Fourier transforms and Dirac delta function

Question

What is the Dirac delta function $\delta(t_1-t_2)$ in Fourier (frequency) space?

Answer 1

1. In this answer we normalize the Fourier transform as $$\tag{1} \hat{\varphi}(\omega)~:=~(2\pi)^{-\frac{n}{2}} \int_{\mathbb{R}^n} \!d^nt~e^{-i\omega\cdot t}\varphi(t). $$ Here the dimension will be $n=2$.

2. The relevant version of the Dirac delta distribution $\delta(t_1-t_2)$ (with two running arguments, so to speak) is here $$\tag{2} \delta[\varphi]~:=~\int_{\mathbb{R}^2} \!d^2t~\varphi(t)~\delta(t_1-t_2)~:=~\int_{\mathbb{R}} \!dt~ \varphi(t,t)$$ for a Schwartz test function $\varphi:\mathbb{R}^2\to \mathbb{C}$.

3. There is a notion of Fourier transform for tempered distributions. The Fourier transformed Dirac delta distribution is $$\hat{\delta}[\varphi]~:=~\delta[\hat{\varphi}]~\stackrel{(2)}{=}~\int_{\mathbb{R}} \!dt~ \hat{\varphi}(t,t) ~\stackrel{(1)}{=}~\int_{\mathbb{R}}\!\frac{dt}{2\pi}\int_{\mathbb{R}^2} \!d^2\omega~e^{-it(\omega_1+\omega_2)}~\varphi(\omega) $$ $$\tag{3} ~=~\ldots ~=~\int_{\mathbb{R}} \!d\omega~ \varphi(\omega,-\omega)~=:~\int_{\mathbb{R}^2} \!d^2\omega~\varphi(\omega)~\delta(\omega_1+\omega_2).$$