Inverse Laplacian

Question

I have seen the following operator somewhere in a paper on cosmology $$ \frac{\partial_i \partial_j}{\nabla^2} - \frac{1}{3} \delta_{ij}. $$ What is the definition of the inverse Laplacian? What is meant by this misleading notation? Is this the inverse Laplacian? If yes, what is then $\frac{1}{\nabla^2}$?

Answer 1

Let's write an arbitrary function $f:\,\Bbb R^3\mapsto\Bbb R$ as a Fourier transform:$$f(\vec{r})=(2\pi)^{-3/2}\int_{\Bbb R^3}\tilde{f}(\vec{k})e^{i\vec{k}\cdot\vec{r}}d^3\vec{k},\,f(\vec{k}):=(2\pi)^{-3/2}\int_{\Bbb R^3}f(\vec{r})e^{-i\vec{k}\cdot\vec{r}}d^3\vec{r}.$$(I've restricted to a $3$-dimensional space because I'm confident that's why $\frac13$ comes up in the expression you encountered.) The most obvious definition of $\frac{1}{\nabla^2}f$ is$$\frac{1}{\nabla^2}f:=(2\pi)^{-3/2}\int_{\Bbb R^n}\frac{\tilde{f}(\vec{k})}{-k^2}e^{i\vec{k}\cdot\vec{r}}d^3\vec{k}=(2\pi)^{-3}\int_{(\Bbb R^3)^2}\frac{f(\vec{R})e^{i\vec{k}\cdot(\vec{r}-\vec{R})}}{-k^2}d^3\vec{k}d^3\vec{R}.$$This coincides with @Vincent's definition of $(\nabla^2)^{-1}$ provided$$(2\pi)^{-3}\int_{\Bbb R^3}\frac{e^{i\vec{k}\cdot\vec{z}}}{k^2+m^2}d^3\vec{k}=\frac{e^{-mz}}{4\pi z}$$is taken as $m\to0^+$ (strictly speaking, this is a distributional limit). By Fourier inversion, this conjecture is equivalent to$$\int_{\Bbb R^3}\frac{e^{-mz-i\vec{k}\cdot\vec{z}}d^3\vec{z}}{4\pi z}=\frac{1}{k^2+m^2}.$$Indeed, spherical polars rewrite the LHS as$$\begin{align}\frac12\int_0^\pi d\theta\sin\theta\int_0^\infty ze^{-(m+ik\cos\theta)z}dz&=\frac12\int_0^\pi\frac{\sin\theta}{(m+ik\cos\theta)^2}d\theta\\&=\frac{1}{2ik}\left[\frac{1}{m+ik\cos\theta}\right]_0^\pi\\&=\frac{1}{k^2+m^2}.\end{align}$$This is quite a complicated manipulation, so it pays to use dimensional analysis as a power-counting sanity check.

Answer 2

Every Laplacian can be inverted using its Green's function. If we have $$\nabla^2V = \rho$$ the inverse is simply $$V(x) = \left(\nabla^2\right)^{-1} (\rho) = -\frac{1}{4\pi}\int \frac{\rho(x')}{|x-x'|} \text{d}^3x'$$ and that's what's meant by $\frac{1}{\nabla^2}$, at least in the places where I've seen it.