# Inverse Laplacian

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 $$\frac{1}{\nabla^2}$$?

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.

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.

• If we define $\Delta:=\nabla^2$, then what is $\Delta^{-1}$? I would say this is what you have defined and not $\frac{1}{\nabla^2}$. Isn't it? Oct 5, 2021 at 15:35
• @Immanuel See the last line in my answer above. In all the papers I've seen, $\frac{1}{\nabla^2}$ means $\Delta^{-1}$. Oct 5, 2021 at 15:37
• @Immanuel Well $\sin^2 x$ does not mean $\sin(\sin(x))$ either Oct 5, 2021 at 15:45
• @Immanuel I won't inveigh on the "should operators' inverses be denoted as reciprocals?" debate, but you might find this an interesting context where notational questions like that are important.
– J.G.
Oct 5, 2021 at 16:05
• @Immanuel I'm working on an edit to show they are.
– J.G.
Oct 5, 2021 at 16:11