# Uncertainty Principle and Fourier Transform

• gareththegeek
In summary, the time dependent wavefunction is related to the Fourier transform of the wavefunction for the angular wavenumber. This allows for analysis of the component frequencies which are related to the angular wavenumber. Taking the Fourier transform also leads to the Heisenberg Uncertainty Principle, where compressing the wavefunction results in a more spread out Fourier transform and therefore limits the ability to know both with 100% accuracy.

#### gareththegeek

I have read that the time dependent wavefunction is related to the Fourier transform of the wavefunction for the angular wavenumber like so

$$\bar{\psi}(k,t) = \frac{1}{\sqrt{2\pi}}\int \psi(x,t)e^{-ikx}dx$$

Can anyone explain why it is relevant to take the Fourier transform of the wavefunction in this case?

Is it the case that the wavefunction is a composite of more than one sinusoidal wave, taking the Fourier transform of which allows analysis of the component frequencies where the component frequencies are related to the angular wavenumber?

I understand that this leads to the Heisenberg Uncertainty Principle since the more you compress the wavefunction the more spread out becomes the Fourier transform, meaning therefore that you cannot know both with 100% accuracy. Is this right?

Thanks,
G

gareththegeek said:
Is it the case that the wavefunction is a composite of more than one sinusoidal wave, taking the Fourier transform of which allows analysis of the component frequencies where the component frequencies are related to the angular wavenumber?

Yes!

I understand that this leads to the Heisenberg Uncertainty Principle since the more you compress the wavefunction the more spread out becomes the Fourier transform, meaning therefore that you cannot know both with 100% accuracy. Is this right?

Yes!

Nice! Perhaps I'm finally beginning to get the hang of this ere quantum stuff then eh, eh?