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Fourier transform of the ground state hydrogen wave function

  1. Jun 21, 2017 #1

    Ado

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    Hi!

    1. The problem statement, all variables and given/known data

    From the website http://www1.uprh.edu/rbaretti/MomentumspaceIntegration8feb2010.htm
    we can see the fourier transform of the ground state hydrogenic wave function :

    Φ(p) = ∫ ∫ ∫ exp(-i p r) (Z3/π )1/2 exp(-Zr) sin(θ) dθ dφ r² dr (1.1)

    After intregation of the variable φ we have :

    Φ(p) = 2π (Z3/π )1/2 ∫ ∫ { exp(-i pr cos(θ ) sin(θ) dθ } exp(-Zr) r² dr (1.2)

    and next :

    Φ(p) = 2π (Z3/π )1/2 ∫ { 2 sin(pr) /pr } exp(-Zr) r² dr (1.3)

    = 4π (Z3/π )1/2 ∫ {sin(pr) /p} exp(-Zr) r dr

    = 8 π1/2/ Z5/2 /( p² + Z² )²

    And I want to understand these different steps..

    2. Relevant equations

    I don't understand the relations used between (1.1), (1.2) and (1.3) :
    (1.1) to (1.2) new cos(θ ) appears
    (1.2) to (1.3) sin(pr)/pr appears


    3. The attempt at a solution

    Can you explain me what happen between (1.1) and (1.3) ?

    Tanks in advance !
     
  2. jcsd
  3. Jun 21, 2017 #2
    In 1.1 we have exp(-ip.r). p.r is the dot product of two vectors, whose value is pr*cosθ, where θ is the angle between the vectors. I suppose in p space, which I'm not familiar with, this angle must be equal to the coordinate θ. There is a bracket missing in 1.2, which should read
    Φ(p) = 2π (Z3/π )1/2 ∫ ∫ { exp(-i pr cos(θ) ) sin(θ) dθ } exp(-Zr) r² dr (1.2)
    This can be integrated over θ by using the substitution u = cosθ and the identity eix = cosx + i*sinx, to give 1.3
     
  4. Jun 22, 2017 #3

    Ado

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    Thanks for your reply mjc123 !

    I had applied your recommendation and I find the result in (1.3) :wink:

    If I used the substitution u = cosθ, I must then integrate between 1 and -1 (cos(0) = 1 and cos(pi) = -1).

    I put du = -sinθdθ, so :

    exp(-i pr cos(θ) ) sin(θ) dθ = -exp(-i pr u) du = - cos(-pru)du + isin(pru)du

    The complex term disappears with the integration and

    1-1 - cos(-pru) du = 2sin(pr)/pr

    I have just a question about the bounds of integration. At the beginning, we integrate between 0 and pi and with the substitution, between 1 and -1. The order of these bounds is important because if we integrate du between -1 and 1 we would have a negative term. My question is probably stupid but why are we constrained at the beginning to integrate between 0 and pi and not between pi and 0 ??

    Thanks in advance !
     
  5. Jun 22, 2017 #4
    A mathematician could no doubt give you a better formal explanation, but I would say basically: because the increment is dθ. You could also integrate from pi to 0 with increment -dθ.
    If you think of it in terms of the area under a curve, for example, suppose you have a function y which is positive over the range x = 0 to 1, and you want the area under this curve, which of course is positive. This is ∫01ydx. We could also go from x = 1 to 0 in increments of -dx, and get ∫10y(-dx) = ∫10(-y)dx = ∫01ydx. Or think of the function y as the differential of the integral, i.e. the rate of change of the area with x. This is obviously positive (or negative if y is negative) with increasing x, i.e. positive dx. So we integrate y by dx over the interval 0 to 1, not 1 to 0. However, if we do a coordinate transformation, we must keep the initial and final limits in their transformed form, and transform dx correctly, so if we made the substitution u = e-x, we would integrate -y(u)/u*du from 1 to e-1.
    Hope this is not too rubbish.
     
  6. Jun 22, 2017 #5

    Ado

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    Thanks a lot for your explication mjc123 and thanks for your help!
     
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