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D'Alambertian operator on the heaviside function?

  1. Nov 2, 2014 #1
    Hey guys,

    How does one compute the following quantity:

    [itex]\Box \theta(x_{0})=\partial_{0}\partial^{0}\theta(x_{0})[/itex]?

    I know that [itex]\partial_{0}\theta(x_{0})=\delta(x_{0})[/itex] which is the Dirac delta, but what about the second derivative?

    Thanks everyone!
  2. jcsd
  3. Nov 2, 2014 #2
    Heaviside and Dirac delta are examples of distributions. They are a kind of generalized function, and only defined by their action on some "test function" in an inner-product. Integrals satisfy the properties of inner products, so choosing some smooth function [itex] \varphi (x) [/itex] that vanishes at infinity, we have,
    [tex] \int \partial_x\delta(x)\varphi(x) dx = -\int \delta(x) \partial_x \varphi(x) dx = -\partial_x\varphi(0) [/tex]
    using integration by parts.

    So the action of the distribution [itex] \partial_x^2 \theta(x) [/itex] on a smooth function [itex] \varphi(x) [/itex] produces [itex] -\partial_x\varphi(0)[/itex]
  4. Nov 3, 2014 #3
    hmm but my situation involves no integrals - basically I'm trying to compute the following:

    [itex]\Box [\theta(x_{0})\phi(x)\phi^{\dagger}(0)][/itex]

    where [itex]\phi, \phi^{\dagger}[/itex] are solutions to the complex Klein-Gordon equation.

    I've been told that [itex]\partial^{0}\theta(x_{0})=\partial_{0}\theta(x_{0})=\delta(x_{0})[/itex]. I'm not sure how this matches up to what you said above?
  5. Nov 5, 2014 #4


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    That product is mathematically ill-defined, because you have a multiplication of 3 distributions. In the so-called classical theory of distributions, there's no way to 'multiply' them.
    As a note (rumor), I heard that Colombeau's theory of distributions somehow addresses this issue.
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