How do I write the 2D Dirac delta in a manifestly rotationally invariant form?

Question

Consider the following integral over a 2D plane,

$$\iint \mathrm{d}^2\mathbf{k}\ e^{i\mathbf{k}\cdot\mathbf{r}} = 4\pi^2\delta^2(\mathbf{r})$$

This is a Fourier transform of a distribution which is rotationally symmetric around the origin, and thus the result should also be rotationally symmetric around the origin. So it should be expressible in terms of the magnitude of $\mathbf{r}$ only, i.e. in terms of $r \equiv \lVert\mathbf{r}\rVert$.

$$\iint \mathrm{d}^2\mathbf{k}\ e^{i\mathbf{k}\cdot\mathbf{r}} = 4\pi^2 f(r)$$

What is the proper mathematical expression for $f(r)$, if it exists?

Clearly $f(r) = \delta(r)$ is unsuitable because it doesn't have the right units. Dimensional analysis suggests that it might be something like $f(r) = \frac{1}{r}\delta(r)$, but I'd like to have some sort of mathematical justification rather than just a guess.

I've looked at Delta function in curvilinear coordinates but that question is somewhat more abstract, plus it doesn't seem to eliminate the dependence on $\theta$ as I would like to do.

Answer 1

Expressions like $\delta(r)/r$ do not generally define a distribution. But if we define a distribution $T$ by setting

$$T(\phi)=\int_0^{2\pi}\int_0^{\infty} \phi(r\cos \theta,r\sin\theta) \delta(r) \mathrm dr \mathrm d\theta=\int_0^{2\pi}\int_0^{\infty} \phi(r\cos \theta,r\sin\theta) (\delta(r)/r) r \mathrm dr \mathrm d\theta$$

for test functions $\phi$, then $T=2\pi \delta(\mathbf x)$. So, informally, $\delta(\mathbf x)=\delta(r)/2\pi r$.

Answer 2

Depending on taste, some pitfalls in the behavior near $0$ can be avoided as follows. Given a rotation-invariant (maybe tempered) distribution $u$ on $\mathbb R^n$, for test function $f$, letting $K=SO(n)$ be the rotation group, $$ u(f) = {1\over |K|} \int_K u(k\cdot f)\;dk = {1\over |K|} u\Big(\int_K k\cdot f\;dk\Big) $$ where $dk$ refers to an invariant ("Haar") measure on $K$, $|K|$ is the total measure of $K$, and the interchange of integral with application of $u$ can be justified at various levels of sophistication. (The key point is that $K$ is compact, and that rotations are nicely-continuous maps on functions... My own blanket preference for such justifications refers to Gelfand-Pettis' "weak" integrals, but there are many alternatives.)

The point I'm aiming at here is that the rotationally-averaged $f$, as a function of radius, is an even test function on $\mathbb R$. Conversely, an even test function on $\mathbb R$ gives a rotationally invariant test function on $\mathbb R^n$. Continuous linear functionals on rotationally-invariant test functions on $\mathbb R^n$ are thus naturally in bijection with continuous linear functionals on even test functions on $\mathbb R$. That is, this avoids talking about an "edge" at "r=0".