Dirac Delta Function - unfamiliar definition

  • Thread starter Orad
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  • #1
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Main Question or Discussion Point

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?
 

Answers and Replies

  • #2
gel
533
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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:
  • #3
Fredrik
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Check out this thread for a very non-rigorous argument.
 
  • #4
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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.
 
  • #5
nicksauce
Science Advisor
Homework Helper
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Lol small world indeed
 

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