5.5. Calculation of scattering cross sections with partial waves

In this section we derive Eqs. 5.19 and 5.20. At a large distance from the scattering centre we can make an Ansatz for the wave function. This consists of the incoming beam and a scattered wave: $$\psi ({\bf r}) \sim e^{i\bf k \cdot r } + f(\theta) \frac{e^{ikr}}{r}. \tag{5.22}\label{eq_psi}$$ $\theta$ is the angle between the incoming beam and the line passing through ${\bf r}$ and the scattering centre. $f$ does not depend on the azimuthal angle $\varphi$ because the incoming wave has azimuthal symmetry, and the spherically symmetric potential will not generate $m\neq 0$ contributions to the scattered wave. $f(\theta)$ is called the scattering amplitude. From the Ansatz it follows that the differential cross section is given directly by the square of this amplitude: $$\frac{d\sigma}{d\Omega} = |f( \theta ) |^2 \tag{5.23}\label{eq_sigma}$$ with the appropriate normalisation.

Beyond ${r_{\text{max}}}$, the solution can also be written in the form $$\psi({\bf r}) = \sum_{l=0}^{\infty} \sum_{m=-l}^l A_{lm} \frac{u_{l}(r)}{r} Y_l^m (\theta,\varphi) \label{eq_psi2}$$ leaving out all $m\neq 0$ contributions because of the azimuthal symmetry: $$\psi({\bf r}) = \sum_{l=0}^{\infty} A_l \frac{u_{l}(r)}{r} P_l (\cos\theta) \tag{5.23}\label{eq_psi3}$$ where we have used the fact that $Y^l_0(\theta, \phi)$ is proportional to $P_l[\cos(\theta)]$. Because the potential vanishes in the region $r>r_{\text {max}}$, the solution $u_{l}(r)/r$ is given by the linear combination of the regular and irregular spherical Bessel functions, and as we have seen this reduces for large $r$ to $$u_l (r) \approx \sin(kr - \frac{l\pi}{2} + \delta_l ).\tag{5.25}\label{bess_approx}$$ We want to derive the scattering amplitude $f(\theta)$ by equating the expressions \eqref{eq_psi} and \eqref{eq_psi3} for the wave function. For large $r$ we obtain, using \eqref{bess_approx}: $$\sum_{l=0}^\infty A_l \left[ \frac{\sin(kr - l\pi/2 + \delta_l )}{kr} \right] P_l (\cos \theta) = e^{i\bf k \cdot r } + f(\theta) \frac{e^{ikr}}{r}. \label{psi3}$$ We write the right hand side of this equation as an expansion similar to that in the left hand side, using the following expression for a plane wave $$e^{i \bf k \cdot r} = \sum_{l=0}^\infty (2l+1) i^l j_l(kr) P_l (\cos \theta).$$ $f(\theta)$ can also be written as an expansion in Legendre polynomials: $$f(\theta) = \sum_{l=0}^\infty f_l P_l (\cos \theta) \tag{5.x}\label{f_exp},$$ so that we obtain: $$\begin{gathered} \sum_{l=0}^\infty A_l \left[ \frac{\sin(kr - l\pi/2 + \delta_l )}{kr} \right] P_l (\cos \theta)= \\ \sum_{l=0}^\infty \left[ (2l+1) i^l j_l (kr) + f_l \frac{e^{ikr}}{r} \right] P_l (\cos \theta). \end{gathered}$$ If we substitute the asymptotic form of $j_l$ in the right hand side, we find: $$\begin{gather} \sum_{l=0}^\infty A_l \left[ \frac{\sin(kr - l\pi/2 + \delta_l )}{kr} \right] P_l (\cos \theta) = \\ \frac{1}{r}\sum_{l=0}^\infty \left[ \frac{2l+1}{2ik} (-)^{l+1} e^{-ikr} + \left(f_l+ \frac{2l+1}{2ik} \right) e^{ikr} \right] P_l (\cos \theta) \tag{5.26}\label{equat}. \end{gather}$$ Both the left and the right hand side of \eqref{equat} contain in- and outgoing spherical waves (the occurrence of incoming spherical waves does not violate causality: they arise from the incoming plane wave). For each $l$, the prefactors of the incoming and outgoing waves should both be equal on both sides in \eqref{equat}. This condition leads to $$A_l = (2l+1) e^{i\delta_l} i^l$$ and $$f_l = \frac{2l+1}{k} e^{i \delta_l} \sin(\delta_l).\tag{5.27}\label{f_delta}$$

Using \eqref{eq_sigma}, \eqref{f_exp}, and \eqref{f_delta}, we can write down an expression for the differential cross section in terms of the phase shifts $\delta_l$: $$\frac{d\sigma}{d\Omega} = \frac{1}{k^2} \left| \sum_{l=0}^\infty (2l+1) e^{i\delta_l} \sin(\delta_l) P_l (\cos \theta)\right|^2.$$ For the total cross section we find, using the orthonormality relations of the Legendre polynomials: $$\label{sigmatotdefrep} \sigma_{\text{tot}} = 2 \pi \int d\theta \sin \theta \frac{d\sigma}{d\Omega}(\theta) = \frac{4\pi}{k^2} \sum_{l=0}^\infty (2l+1) \sin^2 \delta_l.$$