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 with complex arguments

  1. May 13, 2008 #1
    This is probably a silly question to some, but I've been struggling to understand how the delta function behaves when given a complex argument, that is \delta(z), z \in C. I guess the basic definition is the same that the integral over all space is 1, but I'm looking for a more detailed guide on the inner workings. Does someone know of a good reference?

  2. jcsd
  3. May 13, 2008 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    The (Dirac) delta function, which is a "Distribution" or "Generalized Function", rather than a true "function", can be defined in a number of ways. The most common definition is that
    [tex]\int_C f(x)\delta (x)dx[/tex]
    is equal to 1 if 0 is in set C, equal to 0 if 0 is not in set C.

    More generally
    [tex]\int_C f(x)\delta (x-a) dx[/itex]
    is equal to 1 if a is in set C, equal to 0 if 0 is not in set C.

    There is no reason C cannot be a subset of the complex numbers and a a complex number.

    It can be thought of, roughly, as the limit of a sequence of functions, fn, such that fn= n if |z|< 1/n, 0 other wise. Of course, that sequence doesn't actually converge which is why [itex]\delta(z)[/itex] is not a true function!
  4. Apr 8, 2009 #3
    I kind of have a follow-up question to JayFsd (a year later, ha).

    Suppose that [tex]a[/tex] is complex and you have the integral

    [tex]\int_C f\left(x\right) \delta\left(x - a\right)dx[/tex]

    such that [tex]a[/tex] lies on the countour [tex]C[/tex], you would expect the result to be:

    [tex]\int_C f\left(x\right) \delta\left(x - a\right)dx = f\left(a\right)[/tex].

    However, to evaluate this integral for a specific contour, let's say, we do the substitution [tex] x = x\left(t\right),\, 0 <t < 1 [/tex] and get

    [tex]\int_0^1 f\left(x\left(t\right)\right)\delta\left(x\left(t\right) - x\left(t_a\right)\right)\frac{dx}{dt}dt[/tex]

    where [tex] x\left(t_a\right) = a[/tex]

    Ok, so the question I have:

    According to Wikipedia, the dirac delta scales as

    [tex]\delta\left(g\left(x\right)\right) = \sum_i\frac{\delta\left(x - x_i\right)}{|g'\left(x_i\right)}[/tex]

    where the [tex]x_i[/tex] are the roots of [tex]g[/tex].

    So if I use that scaling property in the above integral, then I get

    [tex]\int_0^1 f\left(x\left(t\right)\right)\delta\left(x\left(t\right) - x\left(t_a\right)\right)\frac{dx}{dt}dt = \frac{dx}{dt_a}/\left|\frac{dx}{dt_a}\right|f\left(x\left(t_a\right)\right)[/tex]

    which doesn't agree with the expected result. It seems as if I don't understand the derivation of the scaling property well enough -- that is with real numbers it seems fine, but when you throw complex numbers in the mix I get hopelessly lost.

    Anyone care to chime in and help me?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Dirac delta function with complex arguments