Do somebody knows anything about the Dirac's identity?
\begin{equation} \label{Dirac} \frac{\partial^2}{\partial x_{\mu}\partial x^{\mu}} \delta(xb_{\mu}xb^{\mu}) = -4\pi \delta(xb_0)\delta(xb_1)\delta(xb_2)\delta(xb_3) \end{equation}
here
$xb$, is the 4-vector $x-b$ in Minkowsky spacetime
$\delta$ is the Dirac delta function
$x_0 = -x^0, \quad x_1 = x^1, \quad x_2 = x^2, \quad x_3, = x^3$.
Do you know where can i find some material about it?
Thanks!
UPDATE:
Following Willie's link
i've understood that a solution for the linear wave equation $$ \square \psi(\mathbf{r},t) = g(\mathbf{r},t) $$ for a given $g(\mathbf{r},t)$ is $$ \psi = \int \int g(r',t')G(r,r',t,t')dV'dt' $$ where $$ G(r,r',t,t') = AG^+(r,r',t,t') + BG^-(r,r',t,t') , \qquad A + B = 1 $$ and $$ G^{\pm}(r,r',t,t') = \frac{\delta(t' - (t \mp | \mathbf{r} - \mathbf{r'} | / c))}{4\pi | \mathbf{r} - \mathbf{r'} | } $$
I think Dirac's follow from the solution of
$$ \square \psi(r,t) = \delta(\mathbf{r},t) $$
But i'm not sure of the details. Can you Willie help me?
Thanks
Your identity is in fact the expression for the fundamental solution of the linear wave equation in (1+3) dimensions. This should be in most textbooks on electrodynamics or intro to quantum field theory.
Google also tells me: