%         Created on  11-JANUARY-1996 by Phillip Gutierrez
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\begin{center}
\LARGE\bf Physics \course \\ Lecture \lect \\
\end{center}
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\section{Introduction}
In the previous few lectures, the tools needed to describe a scattering
process of relativistic charged particles has been developed. In this
lecture all the pieces are put together to give the general formula
for the cross section of two charged particles scattering
through the interaction
of an electromagnetic field. In addition, the Feynman rules for the
scattering of spinless charged particles through the
electromagnetic interaction will be
given. The Feynman rules that are derived, are general and can be used
to describe any scattering process and decay process involving photons
and spinless particles.
%%%%%%%%%%%%%%%%%%
\section{Spinless Electron Scattering}
Given that the Klein-Gordon equation has solutions that are similar to those
of the 
Schr\"odinger equation, the transition amplitude $T_{fi}$ for relativistic
particles is the same as for the non-relativistic case. 
The transition amplitude for the non-relativistic case was found to be:
\begin{equation}
  T_{fi} = -i \int^{T/2}_{-T/2} e^{-i (E_i - E_f)t} dt 
       \int_V \phi_f^*(\vec{r}) V(\vec{r},t) \phi_i(\vec{r}) d^3x
\end{equation}
What needs to be done for the relativistic case, 
is to find an appropriate potential to 
substitute in this equation.
Notice that it is being assumed that this equation is valid in 
the relativistic 
case.

First rewrite the Klein-Gordon equation in covariant form:
\begin{equation}
  \left( \partial^\mu \partial_\mu + m^2\right) \phi = 0
\end{equation}
where $\partial^\mu=\left(\partial/\partial t,-\vec{\nabla}\right)$
and $\partial_\mu=\left(\partial/\partial t,\vec{\nabla}\right)$
and like up down indices are summed over. 
Since the interest is in calculating electromagnetic 
interactions, the electromagnetic field must be introduced 
into the problem. This is done by using the minimal substitution
derived in the previous lecture
$mu^\mu\rightarrow p^\mu+eA^\mu$.
Next apply the quantum condition, by converting the canonical momentum
to an operator as $\left(\partial^\mu \rightarrow \partial^\mu
- i e A^\mu\right)$, and substituting into 
the Klein-Gordon equation:
\begin{equation}
  \left( \partial^\mu \partial_\mu + m^2\right) \phi = -V \phi, \qquad
   V= -ie \left(\partial_\mu A^\mu + A^\mu \partial_\mu\right) - e^2 A^2
   \label{eq:pot}
\end{equation}
where the charge of the electron ($q=-e$) has been selected. Note also, that
the sign of $V$ was selected so that the non-relativistic approximation
of the Klein-Gordon equation gives the \Sch equation.

The potential given in equation \ref{eq:pot} is characterized by the
charge $e$; notice that in the early discussions of the Feynman diagrams,
the amplitude was characterized by $e^2$ meaning that there is sill an
addition charge to be found.
As a first approximation to finding the transition amplitude, 
the term with the $e^2$ is ignored. The potential above is substituted 
into the formula for the transition amplitude
where it has been rewritten in covariant form:
\begin{equation}
  T_{fi} =i \int ie\phi^*_f(\vec{r},t) \left(\partial_\mu A^\mu 
     + A^\mu \partial_\mu \right) \phi_i(\vec{r},t) d^4x
\end{equation}
Integrating the first term by parts gives:
\begin{equation}
 T_{fi} =i\int ieA^\mu \left[\phi^*_f(\vec{r},t) \partial_\mu \phi_i(\vec{r},t)
  - \phi_i(\vec{r},t) \partial_\mu \phi^*_f(\vec{r},t) \right] d^4x
\end{equation}
This can be simplified by noting that the probability density and current can 
be written in covariant form as:
\begin{align}
  &\rho = iq \left(\phi^* \frac{\partial^2 \phi}{\partial t^2} -
         \phi \frac{\partial^2 \phi^*}{\partial t^2}\right), \qquad
  \vec{j}=-i q\left(\phi^* \vec{\nabla} \phi - \phi \vec{\nabla} \phi^* 
      \right) \notag \\
  &\Rightarrow j^\mu = i q\left(\phi^* \partial^\mu \phi - 
          \phi \partial^\mu \phi^*\right)
\end{align}
Therefore the transition amplitude can be written in a simplified form as:
\begin{equation}
  T_{fi} = -i \int A^\mu j^{fi}_\mu d^4x \label{eq:trans}
\end{equation}
For a free particle the current density can be written out in a 
simple form. Start with the wave functions for the incoming and 
outgoing particles:
\begin{equation}
  \phi_i = N_i e^{i p^\mu_i x_\mu}, \qquad \phi_f = N_f 
     e^{i p^\mu_f x_\mu}
\end{equation}
Substituting these into the expression for the current density gives:
\begin{equation}
 j^\mu = -e N_i N_f \left(p_i + p_f\right)_\mu e^{i \left(p_f - p_i
  \right)_\mu
   x^\mu} \label{eq:jcur}
\end{equation}

Finally this can be looked at pictorially as a current coming in 
with a fixed momentum interacting with the field and leaving with a 
different momentum.
\begin{figure}[t]
\begin{center}
  \epsfig{file=em_vert.ps}
  \caption{This gives a graphical representation of equation
     \ref{eq:trans}.}
\end{center}
\end{figure}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{The Electromagnetic Field}
Before the transition amplitude 
can be calculated, the form of $A^\mu$ needs to 
be determined. The quickest way to finding what its general form should 
be, is to draw the Feynman diagram for this reaction. As shown below, 
it corresponds to two currents interacting through the exchange of a 
single photon. Therefore $A^\mu$ should be proportional to the current 
of one of 
the two particles times some factor describing the photon. The 
factor describing 
the photon should be a function of the 4-momentum transferred between the 
two particles.
\begin{figure}
\begin{center}
  \epsfig{file=em_vert_2.ps}
  \caption{Feynman diagram depicting a two to two process}
\end{center}
\end{figure}

To arrive at the analytic form of $A^\mu$ start with the following equation:
\begin{equation}
  \partial^\mu \partial_\mu A^\nu=j^\nu_2
\end{equation}
where $j^\nu_2$ is the current that generates the field $A^\nu$.
The form of the current has already given in equation \ref{eq:jcur} as:
\begin{equation}
  j_2^\mu = -e N_B N_D (p_B+p_D)^\mu e^{i(p_D-p_B)^\nu x_\nu}
\end{equation}
where the charge of the electron is used ($q=-e$).
For the equation defining the potential to be valid, it must have the form:
\begin{equation}
  A^\mu = -\frac{1}{q^2} j_2^\mu, \qquad q^2 = (p_D-p_B)^\mu (p_D-p_B)_\mu
\end{equation}
which is as expected. So finally the transition amplitude is given by:
\begin{equation}
  T_{fi}=-i \int j_\mu^1 \left(-\frac{1}{q^2}\right) j_2^\mu d^4x =
  -i N_A N_B N_C N_D (2 \pi)^4 \delta^4(p_D+p_C-p_A-p_B) \mathcal{M}_{fi}
\end{equation}
where the delta function comes from the integration and 
$\mathcal{M}_{fi}$ the invariant 
amplitude is defined as:
\begin{equation}
  -i\mathcal{M}_{fi}=\left(ie (p_A+p_C)^\mu \right) \left(-i 
       \frac{g_{\mu \nu}}{q^2}\right)
       \left(ie (p_B+p_D)^\nu \right)
\end{equation}
%
\subsection{The Feynman Rules}
The invariant amplitude derived above can be used to arrive at rules for 
calculating Feynman diagrams. These rules can then be used whenever a similar 
type of line or vertex occurs. Each external line corresponding to a spin zero 
particle will have a 1 associated with it; in reality an $N_i$, but
the normalization will cancel out. Each vertex between a spin zero 
particle and a photon will have a factor $i e (p_A+p_B)^\mu$, 
while each internal photon will 
have a factor $-ig^{\mu \nu}/q^2$; make sure to include the correct momenta,
in each of the factors.
%
\section{The Transition Rate}
Before the cross section can be calculated, the normalization 
factors $N$ need to be fixed. First of all, recall that the probability 
density of particles described by the wave function is given 
by $\rho=2(-e) E |N|^2$.
Further, recall that the factor $E$ was needed to compensate for the 
Lorentz contraction of the volume $d^3x$ to keep the number of 
particles, $\rho d^3x$, 
constant. Therefore, the normalization is set such that in a 
volume $V$ there are $2E$ particles (this is makes sense, since dividing
through by the charge leaves number of particles), this 
implies that $N=1/\sqrt{V}$.

Keeping in mind that the cross section is related to the square of 
the transition amplitude and that the square of the transition amplitude 
involves the square of a delta function which is infinite, the quantity of 
interest is the transition amplitude per unit volume per unit time; 
in this case the delta 
function is four dimensional and requires dividing by the time and volume 
to be finite. The transition amplitude per unit volume per unit time
is given by:
\begin{equation}
  W_{fi}=\frac{|T_{fi}|^2}{T V} = (2\pi)^4
    \frac{\delta^4(p_C+p_D-p_A-p_B)}{V^4} |\mathcal{M}_{fi}|^2
\end{equation}
%
\section{The Cross Section}
Finally, the cross section can be calculated. It is related 
to the transition rate per unit volume through the density of 
final states and the incident flux:
\begin{equation}
  \text{cross section} = \frac{W_{fi}}{\text{inital flux}}
     (\text{number of final states})
\end{equation}
For a single particle, the number of final states in 
a volume $V$ and momenta in element $d^3p$ is given by:
\begin{equation}
  \text{Number of final states} =\frac{V d^3p}{(2\pi)^3}
\end{equation}
where periodic boundary conditions are used---the allowed states in
one dimension
are given by $L p_x=2\pi n$ with $n$ an integer and if confined to
a momentum range $p_x \rightarrow p_x+dp_x$ the allowed number of states is
$Ldp_x/(2\pi)$ or in three dimensions $Vd^3p/(2\pi)^3$.
Since there are $2E$ particles, the number of final states per particle is:
\begin{equation}
  \text{Number of final states/particle}=\frac{V d^3p}{2 E (2\pi)^3}
\end{equation}
Finally, given that there are two distinct particles in the final state, 
the density of final states is given by:
\begin{equation}
  \text{Number of avalible final states}=\frac{V d^3p}{2 E_C (2\pi)^3}
      \frac{V d^3p}{2 E_D (2\pi)^3} 
\end{equation}

Next the initial flux is calculated. This includes the number of 
particles incident on a given target. This is most easily calculated 
in the frame where the target is at rest and the beam has a 
velocity $\vec{v}_A$.
The number of particles crossing a unit area per unit time is given 
by $2E_A|\vec{v}_A|/V$. 
The number of target particles per unit volume is $2E_B/V$. Therefore 
the initial flux is:
\begin{equation}
  \text{Initial flux} = |\vec{v}_A| \frac{2E_A}{V} \frac{2E_B}{V}
\end{equation} 
Combining the transition rate per unit volume with the density 
of initial and final states gives the differential cross section:
\begin{align}
  d\sigma &= \frac{V^2}{2E_A 2 E_B |\vec{v}_A|} \frac{(2\pi)^4}{V^4}
     \delta^4(p_C+p_D-p_A-p_B) |\mathcal{M}_{fi}|^2 \frac{V^2}{(2\pi)^6}
     \frac{d^3p_C}{2E_C} \frac{d^3p_D}{2 E_D} \notag \\
   \Rightarrow
  d\sigma &= \frac{(2\pi)^4}{2E_A 2 E_B |\vec{v}_A|}
     \delta^4(p_C+p_D-p_A-p_B) |\mathcal{M}_{fi}|^2
     \frac{d^3p_C}{(2\pi)^3 2E_C} \frac{d^3p_D}{(2\pi)^3 2 E_D}
\end{align}
Symbolically the cross section can be written as:
\begin{equation}
  d\sigma = \frac{|\mathcal{M}_{fi}|^2}{F} dQ
\end{equation}
Where the factors are identified as, the Lorentz invariant phase space factor:
\begin{equation}
  dQ=(2\pi)^4 \delta^4(p_C+p_D-p_A-p_B)
    \frac{d^3p_C}{(2\pi)^3 2E_C} \frac{d^3p_D}{(2\pi)^3 2 E_D}
\end{equation}
the incident flux (in the laboratory frame):
\begin{equation}
  F=|\vec{v}_A| 2E_A 2E_B
\end{equation}
and the invariant amplitude $|\mathcal{M}_{fi}|^2$ which contains the 
dynamics of the problem.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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