Uncertainty Principle and Fourier Transform

  • #1

Main Question or Discussion Point

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

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

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
 

Answers and Replies

  • #2
jtbell
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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! :smile:
 
  • #3
Nice! Perhaps I'm finally beginning to get the hang of this ere quantum stuff then eh, eh?
 

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