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From what I understand, the Fourier transform decomposes a function into sines and cosines; and it is up to me to assign original domain and the transformed domain. But from my little experience, I noticed that only domains that are inverse of the other $$[Time]=\frac{1}{[Frequency]}; [Space]=\frac{1}{[wavenumber]}$$ uses Fourier transform.

I see no reason why that would be the case. Or am I misunderstanding something? Why did this happen, or is it just a coincidence? Can I use Fourier transform, say from the space domain to the frequency domain? Thank you!

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    $\begingroup$ Look at the Fourier series. Let $f : \mathbb{R} \to \mathbb{C}$ continuously differentiable and $T$-periodic. Then $f(x) = \sum_{n=-\infty}^\infty c_n e^{2i \pi n x/T}$ where $c_n = \frac{1}{T} \int_0^T f(x) e^{-2i \pi nx/T}dx$ and the series converges absolutely and uniformly. Your problem reduces to the dimension of $x,n,T$ so that $e^{2i \pi nx/T}$ makes sense. Note $\frac{T}{n}$ is a period, $n/T$ is a frequency, $2\pi n/T$ an angular frequency. $\endgroup$ – reuns Dec 26 '17 at 4:13
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    $\begingroup$ Your question and the linked one are mixing up three different things. (1) Mathematically the Fourier transform operates on functions and doesn't care about units. (2) When applied to a function which has a physical meaning (e.g. a time signal like a sound wave, or a spatial signal like an image), the result of the Fourier transform can be interpreted physically as having inverse units (e.g. frequency or wavenumber). (3) In quantum mechanics, the Fourier transform applied to the wavefunction does relate position to momentum; but this is a physical fact not arising purely from the mathematics. $\endgroup$ – Rahul Dec 26 '17 at 4:53
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I believe thanks to @reuns I have found the answer. Let's say you transform from $x$ domain to $y$ domain. Physically, $e^{2i \pi xy}$ is a phase, so it must be dimensionless. Therefore, it can only be that $[x]=\frac{1}{[y]}$.

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