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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


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    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?
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