1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Dirac Delta Function - unfamiliar definition

  1. Aug 8, 2008 #1
    Given:

    [tex]f(x)=\delta(x-a)[/tex]

    Other than the standard definitions where f(x) equals zero everywhere except at a, where it's infinity, and that:

    [tex]\int_{-\infty}^{\infty} g(x)\delta(x-a)\,dx=g(a)[/tex]

    Is there some kind of other definition involving exponentials, like:

    [tex]\int e^{ix(k'-k)}d^3x=\delta^3(k'-k)[/tex]

    I remember learning something about this, but can't find a proof of it in any textbook or online at the moment, and I don't trust my memory enough to know if this is precise. Any thoughts?
     
  2. jcsd
  3. Aug 8, 2008 #2

    gel

    User Avatar

    yes, the dirac delta function is proportional to the fourier transform of the constant function equal to 1. You actually get [itex](2\pi)^3\delta^3(k)[/itex].
    To be rigorous, the fourier transform is defined in terms of distributions and not just standard functions. It is not a standard Riemann integral, although physicists often treat it as such.

    The proof could go like the following. If f is a fourier transformable function with transform [itex]\hat f[/itex]
    [tex]
    \begin{align*}
    \int f(x) \left(\int e^{ixy}\,dy\right)\,dx
    &=
    \int \left(\int e^{ixy}f(x)\,dx\right)\,dy\\
    &= \int \hat f(y) dy
    \end{align*}
    [/tex]
    Using z=0, the last line is
    [tex]
    \int \hat f(y)e^{-iyz}\,dy = 2\pi f(z)=2\pi f(0)
    [/tex]
    -- using the inverse fourier transform. Compare this to [itex]\int f(x)\delta(x)\,dx = f(0)[/itex].

    This is a bit sketchy, because the integral of [itex]e^{ixy}[/itex] doesn't make sense using the Riemann integral.

    Also, the delta function is the derivative of the Heaviside function f(x)=1{x>0}, in the sense of distributions.
     
    Last edited: Aug 8, 2008
  4. Aug 8, 2008 #3

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Check out this thread for a very non-rigorous argument.
     
  5. Aug 8, 2008 #4
    Wow... Irrelevant, but that nicksauce guy who started the other thread is a good friend of mine in RL. I'm gonna call him now!

    Ha, small world.

    Oh, and I understood the (false) justifications, thanks.
     
  6. Aug 10, 2008 #5

    nicksauce

    User Avatar
    Science Advisor
    Homework Helper

    Lol small world indeed
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Dirac Delta Function - unfamiliar definition
  1. Dirac Delta Function (Replies: 2)

  2. Dirac Delta Function (Replies: 0)

  3. Dirac Delta Function (Replies: 1)

  4. Delta Dirac Function (Replies: 4)

Loading...