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Why is Fourier transform of exp(ix) a delta?

  1. Dec 10, 2004 #1
    Why is it that the Fourier transform of [tex] e^{2\pi ikx} [/tex] is equal to [tex] \delta(k) [/tex] ? The delta function is supposed to be zero except at one point. But the integral doesn't converge for [tex] k \ne 0 [/tex]. Yet I see a lot of books on QFT use this identity.
  2. jcsd
  3. Dec 11, 2004 #2

    matt grime

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    delta isn't a function (what is it at the point where it is not zero?)

    It is the delta 'function' because it behaves as the delta function.
  4. Dec 11, 2004 #3


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    Suppose you were to ask for the Fourier Series for f(x)= cos(x)?

    Since the Fourier Series is, by definition, a sum of sines and cosines that add to f(x).
    Since f(x)= cos(x), its Fourier series coefficients are just a1= 1, all other coefficients are 0. The delta "function" (it's really a "distribution" or "generalized function") is the functional version of that.
  5. Dec 11, 2004 #4
    Very interesting. The integral [tex] \int^\infty_{-\infty}e^{2\pi ikx} dx [/tex] does in some ways behave like a delta function. And the delta function is an ideal function. However it's own Fourier transform is an exponential, which is a real funtion. The Fourier transform as an operator on Hilbert space is unitary, and squares to -1. Neither the delta function nor the exponential function are in Hilbert space, the latter because it doesn't satisfy boundary conditions, and the former because it isn't even a funtion. The idea is very informal and lacks rigour. I have never seen it given a rigorous basis.
  6. Dec 12, 2004 #5


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    Then get a book on "distributions" or "generalized functions" everything is done with complete rigor.
  7. Dec 13, 2004 #6
    Thanks for the suggestion. :)
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