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Homework Help: Particle Physics Integral Calculation

  1. May 12, 2010 #1
    1. The problem statement, all variables and given/known data
    In calculating the quantum mechanical amplitude for the Coulomb potential (scattering of say [tex]\alpha[/tex] particle off a massive particle of charge [tex]Ze[/tex]), I came across a fourier transform which I could not calculate.

    [tex] U(r)=\frac{2Ze^2}{4\pi\epsilon_0 r}[/tex]
    [tex]\tilde{U}(k)=\int U(\mathbf{x})e^{i\mathbf{k}\cdot\mathbf{x}}d^3x=\frac{Ze^2}{\epsilon_0 k^2} [/tex]

    Could you please help me calculate this integral

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. May 12, 2010 #2


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    It looks pretty straightforward, just choose your coordinate system so that [itex]\textbf{k}[/itex] points along the z-axis and perform the integration in spherical coordinates.

    Edit: It turns out that the integral doesn't converge directly, so instead you may wish to transform the Yukawa potential [itex]\frac{1}{r}e^{-ur}[/itex], and then take the limit as [itex]u\to 0[/itex]. Strictly speaking, this is only valid if you treat the Coulomb potential as a distribution and not a mathematical function (otherwise, the limit of the integral need not be the same as the integral of the limit).
    Last edited: May 12, 2010
  4. May 12, 2010 #3
    Thanks for your reply, I tried integrating over [tex]\mathbb{R}^3[/tex] but my integral does not converge for [tex]0\leq r<\infty[/tex] i.e. integrating

    [tex]\int_0^{2\pi}\int_0^{\pi}\int_0^{\infty}r e^{ikr \cos\theta} \sin\theta dr d\theta d\phi[/tex]

    Doing the [tex]\theta[/tex] integration first I get

    [tex]\int_0^{2\pi}\int_0^{\infty}\frac{2\sin kr}{k}dr d\phi[/tex]

    But the r integration does not converge. Am I using the wrong limits
  5. May 12, 2010 #4


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    Did you read the edit part of my post?
  6. May 13, 2010 #5
    Yes I read the edit part of your post, but in my lecture notes there was nothing said about approximation by the Yukawa potential, the integral was just given as I stated it above. Is the integral an approximation? It partially works out when integrated. (also is it related to the first born approximation?)
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