Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  Optical theorem and conservation of particle current

+ 7 like - 0 dislike
727 views

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?

This post imported from StackExchange Physics at 2014-04-15 16:37 (UCT), posted by SE-user David Seppi
asked Feb 25, 2013 in Theoretical Physics by David Seppi (35 points) [ no revision ]
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.

This post imported from StackExchange Physics at 2014-04-15 16:37 (UCT), posted by SE-user Luboš Motl

2 Answers

+ 6 like - 0 dislike

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 $\sigma\sim Im(T)$

This post imported from StackExchange Physics at 2014-04-15 16:37 (UCT), posted by SE-user A friendly helper
answered Feb 25, 2013 by A friendly helper (320 points) [ no revision ]
Thanks for the beautiful explanation.

This post imported from StackExchange Physics at 2014-04-15 16:37 (UCT), posted by SE-user David Seppi
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.

This post imported from StackExchange Physics at 2014-04-15 16:37 (UCT), posted by SE-user Vladimir Kalitvianski
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.

This post imported from StackExchange Physics at 2014-04-15 16:37 (UCT), posted by SE-user A friendly helper
+ 4 like - 0 dislike

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
answered Feb 26, 2013 by FabianLackner (40 points) [ no revision ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsO$\varnothing$erflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...