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?


1 Answer 1


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

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

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

\begin{equation} k = \frac{2\pi}{\lambda} \end{equation}

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

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

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

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.


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