# Fourier Transform of a Wave Function relating to Position and Momentum

Why is it that, in quantum mechanics, the fourier transform of a wavefunction of the position of a particle is the wavefunction of it's momentum? I'm trying to learn about the uncertainty principle, and the example was given that if the fourier transform of the wavefunction of a particle's position was very localized (like a single frequency), then the wavefunction is a completely de-localized sine wave. I see why this is true, but how is the frequencies making a particle's wave function determine the momentum of the particle?

A Fourier transform decomposes functions of position into a linear combination of functions of spatial frequency, i.e.

$$$$\phi(k) = \int_{-\infty}^{\infty}\frac{\text{d}k}{2\pi}\psi(x)e^{-ikx}$$$$

where $$\psi(x)$$ is the function in position space and $$k = 2\pi/\lambda$$ is the wavevector. The wavevector is defined as

$$$$k = \frac{2\pi}{\lambda}$$$$

where $$\lambda$$ is the wavelength. $$k$$ can be thought of as a spatial frequency (compare this to the definition of the angular frequency $$\omega = 2\pi/f$$).

Now recall the de Broglie equation relating the momentum of a particle to its wavelength

$$$$\begin{split} p &= h/\lambda \\ \\ &= \hbar k \end{split}$$$$

where $$\hbar = h/2\pi$$. Using the de Broglie equation we can simply perform a change in variables in our original (inverse) Fourier transform:

$$$$\phi(p) = \int_{-\infty}^{\infty}\frac{\text{d}p}{2\pi\hbar}\psi(x)e^{-ipx/\hbar}$$$$

Clearly, if we have a particle very tightly confined in position then we will many momentum components to describe it and vice-versa. I hope this answers your question.