Scattering

Scattering Theory

Classical Scattering Theory

Consider a single incident particle with energy \(E\), collision parameter \(b\), and exit with angle \(\theta\). The basic question of classical scattering theory is to calculate scattering angle with collision parameter.
Generally, incident particles into a cross section area \(d\sigma\) would be scattered into a solid angle \(d\Omega\). The ratio, \(D(\theta) = d\sigma/d\Omega\), is call the differential cross section.

Using the collision parameter, \(d\sigma = bdbd\phi, d\Omega=\sin \theta d\theta d\phi\), so, \begin{equation} D(\theta) = \frac{b}{\sin\theta}|\frac{db}{d\theta}| \end{equation} The total cross section is the integral to solid angle: \begin{equation} \sigma = \int D(\theta) d\Omega \end{equation}

Quantum Scattering Theory

We consider an incident plane wave, \(\psi(z)= Ae^{ikz}\), along \(z\) direction, which meets a scattering potential and produce a spherical wave outwards. So we need to find the solution to the Schrodinger equation with following form: \begin{equation} \psi(r,\theta) \approx A\lbrace e^{ikz}+f(\theta)\frac{e^{ikr}}{r}\rbrace \end{equation} for large \(r\), where the wave number \(k=\frac{\sqrt{2mE}}{\hbar}\). So the question is to find the scattering amplitude \(f(\theta)\), which gives the scatter probability in \(\theta\) direction in order to calculate cross section. \begin{equation} D(\theta) = \frac{d\sigma}{d\Omega}=|f(\theta)|^2 \end{equation} We have two methods to calculate the scattering amplitude: sub-wave method and Born approximation.

Sub-wave Method

Under spherically symmetric potential, the solution to TISE is: \begin{equation} \psi(r,\theta,\phi)=R(r)Y_l^m(\theta,\phi) \end{equation} where \(Y_l^m\) is the spherical harmonic function, and \(u(r)=rR(r)\) obeys the radical equation: \begin{equation} -\frac{\hbar^2}{2m}\frac{d^2 u}{dr^2}+[V(r)+\frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}]u = Eu \end{equation}

Radiation Zone

For large \(r (kr>>1)\) the potential \(V\to 0\) and centrifugal term is ignorable, the equation becomes: \begin{equation} \frac{d^u}{dr^2}\approx -k^2 u \end{equation} with general solution: \begin{equation} u(r) = Ce^{ikr}+De^{-ikr} \end{equation} As we want to find scattered wave, we require \(D=0\). So for large \(r\): \begin{equation} R(r) ~ \frac{e^{ikr}}{r} \end{equation}

Middle Region

In the middle region (where \(V\) is ignorable but the centrifugal term is not), the radical equation becomes: \begin{equation} \frac{d^2 u}{dr^2}-\frac{l(l+1)}{r^2}u=-k^2 u \end{equation} whose general solution is the combination of Bessel functions: \begin{equation} u(r)=ArJ_l(kr)+BrN_l(kr). \end{equation} We rewrite it with the spherical Hankel function (like \(e^{ikr}\) and \(e^{-ikr}\)): \begin{equation} H_l^{(1)}(x)=J_l(x)+iN_l(x) \end{equation} \begin{equation} H_l^{(2)}(x)=J_l(x)-iN_l(x) \end{equation} For the same reason, we choose the first Hankel function \(H_l^{(1)} \to e^{ikr}/r\) for large \(r\). Thus, outside the scattering region (\(V(r)=0\)), the wave function is: \begin{equation} \psi(r,\theta,\phi) = A\lbrace e^{ikz}+\sum_{l,m} c_{l,m} H_l^{(1)}(kr) Y_l^m(\theta,\phi)\rbrace \end{equation} Because we assume the potential has spherical symmetry, only \(m=0\) term exists. So: \begin{equation} \psi(r,\theta) = A\lbrace e^{ikz}+ k\sum_{l=0}^{\infty} i^{l+1}(2l+1)a_l H_l^{(1)}(kr) P_l(\cos\theta)\rbrace \end{equation} where \(a_l\) is called the \(l\)th sub-wave amplitude. Thus, for large \(r\), \begin{equation} \label{11.25} f(\theta) = \sum_{l=0}^{\infty}(2l+1)a_l P_l(\cos\theta) \end{equation} And the total cross section: \begin{equation} \sigma = 4\pi\sum_{l=0}^{\infty} (2l+1)|a_l|^2 \end{equation}