D'Alambertian operator on the heaviside function?

[Dixanadu]

Hey guys,

How does one compute the following quantity:

\Box \theta(x_{0})=\partial_{0}\partial^{0}\theta(x_{0})?

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

Thanks everyone!

 

[dipole]

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 \varphi (x) that vanishes at infinity, we have,
\int \partial_x\delta(x)\varphi(x) dx = -\int \delta(x) \partial_x \varphi(x) dx = -\partial_x\varphi(0)
using integration by parts.

So the action of the distribution \partial_x^2 \theta(x) on a smooth function \varphi(x) produces -\partial_x\varphi(0)

 

[Dixanadu]

hmm but my situation involves no integrals - basically I'm trying to compute the following:

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

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

I've been told that \partial^{0}\theta(x_{0})=\partial_{0}\theta(x_{0})=\delta(x_{0}). I'm not sure how this matches up to what you said above?

 

[dextercioby]

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.