Collisions with Absorption

Elastic Scattering: \(\varphi_{k,\ell,m}(\vec{r})\sim\exp(ikz) + f_k(\theta,\varphi)\exp(ikr)/r \sim \sum_{\ell=0}^\infty i^\ell\sqrt{4\pi(2\ell+1)}Y_\ell^0(\theta)\cdot\frac{1}{2ikr}\left(\exp(i(kr-\ell\pi/2)\exp(2i\delta_\ell) - \exp(-i(kr-\ell\pi/2)))\right)\). Where \(\exp(-i(kr-\ell\pi/2))\) is the incoming wave and \(\exp(i(kr+\ell\pi/2))\) is the outgoing wave.

(Inelastic with absorption) Now we will consider when \(\delta_\ell\) is complex, hence it has a lossy imaginary part. Let \(\eta_\ell \equiv \exp(2i\delta_\ell)\) now \(|\eta_\ell|^2\leq 1 = \exp(-4\Im\delta_\ell)\).

Now we have,

\begin{align} \varphi_{k,\ell,m}(\vec{r}) &\sim\exp(ikz) + f_k(\theta,\varphi)\exp(ikr)/r \nonumber \\ &\sim \sum_{\ell=0}^\infty i^\ell\sqrt{4\pi(2\ell+1)}Y_\ell^0\frac{\sin(kr-\ell\pi/2)}{kr} + \sum_{\ell=0}^\infty i^\ell\frac{\sqrt{4\pi(2\ell+1)}}{2ik}Y_\ell^0(\eta_\ell-1)(-i)^\ell\frac{\exp(ikr)}{r} \nonumber \\ &\sim \sum_{\ell=0}^\infty i^\ell\sqrt{4\pi(2\ell+1)}Y_\ell^0\frac{\sin(kr-\ell\pi/2)}{kr} + \sum_{\ell=0}^\infty \frac{\sqrt{4\pi(2\ell+1)}}{2ik}Y_\ell^0(\eta_\ell-1)\frac{\exp(ikr)}{r} \end{align}

Then,

\begin{equation} f_k(\theta) = \sum_{\ell=0}^\infty\sqrt{4\pi(2\ell+1)}Y_\ell^0(\theta)\frac{\eta_\ell-1}{2ik} \end{equation}

Which gives,

\begin{equation} \sigma_{tot} = \frac{\pi}{k^2}\sum_{\ell=0}^\infty(2\ell+1)|1-\eta_\ell|^2 \end{equation}

For elastic scattering, \(|1-\eta_\ell|^2 = (1-\exp(2i\delta_\ell))(1-\exp(-2i\delta_\ell)) = 2(1-\cos(2\delta_\ell)) = 4\sin^2\delta_\ell\).

This gives us the optical theorem for elastic scattering,

\begin{equation} \sigma_{tot}^{(scat)} = \frac{4\pi}{k}\Im f_k(0). \end{equation}

If we consider a fully absorbing target, then we get \(\sigma_{tot}^{(scat)} = \frac{\pi}{k^2}\sum_{\ell=0}^\infty (2\ell+1)\).

Now we construct the absorbtion cross section: \(\sigma_{abs} = \frac{\Delta\mathcal{P}}{|J_i|} = \frac{\mu\Delta\mathcal{P}}{\hbar k}\). Where \(\Delta\mathcal{P}\) is the total probability current that is absorbed: \(\Delta\mathcal{P} = -\int \vec{J}_{abs}\cdot d\vec{S}\).

\(\vec{J} = \Re\{\varphi_k^*(\vec{r})\frac{\hbar}{i\mu}\nabla\varphi_k(\vec{r})\}\) In the asymtotic limit, \(J_r = \Re\{\varphi_k^*\frac{\hbar}{i\mu}\frac{\partial}{\partial r}\varphi_k\}\). Where \(\varphi_k \sim \sum_{\ell=0}^\infty i^\ell\sqrt{4\pi(2\ell+1)}Y_\ell^0\frac{\exp(i(kr-\ell\pi/2))\eta_\ell - \exp(-i(kr-\ell\pi/2))}{2ikr}\). Then, \((J_{in}^\ell)_r = \frac{\hbar}{\mu k}\frac{\pi(2\ell+1)}{r^2}|Y_\ell^0|^2\) and \((J_{out}^\ell)_r = (J_{in}^\ell)_r|\eta_\ell|^2\). Then, \((J_{abs}^\ell)_r = (J_{in}^{\ell})_r(1 - |\eta_\ell|^2) = \frac{\hbar}{\mu k}\frac{\pi(2\ell+1)}{r^2}|Y_\ell^0|^2(1 - |\eta_\ell|^2)\).

Thus,

\begin{align*} \Delta\mathcal{P} &= -\sum_\ell\int\frac{\hbar}{\mu k}\frac{\pi(2\ell+1)}{r^2}|Y_\ell^0|^2(1 - |\eta_\ell|^2) r^2d\Omega \\ &= -\frac{\hbar}{\mu k}\pi\sum_\ell(2\ell+1)(1 - |\eta_\ell|^2)\int|Y_\ell^0|^2 d\Omega \\ &= -\frac{\hbar}{\mu k}\pi\sum_\ell(2\ell+1)(1 - |\eta_\ell|^2). \end{align*}

Therefore,

\begin{align*} \sigma_{abs} &= \frac{\Delta\mathcal{P}}{|J_i|} \\ &= \frac{\mu}{\hbar k}\frac{\hbar}{\mu k}\pi\sum_\ell(2\ell+1)(1 - |\eta_\ell|^2) &= \frac{\pi}{k^2}\sum_\ell(2\ell+1)(1 - |\eta_\ell|^2). \end{align*}

Our total cross section is then, \(\sigma_{tot} = \sigma_{tot}^{(scat)} + \sigma_{abs} = \frac{4\pi}{k}|\Im f_k(0)|\).