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Fourier Transform of a wavefunction

  1. Aug 10, 2008 #1
    Why shud one take the Fourier transform of a wavefunction and multiply the result with its conjugate to get the probability? Why can't it be fourier transform of the probability directly?

    thank you
     
  2. jcsd
  3. Aug 10, 2008 #2

    Fredrik

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    The wave function can't be a probability (or probability density) since it's complex. A probability must obviously be a real number between 0 and 1.

    Also, you you only start by taking the Fourier transform if you're interested in the probability density of a certain value of the momentum. If you're interested in the probability density of a certain value of the position, you don't have to do a Fourier transform.
     
  4. Aug 10, 2008 #3

    G01

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    As Fredrick said, you don't take Fourier transform of a wave function in the process of finding the probability density. The probability density is given by (in one dimension):

    [tex]P(x)=\int\psi (x)^*\psi (x) dx[/tex]

    which does not involve a Fourier Transform.

    Instead, the Fourier transform of a wave function will give the wave function in momentum space (call it [itex]\phi[/itex]). Again, as Fredrick mentioned, we can use this to find the probability density for the momentum of the particle:

    [tex]P(p)=\int\phi (p)^*\phi (p) dp[/tex]
     
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