{1}{2\pi}\int e^{ip(x-y)}dp\bigg)dy[/tex]

    So if there's some way to justify switching the order of the integrations (which would have to include a new definition of what we mean by an "integral"), the expression in parentheses must be [itex]\delta(x-y)[/itex].
     
  4. Jun 30, 2008 #3

    nicksauce

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    Okay that (sort of) makes sense. Thanks!
     
  5. Jun 30, 2008 #4
    TO be more rigorous you should introduce delta as generalized function and then use a limiting procedure on test-function space D!
     
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