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Dirac function

  1. Nov 11, 2005 #1
    As I read in my quantum mechanics book the delta function is sometimes called the sampling function because it samples the value of the function at one point.
    [tex]\int {\delta (x - x')} f(x')dx' = f(x)[/tex]

    But then I opened a quantum field book and I found equations like that:
    [tex]\phi (x) = \frac{1}{{(2\pi )^{3/2} }}\int {d^4 p\delta (p^2 } - m^2 )A(p)e^{ - ip \cdot x} [/tex]

    [tex](\partial _\nu \partial ^\nu + m^2 )\phi (x) = \frac{1}{{(2\pi )^{3/2} }}\int {d^4 p( - p^2 } + m^2 )\delta (p^2 - m^2 )A(p)e^{ - ip \cdot x} [/tex]

    Can someone explain me what the delta function does here? What it is sampling? How these equations work?

    thank you
  2. jcsd
  3. Nov 11, 2005 #2


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    The delta function does the same thing!

    I don't know if "sampling" is a good adjective, but I'll run with it. In the integral

    [tex]\int d^4 p \, \delta (p^2 - m^2 )A(p)e^{ - ip \cdot x} [/tex]

    you're "sampling" the function [itex]A(p)e^{ - ip \cdot x}[/itex] over the region where [itex]p^2 - m^2 = 0[/itex].

    We can reduce this down to the known case, if we like, as follows: we can split this up into an iterated integral, with [itex]p_0[/itex] being the innermost. Then we have:

    \iiint \int \delta(p_0^2 - \vec{p}\,{}^2 - m^2) A(p) e^{-ip \cdot x} \, dp_0 \, d^3 \vec{p}

    so the innermost integral is of the form:

    \int \delta(x^2 - a^2) f(x) \, dx

    Now, if I make the substitution [itex]u = x^2[/itex], we have [itex]du = 2 x dx[/itex] so that [itex]dx = du / 2 \sqrt{u}[/itex], and the integral becomes

    (yes, I know I'm being lazy with the bounds -- note that when tracing x over [itex](-\infty, +\infty)[/itex], u traces over [itex][0, +\infty)[/itex] twice! I really should break the integral up into two parts)

    \int \delta(u - a^2) f(\sqrt{u}) \frac{1}{2 \sqrt{u}} \, du

    which becomes (one term for each of the times u traces over [itex]0, +\infty)[/itex]:

    f(a) \frac{1}{2a} - f(-a) \frac{1}{-2a}
    = \frac{f(a) + f(-a)}{2a}

    I'll leave it to you to work out the general case when the argument to [itex]\delta[/itex] is an arbitrary function of the dummy variable.

    Anyways, if my intuition about these things is anywhere close to accurate, your integral for [itex]\phi(x)[/itex] should reduce to a surface integral over the hypersurface given by the equation [itex]p^2 - m^2 = 0[/itex].
    Last edited: Nov 11, 2005
  4. Nov 11, 2005 #3
    notice here that the delta function has as its argument quadratic functions. It can be a little confusing about what to do with these. Well just use this formula:
    by the way, the sum is over the roots of g.
    In your equation above, after you use this step everything should be clear- it functions just how you would expect it to.
    and check out this page for help with dirac delta functions
  5. Nov 11, 2005 #4
    thanks Hurykl, thanks Norman, thanks physicsforums, thanks the Internet

    and a great weekend to everybody
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