# D'Alambertian operator on the heaviside function?

1. Nov 2, 2014

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!

2. Nov 2, 2014

### 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 $-\partial_x\varphi(0)$

3. Nov 3, 2014

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?

4. Nov 5, 2014

### 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.