%         Created on  11-JANUARY-1996 by Phillip Gutierrez
%-------------------------------------------------------------------
\tolerance 1000
\documentclass [11pt]{article}
\usepackage{amsmath}
\usepackage{fancyheadings}
\usepackage{epsfig}
\textwidth 6.5in
\textheight 8.5in
\topmargin 0in
\oddsidemargin 0in
\evensidemargin 0in
%\parskip 1.5ex
\baselineskip 3.0ex
%\parindent 0in
%\thispagestyle{myheadings}
%\markright{\today}
%
%
\pagestyle{fancy}
\lhead{\today}
\chead{}
\rhead{P. Gutierrez}
\lfoot{}
\cfoot{Lecture 11-\thepage}
\rfoot{}
%
\def\reff{r_{\text{eff}}}
\newcommand{\nunit}[2]{\ensuremath{#1\,\text{#2}}}
%\input{my_macs.tex}
%\input{d0_style.tex}
\begin{document}

\begin{center}
\LARGE\bf  Physics 4213/5213\\ Lecture 11\\
\end{center}
%
\section{Introduction}
The next few lectures will discuss cross sections in a more quantitative
way than has been done until now. The discussion will start with a of
the wave function in terms of radial states, which allows the introduction
of angular momentum into the problem. This expansion will be used to
describe the incident and scattered particles from a scattering center.
Note that this treatment will not be relativistic, but it will introduce
many of the concepts of scattering cross section. In addition, the idea
of resonance production and why they are identified as particles will
also be examined.
\section{Statement of the Problem}
The problem that is to be examined, is the scattering of particles
off of other particles. The initial state 
particles are in a prepared state,
that is, the energy, momentum and type of particles are known. On the
other hand, the properties of outgoing particles are not known a priori.
Another important point, is that the interaction occurs in a region
of $\approx \nunit{10^{-13}}{cm}$ while the initial state particles
are prepared at a minimum a few cm away and the final state system
is measured at a minimum also a few cm away.
\section{The Incident Beam}
The incident beam of particles is describe by a plane wave wave-function:
\begin{equation}
  \psi_i = e^{ikz}=e^{ikr \cos\theta}
\end{equation}
To both introduce angular momentum and to write the solution in an asymptotic
form (form for large distances compared to the scattering center), the
wave-function is expanded out in terms of the Legendre polynomials;
\begin{equation}
  e^{ikr \cos\theta} = \sum_\ell a_\ell(kr) P_\ell(\cos\theta)
\end{equation}
Calculating the coefficients is achieved by multiplying both sides by
$P_\ell(\cos\theta)$ and integrating over $d\cos\theta$ from $-1$ to 
$+1$. This yields (without proof):
\begin{equation}
  a_\ell(kr) = \int^{+1}_{-1} e^{ikr \cos\theta}  P_\ell(\cos\theta) = 
  i^\ell (2\ell+1) j_\ell(kr)
\end{equation}
This leads to
\begin{equation}
    e^{ikr \cos\theta} = \sum_{\ell=0}^\infty j_\ell(kr) P_\ell(\cos\theta)
    \label{eq:expan}
\end{equation}

Given equation \ref{eq:expan}, the condition that the observation of the wave is
at a large distance ($kr>>1$) is imposed, giving;
\begin{equation}
  \psi_{\text{inc}} = \frac{i}{2 kr} \sum_\ell (2\ell +1) \left[
                      (-1)^\ell e^{-ikr} -e^{ikr}\right] P_\ell(\cos\theta)
\end{equation}
Therefore the plane wave is represented by both an incoming and outgoing 
wave---recall that the
plane wave represents a wave covering all space. The first term in the brackets 
represents the incoming wave, while the second term represents the outgoing
wave. Now consider a scattering potential at $r=0$. In this case, the amplitude 
and phase of the outgoing wave can be altered. The wave-function then becomes
\begin{equation}
  \psi_{\text{total}} = \frac{i}{2 kr} \sum_\ell (2\ell +1) \left[
                      (-1)^\ell e^{-ikr} - \eta_\ell e^{2i\delta_\ell}
                      e^{ikr}\right] P_\ell(\cos\theta) \label{eq:scat}
\end{equation}
where $0\le \eta_\ell \le 1$---$\eta_\ell$ not equal to one represents the case
where part of the beam is absorbed. Since this represents the total wave-function
the scattered part is obtained by subtracting out the incident wave;
\begin{equation}
  \psi_{\text{scat}} = \frac{e^{ikr}}{r} \sum_\ell
     (2\ell+1) \frac{\eta_\ell e^{2i\delta_\ell}-1}{2 ki} P_\ell(\cos \theta)
     = \frac{e^{ikr}}{r} F(\theta) 
\end{equation}
where $F(\theta)$ is referred as the scattering amplitude.

Next, the scattering amplitude needs to be expressed as a cross section. To
do so, the flux first needs to be determined. This is just the current that
crosses some finite area. The current for the Schrodinger equation is defined
as:
\begin{equation}
  \vec{j} = \frac{1}{2im} (\psi^* \vec{\nabla} \psi - \psi \vec{\nabla} \psi^*)
\end{equation}
For the incident wave this is given by:
\begin{equation}
  \vec{j} = \frac{\vec{k}}{m} 
\end{equation}
which corresponds to the flux per unit area across a surface
(it is assumed that the current is normal to the surface, this does not
change any of the results other than to add a factor of $\cos\theta$
where $\theta$ is the angle between the beam and the normal to the surface) 
while the current for the outgoing wave is:
\begin{equation}
  \vec{j} = \frac{\vec{k}}{m} \frac{|F(\theta)|^2}{r^2}
\end{equation}
The differential cross section is the outgoing flux through 
a given differential
area:
\begin{equation}
  \mathcal{F} = \vec{j}\cdot d\vec{A} = 
     \frac{|\vec{k}|}{m} \frac{|F(\theta)|^2}{r^2}
    \frac{d\Omega}{r^2}
\end{equation}
divided by the incoming
flux per unit area ($\vec{j}$):
\begin{equation}
 d\sigma = \frac{|F(\theta)|^2}{r^2} r^2 d\Omega \qquad \Rightarrow
 \qquad \frac{d\sigma}{d\Omega} = |F(\theta)|^2 \label{eq:difs}
\end{equation}
Notice that the magnitude of the momentum $k$ has been given the same
value both before and after the collision. This implies the scattering
is elastic, either off a very heavy particle or in the center of mass
frame. But none the less, this cross section corresponds to elastic
scattering.

The total elastic 
cross section can be derived by integrating out equation \ref{eq:difs}.
This is done by realizing that the Legendre polynomials are orthogonal
to each other:
\begin{equation}
  \int_0^{2\pi}
  \int^1_{-1} P_\ell(\cos \theta) 
              P_{\ell'}(\cos \theta) d\phi d\cos\theta =
  \frac{4\pi \delta_{\ell \ell'}}{2\ell +1}
\end{equation}
Finally, this gives:
\begin{equation}
  \sigma_{\text{elastic}} = 
         \frac{\pi}{k^2} \sum_\ell (2\ell+1) 
    (1+\eta_\ell^2 - 2\eta_\ell \cos 2\delta_\ell(k))
\end{equation}
If $\eta_\ell=1$ then all the particles are scattered elastically and
there is no absorption:
\begin{equation}
  \sigma_{\text{elastic}} = 
         \frac{4\pi}{k^2} \sum_\ell (2\ell+1) sin^2\delta_\ell
\end{equation}
Further, for the case of no absorption
a couple of points need to be made, if $\delta_\ell
=0$ then that particular partial wave does not contribute to the
scattering process. The second point, is that the 
maximum value for any partial wave occurs when $\delta_\ell=\pi/2$.

\section{The Total Cross section}
To determine the total cross section, the inelastic part has to be added to
the elastic part of the cross section. The inelastic part can be derived
by looking at the inward radial flux in equation \ref{eq:scat}
($\psi_{\text{scat}}$) and comparing it to the outward radial flux.
The inward radial flux is given by:
\begin{equation}
  \mathcal{F}_{\text{in}} = \int \vec{j}\cdot r^2 d\Omega =
    \frac{k}{m}\frac{4\pi}{4k^2} \sum_\ell (2\ell+1)
\end{equation}
while the outward flux is given by:
\begin{equation}
  \mathcal{F}_{\text{out}} = \int \vec{j}\cdot r^2 d\Omega =
    \frac{k}{m}\frac{4 \eta_\ell^2 \pi}{4k^2} \sum_\ell (2\ell+1)
\end{equation}
The net 
loss in flux is the difference between the inward and outward
flux and the total inelastic 
cross section is the difference divided by the
incident flux:
\begin{equation}
 \sigma_{\text{inel}} = \frac{\pi}{k^2}\sum_\ell (2\ell +1)(1-\eta^2_\ell)
\end{equation}
Given the elastic and inelastic cross sections, the total cross section
is:
\begin{equation}
 \sigma_{\text{total}} = \sigma_{\text{el}} + \sigma_{\text{inel}}=
   \frac{2\pi}{k^2}\sum_\ell (2\ell + 1) (1-\eta_\ell \cos 2 \delta_\ell)
\end{equation}
Comparing this equation to the scattering amplitude (note that this
is related to the elastic cross section), it is found that
\begin{equation}
  \sigma_{\text{total}} = \frac{4\pi}{k} \text{Im}F(0)
\end{equation}
This is referred to as the optical theorem.
\subsection{Maximum Cross Sections}
From the expressions given above, it is obvious that there are maximum
values that the cross sections can achieve. For the case of no absorption,
$\eta_\ell=1$, the scattering is completely elastic. In this case the
maximum value of the total cross section is
\begin{equation}
  \sigma_{\text{total}}=\sigma_{\text{el}} = \frac{4\pi}{k^2}\sum_\ell
         (2\ell+1) \label{eq:elmax}
\end{equation}
Notice that for the cases of electron-electron and electron-positron
elastic
scattering, the cross sections based on dimensional analysis are
$\sigma \propto 
\alpha_{\text{em}}^2/s$, which is consistent with equation
\ref{eq:elmax}. Also, the scattering is dominated by single photon
exchange, corresponding to $\ell=1$. 
For the scattering of $\nu$ off leptons or quarks, the total
cross section depends linearly on the energy (lab frame)
$\sigma \propto E$. This expression will eventually exceed the limit
given in equation \ref{eq:totmax}.

For the case where the absorption is a maximum $\eta_\ell=0$, the
maximum value of the cross section is
\begin{equation}
 \sigma_{\text{total}}= \frac{2\pi}{k^2} \sum_\ell(2\ell+1) 
 \label{eq:totmax}
\end{equation}
which is simply equal to twice the elastic cross section, therefore
the elastic and inelastic cross sections are equal.
%
\section{Experimental Determination of the Total Cross Section}
To determine the total cross section, all that is needed is to determine
the imaginary part of the forward scattering amplitude. Several problems
exist with this, first it is almost impossible to measure what part of
the particles going forward are scattered an which are beam going straight
through. The second problem is that cross sections are measured and not
amplitudes, so there is no simple method to disentangle the real from the
imaginary parts of the scattering amplitude.

To actually perform such a measurement, the total rate for particles
colliding can be determined. Since this is the imaginary part of the
amplitude, the  real part can be determined by extrapolating the
elastic differential cross section to $\theta=0$. But measuring the total
rate is very difficult experimentally. The method that is used, is a
combination of total rate measurements and measurement of the elastic
differential cross section. In the case of $pp$ and p$\bar{p}$ (in fact
this is true for all hadron-hadron scattering), the differential cross 
section has the functional form of
\begin{equation}
  \frac{d\sigma}{dt} = A e^{-bt}
\end{equation}
where $t=(P_1 - P_3)^2$, $A$ is a normalization factor and $b$ is an
experimentally derived parameter. For very small values of $t$ ($\theta$),
Coulomb scattering starts to dominate over hadronic scattering, in the
region where the two are about equal the two amplitudes interfere with
each other. Since the Coulomb amplitude is known from theory:
\begin{equation}
   f_c(t) = 2\alpha \frac{G^2(t)}{t} e^{i\alpha \phi(t)}
\end{equation}
where $\alpha$ and $G(t)$ come from experiment and $\phi(t)$ is derived
from theory. The nuclear amplitude is written as
\begin{equation}
  f_n(t) = \frac{\sigma_{\text{total}}}{4\pi}(\rho +i)e^{-bt/2}
\end{equation}
where $\rho$ is the ratio of real to imaginary parts of the scattering
amplitude.
By measuring differential cross section into the region where Coulomb
scattering dominates and out to the region where nuclear scattering
dominates, the value of $\rho$ and $\sigma_{\text{total}}$
can be fit, giving the total cross section and ratio of real to imaginary
parts of the amplitude.
\end{document}