Delta function of form: $\frac 1 {4π} \nabla ^2 \frac 1 {\lvert r - {r'} \rvert}$

Question

I was looking at a wikipedia article and saw the following expression given for a delta function in three dimensions: $$\delta (r-r') = \frac 1 {4π} \nabla ^2 \frac 1 {\lvert r -{r'}\rvert}$$ I was hoping someone could show me how this is derived. Intuitively, I thought that the 4π comes about from using the divergence theorem to convert a volume integral over all space into a surface integral, where 4π would be the total solid angle.

Answer 1

It is easy to check that $\frac 1 {4π}\nabla ^2 \frac 1 {\lvert r -{r'}\rvert}$ is zero for $r\neq r'$. All you need to show that it integrates to $1$ on a any volume containing $r=r'.$ For any such volume carve out a small sphere of radius $\varepsilon$ around $r'.$ The integral in the rest of the volume is zero, and the integral on that sphere is given by\begin{align} \frac 1 {4π} \int_{B^3(r',\varepsilon)}\nabla ^2 \frac 1 {\lvert r -{r'}\rvert}dV &= \frac 1 {4π} \int_{\partial B^3(r',\varepsilon)}\nabla \frac 1 {\lvert r -{r'}\rvert}\cdot d\vec A\\ &= \frac 1 {4π} \int_{S^2(r',\varepsilon)} \frac {-1} {\varepsilon^2} dA\\ &= \frac 1 {4π\varepsilon^2} \int_{S^2(r',\varepsilon)} dA = 1\\ \end{align} where I have used divergence theorem in the first line.