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Dirac Delta function as a Fourier transform

  1. Mar 9, 2012 #1


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    It is fairly easy to demonstrate that the Dirac delta function is the Fourier transform of the plane wave function, and hence that:

    [itex]\delta(x)=∫_{-∞}^{∞}e^{ikx}dk[/itex] (eg Tannoudji et al 'Quantum Physics' Vol 1 p101 A-39)

    Hence it should be the case that [itex]∫_{-∞}^{∞}e^{ik}dk = \delta(1) = 0[/itex]

    However the integral on the LHS does not even seem to exist, let alone equal zero. We can write:

    [itex]∫_{-∞}^{∞}e^{ik}dk = ∫_{-∞}^{∞}cos(k)+i sin(k) dk[/itex]

    [itex]= ∫_{-∞}^{∞}cos(k)dk+i ∫_{-∞}^{∞}sin(k)dk[/itex]

    and neither of these integrals exist.

    Yet, based on the Fourier argument, they must both exist and equal zero.

    I have probably done something silly to get myself into this contradiction. Can somebody please help me by showing what it is?
  2. jcsd
  3. Mar 9, 2012 #2
    There are a number of things going on here I don't really understand but chronologically why does delta(1)=0? Also there is nothing trivial about a Dirac delta function, it's always been a trouble maker in terms of rigorous mathematics.
  4. Mar 9, 2012 #3


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    It seems you're missing the crucial piece of information that the Dirac delta is not an ordinary function in the usual sense. Rather it's a generalized function known as a "linear functional" or "distribution".


    It only makes mathematical sense when integrated with another function, e.g.,
    \int_{-\infty}^{\infty}\!\! dx f(x) \delta(x-a) ~=~ f(a)
  5. Mar 9, 2012 #4
    Rigorously, the dirac delta function is actually a distribution, not a function. Even still, you can do a bit of 'physics math' by adding a term like -epsilon*k^2 to the exponential in the integral to help it converge, and then limiting it away after the calculation.
  6. Mar 13, 2012 #5
    That is indeed the Cauchy principle value of the integral, so you can at least make some sense of what you got.
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