# Fourier Transform of a Wave Packet [closed]

In the analysis of coherence and interference, I encountered the following expression: $$F(t)=\Re\int_0^\infty\mathrm d\omega e^{-i\omega t}H(\omega)$$, where $$\Re$$ denotes the real part of the integral, and $$H(\omega)$$ vanishes for $$\omega<0$$. I want to show the Fourier transform of $$F(t)$$ is $$\hat{F}(t)=\frac{\sqrt {2\pi}}{2}[H(\omega)+H^\star(-\omega)]$$ Here is my approach $$\int_0^\infty\mathrm d\omega e^{-i\omega t}H(\omega)=\int_{-\infty}^\infty\mathrm d\omega e^{-i\omega t}H(\omega)=\sqrt{2\pi}H(t)$$ where $$H(t)$$ is the Fourier transform of $$H(\omega)$$ (with out the hat), and I used the fact $$H(\omega)=0$$ for $$\omega<0$$. Hence $$F(t)=\frac{\sqrt {2\pi}}{2}[H(t)+H^\star(t)]$$ It follows that $$\hat F(t)=\frac{\sqrt {2\pi}}{2}\int_{-\infty}^\infty \mathrm d\omega e^{-i\omega t}[H(t)+H^\star(t)]=\pi[H(\omega)+H^\star(-\omega)]$$ Although the form of my result agree with the correct answer, I got a factor of $$\sqrt{2\pi}$$ wrong somewhere. I wonder where did I miss it!

• It's not just a normalization/convention issue? Commented May 9, 2022 at 1:31
• I think so, but in the book the normalization for the transform is always chosen to be $1/\sqrt{2\pi}$, so there is an inconsistency, and I probably done something wrong... Commented May 9, 2022 at 1:37

Mathematicians usually put $$1/\sqrt{2\pi}$$ before each integral. Physicists usually leave it out of the integral like your first equation, but then divide the inverse integral by $$2\pi$$