Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

The optical theorem

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

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?

share|improve this question
1  
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
add comment

2 Answers 2

up vote 9 down vote accepted

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 $\sigma\sim Im(T)$

share|improve this answer
    
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
add comment

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.