# 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,

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

Which gives,

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

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,

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

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, $$\sigma_{tot} = \sigma_{tot}^{(scat)} + \sigma_{abs} = \frac{4\pi}{k}|\Im f_k(0)|$$.

## The Attractive Potential

### Slow Particle

\begin{align*} u_{k,0}(r) &= \begin{cases}A\sin(kr+\delta_0) & r>r_0 \\ B\sin(Kr) & r Imposing continuity, $$\left(\frac{B}{A}\right)^2 = \frac{k^2}{k^2 + k_0^2\cos^2(Kr_0)}, k_0 = \sqrt{\frac{2mV_0}{\hbar^2}}$$. If $$Kr_0 = \frac{\pi}{2} + \pi n$$ then $$\left(\frac{B}{A}\right)^2\to 1$$ and $$\delta_0 = \frac{\pi}{2} - kr_0 \to \frac{\pi}{2}$$. If $$Kr_0 = \pi + \pi n$$ then $$\left(\frac{B}{A}\right)^2\to \frac{k^2}{k^2 + k_0^2}$$ and $$\delta_0 \to -kr_0$$. Further, this goes to zero for slow particles. This is called Ramsauer-Townsend effect - it is as if the particle does not see the target.

The maximal cross section for a particular $$\ell$$ is expected to be $$\frac{4\pi}{k^2}(2\ell+1)$$. In general, $$\frac{4\pi}{k^2}\frac{\Gamma^2(E_R)/4}{(E-E_R)^2 + \Gamma^2(E_R)/4}$$.

Now if we analyze higher angular momentums, $$V_{eff} = V + \frac{\hbar^2\ell(\ell+1)}{2\mu r^2}$$. Then we get metastable bound states in our attractive potential with lifetimes $$\tau = \frac{\hbar}{\Gamma}$$.

Created: 2024-05-30 Thu 21:19

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