To make the optical theorem more apparent one can think of a clever experimentalist who does not arrange detectors in all possible directions from the target to determine the total cross section but only one detector with area $S_D$ in flight direction of the incoming particles. This detector should be small and far away from the target to make sure that only non-scattered particles are detected. For the calculation we take the incoming plane wave to enter in x-direction and the target to be located at the origin. Far away from the target the scattering state can be written as $$\psi(\vec{r})=e^{ikx}+f(\theta,\phi)\frac{e^{ikr}}{r}$$ and the particle current is given by $\vec{j}=\frac{\hbar}{m}\text{Im}(\psi^*\vec{\bigtriangledown}\psi) $. The detected particles per second are given by $$ \dot{N}=\int_{S_D} \vec{j} \; \text{d}\vec{A} =\frac{\hbar}{m} \int_{S_D} \text{Im}(\psi^*\vec{\bigtriangledown}\psi) \; \text{d}\vec{A} \\= \frac{\hbar}{m} \int_{S_D} \text{Im}(ik+f^*\frac{e^{-ikr}}{r} i \vec{k} e^{ikx} + e^{-ikx} \vec{\bigtriangledown}(f\frac{e^{ikr}}{r}) + f^*\frac{e^{-ikr}}{r} \vec{\bigtriangledown}(f\frac{e^{ikr}}{r}) )\; \text{d}\vec{A}. $$
If $f(\theta,\phi)=0$ (no scatterer) the detector finds $\frac{\hbar k}{m}S_D$ particles per second. The presence of the scatterer reduces the number of particles where the difference is given by the total number of scattered particles $\frac{\hbar k}{m}\sigma_{\text{tot}}$, where $\sigma_{\text{tot}}$ is the total cross section. The crucial point for this statement is particle conservation. With the detector area located at $x=x_0$ and radius $\rho_0$ facing in x-direction $\text{d}\vec{A}=\vec{e}_x\text{d}A$ we can write:
$$\frac{\hbar k}{m}\sigma_{\text{tot}}=\frac{\hbar k}{m}S_D-\dot{N} \\
=-\frac{\hbar}{m} \int_{S_D} \text{Im}(f^*\frac{e^{-ikr}}{r} i k e^{ikx} + e^{-ikx} \partial_x(f\frac{e^{ikr}}{r}) + f^*\frac{e^{-ikr}}{r} \partial_x(f\frac{e^{ikr}}{r}) )\; \text{d}A=\frac{\hbar}{m}(T_1+T_2+T_3).$$
In general this integral is very complicated but we can use the fact that the detector area is far away from the target. A first guess for this limit would be to take a fixed detector radius $\rho_0$ and move the detector far away $x_0 \rightarrow \infty$. However, in this limit we have $\sigma_{\text{tot}}=0$ since the scattered wave drops with $\frac{1}{r}$. To obtain a finite value for $\sigma_{\text{tot}}$ one has to keep the ratio $\frac{\rho_0}{x_0}=\tan(\theta_0)$ fixed as $x_0 \rightarrow \infty$ and then perform the limit $\theta_0 \rightarrow 0$ afterwards. The actual calculation is a bit tricky but I will show it for the first term:
$$T_1= -\int_{S_D}\text{Im}(ikf^*\frac{e^{ik(x-r)}}{r})\text{d}A \\
=-\text{Im}\int_{\phi=0}^{2\pi}\int_{\rho=0}^{\rho_0}ikf^*e^{ik(x_0-\sqrt{x_0²+\rho²})}\frac{\rho}{\sqrt{x_0²+\rho²}}\text{d}\rho\text{d}\phi .$$
Now we use $\frac{\rho}{x_0}\ll 1$:
$$T_1=-2\pi \text{Im}(ik f^*(0)\int_{\rho=0}^{\rho_0} e^{-ik\frac{\rho^2}{2x_0}} \frac{\rho}{x_0} \text{d}\rho)\\
= -2\pi\text{Im}(f^*(0)(1-e^{\frac{ik}{2}\tan^2(\theta_0)x_0})). $$
To perform the limit $x_0 \rightarrow \infty$ we add a small imaginary part to $k\rightarrow k+i\epsilon$ then perform $x_0 \rightarrow \infty$ and let $\epsilon \rightarrow 0$ afterwards. Fees so good to be a physicist :) Therefore, the first contribution to the total cross section is $T_1=2\pi \text{Im}(f(0))$. It turns out that the second term $T_2$ gives $T_1$ as well and the third term $T_3$ gives zero because it drops faster than $\frac{1}{r}$. Altogether this gives the optical theorem $\frac{\hbar k}{m}\sigma_{\text{tot}}=2\frac{\hbar}{m}T_1$.
This post imported from StackExchange Physics at 2014-04-15 16:37 (UCT), posted by SE-user FabianLackner
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