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Homework Help: F Transform

  1. Dec 10, 2007 #1


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    I am studying Quantummechanics, but I don't see how Fourriertransforms in quantum mechanics work

    I want to know how I can Fourrier Transform the Hydr. ground state, so the transform of
    [tex] \phi\left(r\right)=\left(\frac{1}{\pi a_{0}^3}\right)^\frac{1}{2} e^\left-(\frac{r}{a_{0}}\right) [/tex]

    Does someone knows the answer


  2. jcsd
  3. Dec 11, 2007 #2
    A Fouriertransform is used to convert one function in a certain space into the same function in one other space.

    To represent the hydrogen ground state in p-space (momentum space) you evaluate the following integral over the whole space:

    psi(p) = (2 pi h)^{-3/2} \int {exp(ip.r/h)*phi(r) dr}.

    Where the expression for the hydrogen ground state is to be inserted at phi, h represent the reduced placks constant and the expression p.r is a scalar product between the two vectors p and r.

    Hope you find yourself wiser after reading this!

    /The Latex-rookie
  4. Dec 11, 2007 #3


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    He Reid thanks

    but should it not be (2 pi h)^{-1/2}?
    Do you know where on the net I can find this formula

    Thanks for the help
  5. Dec 11, 2007 #4


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    It doesn't really matter which factor you put in front of the transform, as long as you make sure that the factor of the inverse transform is such that transforming forth and back subsequently yields the same answer. It's a matter of convention and convention differs in different areas of physics. In this case, Reid suggested ( (2 pi h)^{-1/2} )^3 because you are working in three dimensions. Other common factors include (2pi)^{-1/2}, (2 pi i)^{-1/2}.
  6. Dec 12, 2007 #5
    Hi again,
    sorry I don't know where to find it on the net. It is the eq. (2.59) in the book Quantum mechanics, second edition by Bransden & Joachain.

    As CompuChip writes, the coefficient in front of the integral is just for later convinience.

    I must however confess to a mistake: the exponent should be negative, i.e.
    psi(p) = (2 pi h)^{-3/2} \int {exp(-ip.r/h)*phi(r) dr}. I hope I haven't mislead you too bad.

    It is the in the exponent of the Fouriertransform from the p-space to r-space where the exponent should be positive, i.e.
    phi(r) = (2 pi h)^{-3/2} \int {exp(ip.r/h)*psi(r) dr}.

    Have a nice day! :)
  7. Dec 12, 2007 #6


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    The constant infront you can always put in there afterwards, since you demand the wavefunctions to be normalised. The constant depends on what dimensions you have. For 1-dim sqrt(2pi*hbar) is mostley used for example.

    The momentum-wave function is the Fourier transform of the position-wave function, so the exponential should have minus sign.

    And vice versa, the position-wave function is the inverse Fourier transform of the momentum-wave function, so no minus sign in the exponential.

    So what I told you here is just a confirmation on what CompuChip and Reid told you =) Go a head and try, if you dont get the correct answer, post your attempt here and we'll try to help you.
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