# Coulomb Potential Scattering

We now consider the scattering of an electron from the Coulomb potential. This problem is important because it is relevant to the famous scattering experiment by Rutherford that showed that the atomic nucleus only makes up a very small fraction of the total size of an atom. We will first use the Born approximation to find the cross section for a Yukawa potential,

$V(r)=V_0\frac{e^{-\alpha r}}{r},$

which reduces to the Coulomb potential when $\alpha\to 0.$ We will then find the exact cross section for the Coulomb potential, starting with the Schrödinger equation.

## Scattering From a Yukawa Potential in the Born Approximation

Comparison the differential cross sections for various potential type
Here we compare the differential cross sections for the Yukawa, Gaussian and exponential potential. We can see that for small $q\alpha \!$ all differential cross sections are similar. The differential cross section of the exponential potential decreases faster than the others and that of the Yukawa potential decreases more slowly than the others.

As alluded to earlier, let us first consider the Yukawa potential,

$V(r)=V_0\frac{e^{-\alpha r}}{r}.$

Within the Born approximation, the scattering amplitude is

\begin{align} f(\theta) &= -\frac{m}{2\pi\hbar^2}2\pi\int_0^\infty dr'\,r'^2 \frac{e^{-i|\mathbf{k}'-\mathbf{k}|r'}-e^{i|\mathbf{k}-\mathbf{k}'|r'}}{-i|\mathbf{k}'-\mathbf{k}|r'}\frac{V_0e^{-\alpha r'}}{r'} \\ &= -\frac{2mV_0}{\hbar^2}\int_0^\infty dr' r'^2 \frac{\sin\left(|\mathbf{k}'-\mathbf{k}|r'\right)}{|\mathbf{k}'-\mathbf{k}|r'}\frac{e^{-\alpha r'}}{r'} \\ &= -\frac{2mV_0}{\hbar^2}\frac{1}{\left(\left|\mathbf{k}'-\mathbf{k}\right|\right)^2+\alpha^2} \end{align}

For the elastic scattering $\left| \mathbf{k} \right| = \left| \mathbf{k'} \right| = k \!$. Therefore,

$(\left|\mathbf{k}'-\mathbf{k}\right|)^2 = 2k\sin\frac{\theta}{2}$

and thus

$f = -\frac{2mV_0}{\hbar^2}\frac{1}{4k^2\sin^2\frac{\theta}{2}+\alpha^2}.$

The differential cross section is then

$\frac{d\sigma}{d\Omega}= \left( \frac{2mV_0}{\hbar^2} \right) ^2 \left( \frac{1}{4k^2\sin^2\frac{\theta}{2}+\alpha^2}\right)^2.$

## Exact Coulomb Scattering Cross Section

Parabolic coordinates
$\phi\!$ represents rotation about the z-axis, $\xi\!$ represents the parabolas with their vertices at a minimum, and $\eta\!$ represents parabolas with their vertices at a maximum.

When we are considering scattering due to the Coulomb potential, we can not neglect the effect of this potential at large distances because it is only a $\frac{1}{r}$ potential.

We will work in parabolic coordinates, which are related to Cartesian coordinates by

$\xi=\sqrt{x^2+y^2+z^2}-z$

$\eta=\sqrt{x^2+y^2+z^2}+z$

$\phi=\tan^{-1}\left (\frac{y}{x}\right )$

The Schrödinger equation in parabolic coordinates is

$\left[ - \frac{\hbar^2}{2\mu}\frac{4}{\xi+\eta} \left( \frac{\partial}{\partial \xi}\xi\frac{\partial}{\partial \xi}+\frac{\partial}{\partial \eta}\eta\frac{\partial}{\partial \eta}+\frac{\xi+\eta}{4\xi\eta}\frac{\partial^2}{\partial \phi^2} \right) -\frac{2Ze^2}{\xi+\eta} \right] \psi=\frac{\hbar^2k^2}{2\mu}\psi.$

Recall that, for a spherically symmetric potential, the scattering amplitude is a function of $\theta\!$ only; therefore, we will seek solutions that are independent of $\phi.\!$

We will look for solution of the form,

$\psi(\xi,\eta,\phi)=e^{\frac{i}{2}k(\eta-\xi)}\Phi(\xi)=e^{ikz}\Phi(r-z).$

The Schrödinger equation then becomes

$\left[ \xi\frac{\partial^2}{\partial \xi^2}+(1-ik\xi)\frac{\partial}{\partial \xi}+\frac{Ze^2\mu}{k\hbar^2}k \right]\Phi(\xi)=0.$

We now assume a series solution,

$\Phi(\xi)=\sum_{n=0}^\infty a_n \xi^n.$

The recursion relation for the coefficients is then

$\frac{a_{n+1}}{a_{n}}=\frac{in-\lambda}{(n+1)^2}k=\frac{n+i\lambda}{(n+1)^2}ik,$

where

$\lambda=\frac{Ze^2\mu}{k\hbar^2}.$

Recall that the confluent hypergeometric function 1F1 is given by

$_1F_1(a,c,z)=1+\frac{a}{c}z+\frac{a(a+1)}{c(c+1)}\frac{z^2}{2!}+\dots+\frac{a(a+1)\dots(a+n-1)}{c(c+1)\dots(c+n-1)}\frac{z^n}{n!}+\dots$

The recursion formula for its coefficients is

$\dfrac{\alpha_{n+1}}{\alpha_n}= \left( \dfrac{a+n}{c+n} \right) \dfrac{1}{n+1}.$

Comparing this to what we obtained earlier, we find that Φ(ξ) is

$\Phi(\xi)=A{}_1F_1(i\lambda,1,ik\xi),\!$

where $A\!$ is a c-number.

The full wave function is thus

$\psi(r,z)=A{}_1F_1(i\lambda,1,ik(r-z))e^{ikz}.\!$

In the limit, $z\to\infty,$ the confluent hypergeometric function is approximately

${}_1F_1(a,c,z)=\frac{\Gamma(c)(-z)^a}{\Gamma(c-a)} \left[ 1-\frac{a(a-c+1)}{z} \right] +\frac{\Gamma(c)}{\Gamma(a)}e^zz^{a-c}.$

We may use this to rewrite $\Phi\!$ in the limit of large ξ as

$\Phi(\xi)=Ae^{\frac{-\pi}{2}\lambda} \left[ \frac{e^{i\lambda \ln(k\xi)}}{\Gamma(1-i\lambda)} \left( 1-\frac{\lambda^2}{ik\xi} \right) +\frac{i\lambda}{ik\xi}\frac{e^{ik\xi+i\lambda \ln(k\xi)}}{\Gamma(1+i\lambda)} \right].$

In the same limit, the full wave function $\psi\!$ is

$\psi(r,\theta)=\frac{Ce^{\frac{-\pi\lambda}{2}}}{\Gamma(1-i\lambda)} \left[ \left( 1-\frac{\lambda^2}{2ikr}\frac{1}{\sin^2\frac{\theta}{2}} \right) e^{ikz}e^{i\lambda \ln(k(r-z))}+\frac{f(\theta)}{r}e^{ikr+i\lambda \ln(2kr)} \right],$

where $f(\theta)\!$ is

$f(\theta)=\frac{\lambda\Gamma(1-i\lambda)}{2k\Gamma(1+i\lambda)}\frac{1}{\sin^2\frac{\theta}{2}}e^{i\lambda \ln(\sin^2\frac{\theta}{2})}=\frac{\lambda\Gamma(1-i\lambda)}{2k\Gamma(1+i\lambda)} \left( \sin^2\frac{\theta}{2} \right) ^{i\lambda-1}.$

The differential cross section is thus

$\frac{d\sigma}{d\Omega}=|f(\theta)|^2=\frac{\lambda^2}{4k^2\sin^4(\frac{\theta}{2})}=\frac{(Ze^2)^2}{16E^2}\frac{1}{\sin^4(\frac{\theta}{2})}.$

Note that this is the same result that we would obtain in the $\alpha\to 0$ limit of the Yukawa potential using the Born approximation, as well as from a classical calculation.

## The Gamow Factor

We will now determine the relative probability of finding a particle at the origin to that of finding a particle in the incident beam, simply given by

$\left |\frac{\psi(0)}{\psi(\infty)}\right |^2.$

Since ${}_1F_1(a,c;0)=1,\!$ the probability density at the origin is just

$|\psi(0)|^2=|C|^2.\!$

At large distances, on the other hand, the probability density is

$|\psi(\infty)|^2=\frac{|C|^2 e^{-\pi\lambda}}{|\Gamma(1-i\lambda)|^2}.$

Using the fact that

$|\Gamma(1-i\lambda)|^2=\frac{2\pi|\lambda|e^{-\pi\lambda}}{e^{2\pi\lambda}-1},$

this becomes

$|\psi(\infty)|^2=\frac{|C|^2}{2\pi|\lambda|}(e^{2\pi\lambda}-1).$

We therefore obtain

$\left |\frac{\psi(0)}{\psi(\infty)}\right |^2=\frac{2\pi|\lambda|}{e^{2\pi\lambda}-1}.$

If we now define the velocity,

$\frac{\hbar k}{\mu}=v,$

then this becomes

$|\psi(0)|^2=\frac{2\pi Ze^2}{\hbar v}e^{-\frac{2\pi Ze^2}{\hbar v}}.$

The quantity $G=\frac{\pi Ze^2}{\hbar v}, \!$ that appears in the exponential is known as the Gamow factor. The Gamow factor (or Gamow-Sommerfeld factor), named after its discoverer George Gamow, is measure of the probability that two nuclear particles will overcome the Coulomb barrier in order to undergo nuclear reactions, such as nuclear fusion. Classically, there is almost no possibility for protons to fuse by overcoming the Coulomb barrier, but, when George Gamow instead applied quantum mechanics to the problem, he found that there was a significant chance for the fusion due to tunneling.