# Intuition behind Fourier transformed spaces

Intuitively I've been able to understand a Fourier transform a change-of-basis formula - you're basically moving from position to momentum basis or from time to frequency - but what does it mean that these spaces are 'conjugate' to each other? Does this have to do with them being complete bases?

A related question comes from considering the electric field generated by a travelling electron, $\textbf{E}(\textbf{r},t)$. If we consider sending $\textbf{E}$ to position-frequency space $\tilde{\textbf{E}}(\textbf{r},\omega)$, I find it weird that there is no longer a time dependency. Have we 'smeared' the electron across its trajectory and computed some quasi-average electric field? Is there some other interpretation that might make more sense?

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More on Fourier transform and intuition: physics.stackexchange.com/q/39442/2451 – Qmechanic Dec 21 '12 at 14:06
Its not weird that time doesn't appear in the position-frequency field because you integrate over time to get there. The $\tilde{\textbf{E}}(\textbf{r},\omega)$ gives you the frequency content of the field. To capture lower and lower frequencies, you need to integrate over more time. The wiki page on conjugate variables explains it pretty well. – xxx Dec 21 '12 at 14:08
Yeah - I understood mathematically why there wasn't a time dependence, but I guess I'm looking for intuition - it's easy to understand time/space dependence, but what is a momentum/wavelength dependent electric field? – alexvas Dec 22 '12 at 0:38

$t$ does not appear in $E(r,\omega)$ this is just an expansion coefficient in time dependent basis. You have moved the time dependence from the coefficients $E(r,t)$ to the basis functions $exp(i \omega t).$