- #1
Xyius
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Homework Statement
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##
Homework 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.
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.
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