Optical theorem and conservation of particle current

$$\sigma_{tot} = \frac{4\pi}{k} \text{Im}(f(0))$$

links the total cross section with the imaginary part of the scattering amplitude.

My lecture notes say that this is a consequence of the conservation of the particle current. How do I get to this consequence?

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Just look at a derivation en.wikipedia.org/wiki/Optical_theorem#Derivation - Because $|\psi|^2$ is proportional to the particle current and it may be calculated in two ways, the right verbal description is exactly what you said - the optical theorem follows from the conservation of the particle current. –  Luboš Motl Feb 25 '13 at 15:43

Conservation of particle current is nothing but the statement that a theory has to be unitary. In other words the scattering matrix $S$ has to obey

$SS^\dagger=1$

Defining $S=1+iT$ i.e. rewriting the scattering matrix as a trivial part plus interactions (encoded in $T$ which corresponds to your $f$) one finds from the unitarity condition:

$iTT^\dagger=T-T^\dagger=2Im(T)$

$TT^\dagger$ is nothing but the crosssection (I suppressed some integral signs here for brevity) the optical theorem is right there. Hence one finds $\sigma\sim Im(T)$

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Thanks for the beautiful explanation. –  David Seppi Feb 25 '13 at 16:53
Although this resembles the optical theorem, here $T$ and $TT^+$ are a matrices with generally non diagonal matrix elements. In order to obtain the optical theorem, one has to work with it little more. –  Vladimir Kalitvianski Feb 25 '13 at 18:05
Yes, you're right of course. I was a bit sloppy in my derivation and did not care about indices. Nevertheless, it is true that the optical theorem is nothing but a consequence of unitarity. –  A friendly helper Feb 25 '13 at 20:54

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 $\frac{\hbar k}{m}\sigma_{\text{tot}}=2\frac{\hbar}{m}T_1$.

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