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Explicitly Deriving the Delta Function

  1. Oct 29, 2008 #1
    When working with fourier transforms in Quantum mechanics you get the result that
    [tex]\int_{-\infty}^{\infty}e^{-ikx}e^{ik'x} = \delta(k-k')[/tex]

    I understand conceptually why this must be true, since you are taking the fourier transform of a plane wave with a single frequency element.

    I have also seen it sort of derived by looking at the formula for the fourier series and tracking its components in the limit that it becomes a continuous fourier transorm (letting the period go to infinity and [tex]\Delta\omega[/tex] go to 0)

    But I really want to come up with some explicit expression, from doing the integral that behaves like a delta function. I have tried messing around with it, by sticking it inside of another integral and multiplying it by a test function etc. Is there a way to do this?
  2. jcsd
  3. Oct 29, 2008 #2


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    Not sure exactly what you're looking for here. How about this? Insert a "convergence factor" [itex]\exp[-\varepsilon x^2/2][/itex] in the integrand; call the resulting function [itex]\delta_\varepsilon(k)[/itex]:
    [tex]\delta_\varepsilon(k)\equiv \int_{-\infty}^{+\infty}dx\,e^{-\varepsilon x^2/2}e^{ikx}.[/tex]
    The result is
    [tex]\delta_\varepsilon(k)= (2\pi/\varepsilon)^{1/2}\exp[-k^2/2\varepsilon].[/tex]
    Now we want to take the limit [itex]\varepsilon\to 0[/itex]. For [itex]k\ne 0[/itex], the limit is zero, and for [itex]k=0[/itex], the limit is infinity. Furthermore, the integral of [tex]\delta_\varepsilon(k)[/itex] from minus to plus infinity is one. So the function has all the properties of the delta function in this limit.
  4. Oct 29, 2008 #3


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    I wrote an extremely non-rigorous argument in this thread. If you want to learn to deal with delta functions more rigorously, this is probably a better place to start.

    Do you already know that stuff about how to define the delta function as a distribution? I'm just asking because it might help other people to give you a better answer.
  5. Oct 29, 2008 #4
    The convergence factor seems to work, except when I integrate the delta function you defined I don't get 1, I get some constant with pi. Maybe I am doing the integral wrong.
  6. Oct 29, 2008 #5

    Ben Niehoff

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    The delta function is defined by

    [tex]\int_{-\infty}^{\infty} f(k) \delta(k - k') \; dk = f(k')[/tex]

    So, try plugging in your expression into the above integral, and verify that it gives the correct answer. (Exchange the order of integration, and you should just get the Fourier transform of the Fourier transform of f.)

    Note: I think your expression is off by a factor of [itex]2 \pi[/itex].
    Last edited: Oct 30, 2008
  7. Oct 29, 2008 #6


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    Oops, I missed the [itex]2\pi[/itex] also. I should have written
    [tex]\delta_\varepsilon(k)\equiv {1\over2\pi}\int_{-\infty}^{+\infty}dx\,e^{-\varepsilon x^2/2}e^{ikx}.[/tex]
    \delta_\varepsilon(k)= (1/2\pi\varepsilon)^{1/2}\exp[-k^2/2\varepsilon].[/tex]
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