Fourier Transform of a wavefunction

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SUMMARY

The Fourier transform of a wavefunction is essential for obtaining the probability density in momentum space, not position space. The wavefunction, being complex, cannot represent probability directly, which must be a real number between 0 and 1. To find the probability density for position, the integral P(x) = ∫ψ(x)*ψ(x) dx is used, while for momentum, the integral P(p) = ∫φ(p)*φ(p) dp is applied after transforming the wavefunction into momentum space.

PREREQUISITES
  • Understanding of wavefunctions in quantum mechanics
  • Knowledge of Fourier transforms and their applications
  • Familiarity with probability density functions
  • Basic calculus for evaluating integrals
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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
 
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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.
 
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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