# Homework Help: Eikonal Approximation: Find total Scattering Cross Section

1. Apr 13, 2014

### Xyius

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.

$$V(r)= \begin{cases} V_0, \text{for } r < a \\ 0 , \text{for } r >a \end{cases}$$

Where $V_0 > 0$

2. Relevant equations

The differential cross section is given by..

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

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..

$$\text{Im}\left(f(\theta=0)\right)=\frac{\sigma_{tot}}{4\pi}$$

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

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

Where..

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

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..

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

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..

$$\Delta(b)=\frac{-m}{2k\hbar^2}\int_{-\infty}^{\infty}V_0dz$$

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. Apr 14, 2014

### maajdl

The range of integration should be: b²+z²<a² .

3. Apr 14, 2014

### Xyius

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 $z=\sqrt{a^2-b^2}$?

4. Apr 15, 2014

### maajdl

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