two dimensional delta function

Question

Is it correct to write

$\delta(x,y)=\delta(x)\delta(y)$

where $\delta(x,y)$ is the delta function in two dimensions? Or are there some cases where the above fails to give the correct results when integrated over?

Thanks a lot

Answer 1

From a theoretical point of view, there is a couple of minor issues: first, the domain of definition: $\delta_{(0,0)}$ belongs to $D'(\Omega)$ with $\Omega\subset\Bbb R^2$ open set. But $\delta_0\in D'(\Xi)$ with $\Xi$ - open subset of $\Bbb R$, therefore in order to define the " product" of $\delta_0(x)$ and $\delta_0(y)$ you need to use the tensor product of distributions, namely the following:

let $$T\in D'(\Omega_1),\quad S\in D'(\Omega_2),$$ then we can define $$T\otimes S\in D'(\Omega_1\times\Omega_2)$$ by $$\langle T\otimes S,\phi(x,y)\rangle=\langle T(x),x\to\langle S(y),\phi(x,y)\rangle\rangle$$ (I took the liberty of noting $T(x)$ to indicate in which variable $T$ works) for a test function $\phi$. Now not all open sets in $\Bbb R^2$ are descartes products of open sets in $\Bbb R$.

Therefore, we can say that, $$\delta_0(x)\otimes \delta_0(y)\in D'(\Bbb R^2)$$ and $$\delta_0(x)\otimes \delta_0(y)=\delta_{(0,0)}(x,y).$$

However, in the case of $\delta$-distributions for all practical purposes this formalism in rather unnecessary.