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Eikonal Approximation: Find total Scattering Cross Section

  1. Apr 13, 2014 #1
    1. The problem statement, all variables and given/known data
    Using the Eikonal approximation
    (1) Determine the expression for the total scattering cross section of a particle in a potential V(r)
    (2) Using this result, compute the total scattered cross section for the following potential.

    [tex]
    V(r)=
    \begin{cases}
    V_0, \text{for } r < a \\
    0 , \text{for } r >a
    \end{cases}
    [/tex]

    Where ##V_0 > 0##



    2. Relevant equations

    The differential cross section is given by..

    1.) [tex]\frac{d\sigma}{d\Omega}= |f(\vec{k}',\vec{k})|^2[/tex]

    Where ##f(\vec{k}',\vec{k})## is the scattering amplitude.

    2.) The optical theorem is a easy way to find the total scattered cross section from a potential. It is given by..

    [tex]\text{Im}\left(f(\theta=0)\right)=\frac{\sigma_{tot}}{4\pi}[/tex]

    3.) The expression for the Eikonal approximation is..

    [tex]f(\vec{k}',\vec{k})=-i k \int_0^{\infty}db b J_0(kb\theta)[e^{2 i \Delta(b)}-1][/tex]

    Where..

    [tex]\Delta(b)=\frac{-m}{2k\hbar^2}\int_{-\infty}^{\infty}V(\sqrt{b^2+z^2})dz[/tex]

    Where ##V(\sqrt{b^2+z^2})## means ##V## OF ##\sqrt{b^2+z^2}##, not times.

    ##b## is the impact parameter, and in the book they say that ##l## can be treated as a continuous variable since we are at high energies, and they say that ##l = bk ##. Not sure if this helps.

    3. The attempt at a solution

    I used the optical theorem (equation 2) to get..

    [tex]\sigma_{tot}=4 \pi \text{Im} \left(f(\vec{k}',\vec{k})\right)=-4 \pi k \int_0^{\infty}db b J_0(kb\theta)[\text{Re}(e^{2i\Delta(b)})-1]_{\theta=0}[/tex]

    So it seems that all I would need to do is calculate the integral, but I am having trouble finding ##\Delta(b)## because the limits are infinity. When I plug in the potential I get..

    [tex]\Delta(b)=\frac{-m}{2k\hbar^2}\int_{-\infty}^{\infty}V_0dz[/tex]

    Which diverges? Clearly the limits simplify to something that's not infinity on both ends. But the variable of integration is z, which has no limit in either direction. I found some examples, but the ones I found use the gaussian potential and do not change the limits of integration.
     
    Last edited: Apr 13, 2014
  2. jcsd
  3. Apr 14, 2014 #2

    maajdl

    User Avatar
    Gold Member

    The range of integration should be: b²+z²<a² .
     
  4. Apr 14, 2014 #3
    So I guess the integration is only over the scatterer?

    So if I have ##b^2+z^2=a^2 \rightarrow z=\sqrt{a^2-b^2}##. So the limits of integration should be from ##z=-\sqrt{a^2-b^2}## to ##z=\sqrt{a^2-b^2}##?
     
  5. Apr 15, 2014 #4

    maajdl

    User Avatar
    Gold Member

    The limits are over the whole space.
    But your potential V(r) is not.
     
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