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\title{Making sense of field-theory Lagrangians for $\eta$ and $\eta^{\prime}$ photoproduction}

\author{M. Dugger}
\affiliation{Arizona State University, Tempe, AZ  85287-1504}

\begin{abstract}
Interaction Lagrangians for $\eta$ and $\eta^{\prime}$ photoproduction are discussed.
\end{abstract}

\maketitle
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Introduction}

The purpose of this document is to help the student of nuclear/particle physics
obtain an intuitive feel for the terms seen in Lagrange densities for 
$\eta$ and $\eta^{\prime}$ photoproduction.
For this reason, this document is designed to only show the most relevant 
aspects of field theory. Along the way I will show how to obtain a non-relativistic potential
from a relativistic Lagrange density. 
Once the connection from relativistic Lagrange densities
to non-relativistic potentials is made, we will see how the 
nucleon-nucleon potential can be related to a single pion exchange model. 
We will then piece together interaction terms commonly found in
Lagrange densities for $\eta$ and $\eta^{\prime}$ photoproduction.

%Since it is impracticle to give a thourough treatment of field theory
%in a short document 

\section{Review of Lagrangians}

\subsection{The classical discrete Lagrangian}

It is assumed that the concept of generalized coordinates are understood.
Also, in all that follows, generalized forces are assumed to be
are derivable from potentials. This allows us to write the Lagrangian as
\begin{eqnarray}
L = T - V, \nonumber
\end{eqnarray}
where $T$ is the kinetic energy and $V$ is the potential energy.
It is important to note that
the Lagrangian is a function of the generalized coordinate $q$, and the
time derivative of the generalized coordinate, $\dot{q}$. To make 
this statement explicit 
the Lagrangian is often written as $L(q, \dot{q})$. 

The equations of motion are derived using the Euler-Lagrange equation:
\begin{eqnarray}
\frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q}_k} \right) - \frac{\partial L}{\partial q_k} = 0 . \nonumber
\end{eqnarray}
The Euler-Lagrange equation may be derived by taking the extremum $\delta S = 0$, where $S = \int L dt$.
The generalized momentum $p_k$ is given by
\begin{eqnarray}
p_k \equiv \frac{\partial L}{\partial \dot{q}}. \label{eq:pClassical}
\end{eqnarray}
The Euler-Lagrange equation now implies that any coordinate $q_k$ that does not occur explicitly within $L$
has a corresponding momentum $p_k$ that is a constant of the motion.

For the very simple case where $T = \frac{1}{2} m \dot{q}^2$, and $V = m g q$, we get 
$L = \frac{1}{2} m \dot{q}^2 - m g q$, and the equation of motion becomes
\begin{eqnarray}
m \ddot{q} = m g. \nonumber
\end{eqnarray}
While the generalized momentum $p_q = \frac{\partial L}{\partial \dot{q}} = m \dot{q}$.
We have just found that the acceleration is equal to $g$, $p_q = m \dot{q}$, and that all momenta
not associated with the generalized coordinate $q$ are constant. 
%Told you this would be a very simple example!



\subsection{The relativistic field Lagrangian}
For a field, the Lagrangian is defined over the region of space occupied by the field.
For this reason, it is more convenient to work with the ``Lagrange density'' $\mathcal{L}$, than
with $L$ directly. This Lagrange density
is related to the Lagrangian in the following manner
\begin{eqnarray}
L = \int^{+ \infty}_{- \infty} d^3 x \mathcal{L} . \nonumber
\end{eqnarray}
In this case, the Euler-Lagrange equation is modified to read
\begin{eqnarray}
\partial^{\mu} \left( \frac{\partial \mathcal{L}}{\partial (\partial^{\mu} \phi)} \right) - 
\frac{\partial \mathcal{L}}{\partial \phi} = 0 , \nonumber
\end{eqnarray}
where $\partial^{\mu} \equiv \partial/\partial x^{\mu}$, $\phi \equiv \phi(x)$,
and as before, is constructed to minimize the action $S = \int L dt = \int \mathcal{L} d^4 x$.
Also, in all that follows, any upper index that
matches a lower index is to be summed (i.e. $A^{\mu} B_{\mu} \equiv \sum_{\mu} A^{\mu} B_{\mu}$). 
It is important to note that the Lagrangian has units of energy and must be a scalar.

After looking at the Euler-Lagrange equation, it might be tempting to write $\mathcal{L}(\phi, \partial^{\mu} \phi)$, but
that would be wrong.
For a field, the Lagrange density $\mathcal{L}$ is a function of the field amplitude $\phi$ and the 
time-derivative of that field $\dot{\phi}$. This is often written explicitly as
$\mathcal{L}(\phi, \dot{\phi})$.

The ``momentum density'' $\pi(x)$ of the field is given by
\begin{eqnarray}
\pi(x) \equiv \frac{\partial \mathcal{L}}{\partial \dot{\phi}(x)},  \nonumber
\end{eqnarray}
a direct analogy to the classical case shown in Eqn. \ref{eq:pClassical}.

It is important to note that the generalized coordinate is an amplitude $\phi$
evaluated at some point in space $x$, and is written $\phi(x)$. If we were
to analyze a string under tension as a one dimensional field, 
we could write the amplitude as a physical displacement. However, if we analyze 
the electromagnetic four-vector potential $A^{\mu}(\vec{x})$ as a field, we would have 
the amplitude of $A^{\mu}$ at some position $\vec{x}$ as the generalized
field coordinates. In the case of 
the $A^{\mu}(\vec{x})$ field, there is no physical coordinate displacement associated
with the generalized coordinate. We could, however, if so desired,  find the electric and 
magnetic fields associated with $A^{\mu}(\vec{x})$ through the relations:
\begin{eqnarray}
\vec{E} & = & -\vec{\nabla} A^0 - \frac{\partial \vec{A}}{\partial t} ,  \nonumber \\
\vec{B} & = & \vec{\nabla} \times \vec{A} \nonumber .
\end{eqnarray}

For our purposes, we wish to look at the scalar field $\phi(x)$ that has the free Lagrangian
density $\mathcal{L}_0$
\begin{eqnarray}
\mathcal{L}_0 = \frac{1}{2} [ (\partial^{\mu} \phi)^2 - m^2 \phi^{2} ], \nonumber
\end{eqnarray}
where for any four-vector quantity $(Z^{\mu})^2 \equiv Z^{\mu} Z_{\mu}$.
After solving the Euler-Lagrange equation, we find the equation of motion
\begin{eqnarray}
(\partial^{\mu} \partial_{\mu} + m^2) \phi = 0. \nonumber
\end{eqnarray}
This is simply the Klein-Gordon equation. This means that our free Lagrangian
density $\mathcal{L}_0$ describes a relativistic spin-zero field. 
(Note: Throughout this document, the units are taken such that $\hbar = c = 1$.)

Sometimes, the determination of the equation of motion is described as first
quantization. However, since we have not quantized anything yet, this is somewhat
a misnomer. To quantize the field, we need to impose a quantization
scheme. The confusion regarding first quantization comes from the fact that
the Klein-Gordon equation applied to a particle is, in fact, quantum mechanics.
Yet, if the Klein-Gordon equation is applied to a field, we are still doing
classical relativistic physics. 
%The easiest example to see why this should 
%be true is the free classical electromagnetic four-vector potential field $A^{\mu}(\vec{x})$.
%We could derive $A^{\mu}(\vec{x})$ by purely classical means. In fact, there are
%different ways to derive $A^{\mu}(\vec{x})$, with one of these methods
%being the use of classical relativistic Lagrangian dynamics. 
%When we define a field as the starting point in an analysis, it is not
%surprising at all to find that the resulting field has wave-like properties.
%It is expected. We start with a field and end up with a field. For quantum
%mechanics of particles, we start with particles and end up with 
%wave equations. 
%Anyway, I only mention this because some people like the
%terms ``first-'' and ``second-quantization'', and other do not.

\subsection{Quantizing the field}
The position-momentum commutators for the quantum mechanics of point particles are as follows:
\begin{eqnarray}
[x,p_x]  & = & i  \nonumber \\
\!\![x,y] & = & [p_x,p_y] = 0. \nonumber
\end{eqnarray}
For quantum fields it seems reasonable that similar relations should hold true.
However, since the field is located over a region, we expect that the 
commutation relations for fields will take this into account.
 
One canonical quantization scheme is to force the commutation relations
between the generalized coordinate and the generalized momentum-density to be
\begin{eqnarray}
[\phi(\vec{x},t_0),\pi(\vec{y},t_0)]  & = & i \delta^{(3)}(\vec{x} - \vec{y}), \nonumber \\
\:[\phi(\vec{x},t_0),\phi(\vec{y},t_0)] & = & [\pi(\vec{x},t_0),\pi(\vec{y},t_0)] = 0, \nonumber 
\end{eqnarray}
where it is important to note that each of these field operators are taken to be at the same time $t_0$.
These are commonly referred to as ``equal time commutation relations''. If the time of the fields
are not identical, then these commutation relations do not hold.

After doing a bunch of math we would end up with (for the Klein-Gordon field):
\begin{eqnarray}
\phi(\vec{x}) & = & \int \frac{d^3 p}{(2 \pi)^3\sqrt{2 E}}
\left(a_{\vec{p}}\: e^{-i p^{\mu} x_{\mu}} + a^{\dagger}_{\vec{p}}\: e^{i p^{\mu} x_{\mu}}\right) , \nonumber 
\end{eqnarray}
where
\begin{eqnarray}
\: [a_{\vec{p}},a^{\dagger}_{\vec{p}\:'}] & = & (2 \pi)^3 \delta^{(3)}(\vec{p} - \vec{p}\:') \nonumber .
\end{eqnarray}

We have fields $\phi(\vec{x})$ with creation and annihilation operators. This means that these fields do NOT
represent quantum-mechanical states. These fields, instead, are themselves operators.
This is one of the key differences between regular quantum mechanics of point particles, 
and quantum field theory. When you solve the Klein-Gordon equation for a point particle, you
end up with a state vector. However, when you solve (and quantize) the Klein-Gordon equation 
in field theory, you end up with an operator. 

What does this field operator do?
The field operator $\phi(\vec{x})$ will create or annihilate a particle at position
$\vec{x}$. That is pretty much it.

In particular, if a state of momentum $\vec{p}$ is defined so that
\begin{eqnarray}
%\frac{|\vec{p}\rangle}{ \sqrt{2 E_{\vec{p}}}}  = a^{\dagger}_{\vec{p}} | 0 \rangle , \nonumber
|\vec{p}\rangle =  \sqrt{2 E_{\vec{p}}}\; a^{\dagger}_{\vec{p}} | 0 \rangle , \nonumber
\end{eqnarray}
then
%\begin{eqnarray}
%\langle \vec{p} | \vec{q} \rangle = 2 E_{\vec{p}} (2 \pi)^3 \delta^{(3)}(\vec{p} - \vec{q}), \nonumber
%\end{eqnarray}
%and
\begin{eqnarray}
\langle \vec{p} | & = & \langle 0 |  a_{\vec{p}} \sqrt{2 E_{\vec{p}}}  , \nonumber \\ 
a_{\vec{p}\:'} |\vec{p}\rangle & = & (2 \pi)^3 \delta^{(3)}(\vec{p} - \vec{p}\:') \sqrt{2 E_{\vec{p}}}\; | 0 \rangle , \nonumber \\
\langle \vec{p} | a^{\dagger}_{\vec{p}\:'}  & = & \langle 0 |(2 \pi)^3 \delta^{(3)}(\vec{p} - \vec{p}\:') \sqrt{2 E_{\vec{p}}}, \nonumber \\
a_{\vec{p}} | 0 \rangle & = & \langle 0 | a^{\dagger}_{\vec{p}} \; = \; 0 \nonumber
\end{eqnarray}
and
\begin{eqnarray}
\phi(\vec{x}) | 0 \rangle = \int \frac{d^3 p}{(2 \pi)^3 2 E} e^{- i p^{\mu} x_{\mu}} | \vec{p}\rangle \nonumber
\end{eqnarray}
and
\begin{eqnarray}
\langle 0 |\phi(\vec{x}) | \vec{p} \rangle = e^{i p^{\mu} x_{\mu}} \nonumber .
\end{eqnarray}
%Don't worry too much about the factors of $\sqrt{ 2 E}$. 
%I am following the convention
%found in Peskin and Schroeder's book ``An Introduction to Quantum Field Theory''.
The important thing is to notice that the field operators create and annihilate states.

\section{The propagator}
So that we can make sense of what propagators are and how they are related to
the Lagrange density $\mathcal{L}$, I will show how the propagator is found using
two different methods. The first method will clearly show that the propagator
describes a state being created at some point in space-time and then annihilated
at some other space-time point. The second method will clearly show how the propagator
is related to the free Lagrange density $\mathcal{L}_0$.

\subsection{Propagator part 1}
In this method we will define the propagator $D(x_1 - x_2)$ as
\begin{eqnarray}
D(x_1 - x_2) = \langle 0 | \phi(\vec{x_2},t) \phi(\vec{x_1},0) | 0 \rangle \nonumber,
\end{eqnarray}
where, in this case, I have taken $t > 0$.

The propagator, as defined above, says that a state is created as position $x_1$ at time $t = 0$,
and is annihilated at position $x_2$ at time $t$. Working out the equation, we find that
the only term that can contribute contains 
$\langle 0 | a_{\vec{p}} \; a^{\dagger}_{\vec{q}} | 0 \rangle = (2 \pi)^3 \delta^{(3)} (\vec{p} - \vec{q})$
and leads to
\begin{eqnarray}
D(x_1 - x_2) = \int \frac{d^3 p}{(2 \pi)^3 2 E} e^{-i p \cdot (x_1 - x_2)} , \label{eq:prop1} 
\end{eqnarray}
where the dot product of two four-vectors $A^{\mu}$, $B_{\mu}$ is defined such that $A \cdot B = A^{\mu} B_{\mu}$.

We are now ready to see our very first Feynman diagram in Fig. \ref{fig:f1}! We can see that
the figure makes a lot of sense. The figure says that a field propagates from point $x_1$ to
point $x_2$. We can even associate an equation with this propagation.

\begin{figure}
\includegraphics[scale=0.45]{f1.eps}
\caption{ Feynman diagram for a propagator from point $x_1$ to point $x_2$.
\label{fig:f1}}
\end{figure}

\subsection{Propagator part 2}

Another way to define the propagator for our scalar field theory 
is by taking the equation of motion and ``inverting it''.
The equation of motion for our scalar field theory is the Klein-Gordon equation:
\begin{eqnarray}
(\partial^{\mu} \partial_{\mu} + m^2) \phi = 0. \nonumber
\end{eqnarray}
If I were to write this as $A \phi = 0$, I could define $A^{-1}$ as such:
\begin{eqnarray}
A A^{-1} = \delta^{(4)}(x_1 - x_2). \nonumber
\end{eqnarray}
That is what I mean by inverting the equation of motion. For the field of interest we define
the propagator $D(x_1 - x_2)$ through the relation
\begin{eqnarray}
-i (\partial^2 + m^2) D(x_1 - x_2) = \delta^{(4)}(x_1 - x_2). \nonumber
\end{eqnarray}

To find $D(x_1 - x_2)$ it is helpful to remember that the $\delta$-function can be written
\begin{eqnarray}
\delta^{(4)}(x_1 - x_2) = \int \frac{d^4 k}{(2 \pi)^4} e^{i k \cdot (x_1 - x_2)} ,\label{eq:deltaFunction}
\end{eqnarray}
and that
\begin{eqnarray}
(\partial^2 + m^2) e^{i k \cdot (x_1 - x_2)} = 
(-k^{\mu}k_{\mu} + m^2) e^{i k \cdot (x_1 - x_2)} \nonumber.
\end{eqnarray}
By inspection we can now write the propagator as
\begin{eqnarray}
D(x_1 - x_2) = i \int \frac{d^4 k}{(2 \pi)^4} \frac{e^{i k \cdot (x_1 - x_2)}}{k^2 - m^2 + i \epsilon}, \label{eq:prop2} 
\end{eqnarray}
where $\epsilon$ represent an infinitesimal shift in the imaginary part of the denominator and is a technical
detail we will not worry about.
But wait! This does not look like our previous result. How is this propagator in Eqn. \ref{eq:prop2} the same as the one
in Eqn. \ref{eq:prop1}?

To see that Eqn. \ref{eq:prop2} might represent Eqn. \ref{eq:prop1}, we should notice that Eqn. \ref{eq:prop1}
does not have the zero component of $k$ integrated over. Perhaps, if we integrate Eqn. \ref{eq:prop2} over
the variable $k^0$ we will obtain Eqn. \ref{eq:prop1}. To do this integration requires that we use the method
of contours of complex analysis. We don't want to get bogged down in too much math, so I'll just state the result for the case when $t_2 > t_1$
\begin{eqnarray}
D(x_1 - x_2) = \int \frac{d^3 p}{(2 \pi)^3 2 E} e^{-i p \cdot (x_1 - x_2)} \nonumber ,
\end{eqnarray}
which is precisely Eqn. \ref{eq:prop1}.

Another issue that might give you pause, is that the integrand of the propagator given in 
Eqn. \ref{eq:prop2} is infinite when $k^2 = m^2$! Isn't the invariant quantity for 
a four-momentum the mass? That is true, however, this four-momentum is said to be off the 
``mass shell''. In classical quantum mechanics of point particles, the energy is not
always conserved. For quantum field theory, the 4-momentum is conserved (working with plane
waves $\Rightarrow \Delta p^{\mu} = 0$ and $\Delta x \rightarrow \infty$ ),
but the mass of the particles will vary from that of ``free particles''. This variation
of the mass is called being off the mass shell. 

In the field theory that we are interested in, we are more concerned with the form of the 
propagator 
given in Eqn. \ref{eq:prop2}. Note: it is possible for the propagator to be defined such that
the factor of $i$ is removed from Eqn. \ref{eq:prop2}. I have elected to keep the factor of $i$ so
that our two methods of determining the propagator are consistent.
%Moreover, we really don't even want that equation in terms
%of space-time. Instead, we want the 
%fourier transform $\tilde{D}(x_1 - x_2)$ of Eqn. \ref{eq:prop2} given simply as
%\begin{eqnarray}
%\tilde{D}(k) = \frac{i}{k^2 - m^2 + i \epsilon} . \nonumber
%\end{eqnarray}
%This is the factor that we would associate with Fig. \ref{fig:f1} when trying to determine
%a quantum mechanical amplitude using the Feynman rules.
Also, it is useful to note that the Fourier transform of Eqn. \ref{eq:prop2} to momentum-space
can be written as
\begin{eqnarray}
\tilde{D}(k) =  \frac{i}{k^2 - m^2 + i \epsilon}.
\nonumber
\end{eqnarray}

 

I want you to notice that we only needed the free Lagrange density $\mathcal{L}_0$
to determine the propagator. When you see a Lagrange density  $\mathcal{L}$, you should mentally separate
the free part $\mathcal{L}_0$ from the interaction part $\mathcal{L}_{int}$, 
so that $\mathcal{L} = \mathcal{L}_0 + \mathcal{L}_{int}$, and notice
that the free part $\mathcal{L}_0$ determines the field propagators.

%Aside: I want to make something clear regarding terminology. Sometimes you will here me
%refer to the free Lagrange density $\mathcal{L}_0$ as being the kinetic part 
%of the Lagrangian. This is a misnomer. The kinetic part of $\mathcal{L}$
%for our simple theory is $(\partial \phi)^2/2$ and not $[(\partial \phi)^2 - m^2 \phi^2]/2$.
%The mass term is more accurately associated with potential energy. So, please
%excuse me if I say that  $\mathcal{L}_0$ is the kinetic part of the 
%Lagrange density. Some habits are hard to break.

Now that we have associated the propagator with the free part of $\mathcal{L}$, it
is time to ask about the interaction part of the Lagrange density.

\section{The vertex}

I will describe a first order process that has as its Lagrange density
\begin{eqnarray}
\mathcal{L} = 
& \frac{1}{2} [ (\partial \phi_a(\vec{x}))^2 - m_a^2 \phi_a^2(\vec{x})] + & \nonumber\\
& \frac{1}{2} [ (\partial \phi_b(\vec{x}))^2 - m_b^2 \phi_b^2(\vec{x})] + &\nonumber\\
& \frac{1}{2} [ (\partial \phi_c(\vec{x}))^2 - m_c^2 \phi_c^2(\vec{x})] - &
g  \phi_a(\vec{x}) \phi_b(\vec{x}) \phi_c(\vec{x}) \nonumber .
\end{eqnarray}
%You might be saying to yourself: This equation is too long! If I know that
%the fields are scalar (Klein-Gordon) then the $\mathcal{L}_0$ part is not even needed.
%We could get by just as well by specifying the interaction term and stating
%that the fields $\phi_a \phi_b \phi_c$ are scalar. You would be right in 
%making such a statement. This is why, in many cases, the free part
In many cases, the free part
is not even written. It is easier just to write
\begin{eqnarray}
\mathcal{L}_{int} =
- g  \phi_a \phi_b \phi_c \nonumber .
\end{eqnarray}
You will also often encounter the case where the subscript $int$ is removed, 
and see things like
\begin{eqnarray}
\mathcal{L} =
- g  \phi_a \phi_b \phi_c \nonumber ,
\end{eqnarray}
to describe a Lagrange density. 
%So, if you see a theorist write down a field that
%only contains interaction parts, it is assumed that the free part
%has been conveniently suppressed to save on writing a bunch of free field Lagrangians.

To get an intuitive idea of what an interaction term in the Lagrange density tells us,
consider
three different fields $\phi_a(\vec{x})$, $\phi_b(\vec{x})$, and $\phi_c(\vec{x})$, 
representing three different particles (denoted as $A$, $B$, and $C$) with the mass of $A$ greater
than the mass of $B + C$. Each of these fields have their own
$\mathcal{L}_0$, but have a common interaction term 
$\mathcal{L}_{int} = - g \phi_a(\vec{x}) \phi_b(\vec{x}) \phi_c(\vec{x})$.
Lets just naively find $\langle \vec{p_b}, \vec{p_c} |\: L_{int} | \vec{p_a} \rangle$:
\begin{eqnarray}
\langle \vec{p_b}, \vec{p_c} |\: L_{int} | \vec{p_a} \rangle = 
\langle \vec{p_b}, \vec{p_c} |\: \int d^4x \mathcal{L}_{int} | \vec{p_a} \rangle = \nonumber \\
-g \langle \vec{p_b}, \vec{p_c} |\:\int d^4x \phi_b(\vec{x}) \phi_c(\vec{x}) \phi_a(\vec{x}) \: | \vec{p_a} \rangle \nonumber
\end{eqnarray}
When we multiply the fields together the only term that will survive will have a factor of 
$b^{\dagger}_{\vec{q}} \; c^{\dagger}_{\vec{r}} \; a_{\vec{s}}$ . 
This will give us, after a bit of math, 
\begin{eqnarray}
& -g & \langle \vec{p_b}, \vec{p_c} |\:  \int d^4x \phi_b(\vec{x}) \phi_c(\vec{x}) \phi_a(\vec{x}) \: | \vec{p_a} \rangle \nonumber \\ 
= & -g & \delta^{(4)}(p_a - (p_b + p_c)) ,\label{eq:v1} 
\end{eqnarray}
where the $\delta$-function tells us to conserve momentum and energy at the vertex.

I will now try and make the same sort of statement in a different way. The method I am about
to describe will involve more math but better describes how to calculate something called 
the $\mathcal{M}$ amplitude that we associate with Feynman diagrams.

%A technical detail:
%I will define an amplitude for a processes (e.g. $A \rightarrow B + C$) as $Z_J$,
%and $Z_0$ for all process that contain field excitations that are not connected to external particles. 
%For what follows, we will be interested in $Z$ defined as $Z \equiv Z_J/Z_0$.

We will now draw our next Feynman diagram. This Feynman diagram can be found in Fig. \ref{fig:f2} and
says that particle $A$ decays into particles $B$ and $C$ at a vertex.

\begin{figure}
\includegraphics[scale=0.45]{f2.eps}
\caption{ Vertex at point $x$.
\label{fig:f2}}
\end{figure}

I will define an amplitude for a processes (e.g. $A \rightarrow B + C$) as $Z$.
To determine $Z$ we associate a factor of $- i g$ with the vertex, propagate all of the lines,
and integrate over the vertex position (remember we are dealing with plane waves $\Rightarrow$
vertex can be anywhere):
\begin{eqnarray}
Z  =  (- i g) \int d^4x D(a - x) D(x - b) D(x - c) = \nonumber \\
-ig \!\! \int\!\!\!\!\! \int\!\!\!\!\! \int\!\!\!\!\! \int\!\!
\frac{d^4k_a d^4k_b d^4k_c i^3 e^{-ix \cdot (k_a - k_b -k_c)} e^{i(k_a\cdot a - k_b \cdot b -k_c \cdot c)}}
{(2 \pi)^{12} (k_a^2 - m_a^2)(k_b^2 - m_b^2)(k_c^2 - m_c^2)
} d^4x , \nonumber 
\nonumber
\end{eqnarray}
where I have collected terms in the exponent common in $x$. When we integrate over the
$x$ variable and notice that this is just a definition of the $\delta$-function (see Eqn. \ref{eq:deltaFunction}), we obtain
\begin{eqnarray}
Z  = -i g (2 \pi)^4\delta^{(4)}(k_a - k_b - k_c) D(a)D(b)D(c). \nonumber
\end{eqnarray}

To obtain $ -i \mathcal{M}$ from $Z$, we just throw away the overall momentum-energy conserving
$\delta$-function (along with the $(2 \pi)^4$) and all propagators associated with external lines. 
For our simple example we get
\begin{eqnarray}
-i M  = -i g . \nonumber
\end{eqnarray}

We have just seen that it is natural to associate the coupling constant and momentum-energy conserving
$\delta$-function at an interaction vertex, with the interaction part of a Lagrange density $\mathcal{L}_{int}$
given as the product of fields, $\mathcal{L}_{int} = -g \phi_a(\vec{x}) \phi_b(\vec{x}) \phi_c(\vec{x})$.
Moreover, we have seen how the diagram is associated with something called the $\mathcal{M}$-amplitude, and that
the coupling constants of the fields gives a measure of the reaction strength within $\mathcal{M}$.


Now when you see a Lagrange density, you should think of the interaction terms as describing
several fields being created and annihilated at some point in space-time.

\subsection{Identical particles at the vertex}

In the previous subsection we had three different fields ($\phi_a$, $\phi_b$, $\phi_c$)
that interacted at a vertex. If, instead, we had one field with the
Lagrange density
\begin{eqnarray}
\mathcal{L} =
 \frac{1}{2} [ (\partial \phi)^2 - m^2 \phi^2] + g \phi^3 \nonumber ,
\end{eqnarray}
then, when we form
\begin{eqnarray}
\langle \vec{p_1}, \vec{p_2} |\: L_{int} | \vec{p_3} \rangle =
\langle \vec{p_1}, \vec{p_2} |\: \int d^4x \mathcal{L}_{int} | \vec{p_c} \rangle = \nonumber \\
-g \langle \vec{p_1}, \vec{p_2} |\: \int d^4x  \phi(\vec{x}) \phi(\vec{x}) \phi(\vec{x}) \: | \vec{p_3} \rangle, \nonumber
\end{eqnarray}
we would not be able to distinguish which field creates or annihilates any of the particular $p_1$, $p_2$, $p_3$ states.
We would have three ways to choose the first field, two ways to choose the second field, and one way
to choose the last field. For this reason, we obtain a factor of $3!$ when compared to Eqn. \ref{eq:v1}:
\begin{eqnarray}
& -g & \langle \vec{p_1}, \vec{p_2} |\:  \int d^4x \phi(\vec{x}) \phi(\vec{x}) \phi(\vec{x}) \: | \vec{p_3} \rangle \nonumber \\
  = & -g & 3! \: \delta^{(4)}(p_3 - (p_1 + p_2)) .\nonumber 
\end{eqnarray}
Many times you will see $\mathcal{L}_{int} = - g \phi^3/(3!)$, where the factor $1/(3!)$ is put in
so that we don't need to worry about pesky factors of $3!$ in the Feynman rules.

A more complete description of how $Z$ is obtained in terms of the propagators for the 
$\mathcal{L}_{int} = - g \phi^3/(3!)$ theory
can be found in the appendix.


\section{Second order interaction}
I will describe a second order process (each vertex represents an order,
see appendix) describing the elastic scattering 
of two scalar particles $A + B$ with the interaction Lagrange density given as
$\mathcal{L}_{int} = -g \phi_a \phi_b \phi_c$.
The process we will investigate is typically written
$A + B \rightarrow A + B$.

\begin{figure}
\includegraphics[scale=0.45]{f3.eps}
\caption{ Feynman diagram for $A + B \rightarrow A + B$. The initial and final
positions are labeled: $A_I$ (initial position for particle $A$); 
$B_I$ (initial position for particle $B$);
$A_F$ (final position for particle $A$), and
$B_F$ (final position for particle $B$).
\label{fig:f3}}
\end{figure}

The second-order Feynman diagram for $A + B \rightarrow A + B$ is shown
in Fig. \ref{fig:f3}, where I have labeled the various fields in terms of initial and 
final states.

As was done previously, I will associate a factor of $- i g$ with each vertex, propagate 
all of the lines, and integrate over the vertex positions to obtain:
\begin{eqnarray}
Z  =  (- i g)^2 \!\! \int\!\!\!  d^4x_1 \!\!\! \int \!\!\! 
d^4x_2 D_{a_I x_1} D_{x_1 b_F} D_{x_1 x_2} D_{b_I x_2} D_{x_2 a_F},  \nonumber \\
\end{eqnarray}
where I have defined $D(x - y) \equiv D_{x y}$ to save space.
Integrating over $x_1$ we get
\begin{eqnarray}
Z  =  (- i g)^2 \!\!  \int \!\!\! 
d^4x_2 D(a_I) D(b_F) D(x_2) D_{b_I x_2} D_{x_2 a_F} \times  \nonumber \\
(2 \pi)^4 \delta^{(4)}(p_{a_I} - p_{b_F} - p_c). \nonumber
\end{eqnarray}
Now integrating over $x_2$ to obtain 
\begin{eqnarray}
Z   =  (- i g)^2 \!\!  \int \!\!\!
\frac{d^4k}{(2 \pi)^4} \frac{i D(a_I) D(b_F) D(b_I) D(a_F)}
{k_c^2 - m_c^2} \times  \nonumber \\
(2 \pi)^4 \delta^{(4)}(p_{a_I} \!-\! p_{b_F} \!-\! p_c) (2 \pi)^4  
\delta^{(4)}(p_{b_I}\! -\! p_{a_F}\! -\! p_c).  \nonumber 
\end{eqnarray}
After performing the last integral:
\begin{eqnarray}
Z   =  -i g^2 
\frac{D(a_I) D(b_I) D(b_F) D(a_F)}
{(p_{a_I} - p_{b_F})^2 - m_c^2} \times  \nonumber \\
(2 \pi)^4 \delta^{(4)}(p_{a_I} \!+\! p_{b_I} \!-\! p_{a_F} \!-\! p_{b_F}) . \nonumber
\end{eqnarray}
Now, throw away the external field propagators and the factor $(2 \pi)^4 \delta$-function
and set to $-i \mathcal{M}$ to get
\begin{eqnarray}
\mathcal{M}   =  \frac{g^2}{(p_{a_I} - p_{b_F})^2 - m_c^2} .\nonumber 
\end{eqnarray}

From our examples of how to calculate Feynman diagrams, it appears that there
are common physical situations that might give the same general results. This is, in fact, the case.
We don't need to calculate all of the integrals. Instead, we can create a list
of rules that are called ``Feynman rules'' for dealing with Feynman diagrams
and make our lives a little bit easier. Please see the handout that goes with
this document. It shows the Feynman rules for the theory presented here, along
with the $\mathcal{M}$-amplitude calculations I have shown, along with
other Feynman diagram calculations I have not shown.

Disclaimer: The method I used in this document to find the $\mathcal{M}$-amplitude are only
guaranteed to work
for the specific calculations shown. To obtain $\mathcal{M}$ for 
Feynman diagrams different from what I have shown, you will have to use established Feynman rules. 

Now that we have the $\mathcal{M}$-amplitude for two-body scattering we need to see how this is related
to the physical measurements of lifetime and cross-section. The connection will be shown in the next
section. After which, we will be concerned with how a potential energy term like 
$V \propto e^{- m r}/r$ comes from the interaction Lagrange density.

\section{The $\mathcal{M}$-amplitude connection to the physical observables of lifetime and cross-section}

In what follows, I will show the relation between $\mathcal{M}$ and physical observables
for some simple reactions. I will state these relations without proof.

First we need to define a few quantities:
\begin{itemize}
\item $\tau \equiv$ the mean lifetime of a particle,
\item $\Gamma \equiv$ the decay rate of a particle,
\item $\sigma \equiv$ the cross section for a reaction,
\item $d\sigma/d\Omega \equiv$ the differential cross section.
\end{itemize}

\subsection{Decay rate}
The relationship between the decay rate and lifetime is given simply as
\begin{eqnarray}
\Gamma   =  \frac{1}{\tau} ,\nonumber
\end{eqnarray}
and is related to the number of particles $N$ through the relation $N(t) = N(0) e^{-\Gamma t}$.

We now want to relate the decay rate to the $\mathcal{M}$-amplitude for a two-body
decay $1 \rightarrow 2 + 3$ in the center-of-mass frame:
\begin{eqnarray}
\Gamma   =  \frac{S |\vec{p_f}|}{8 \pi m_1^2} |\mathcal{M}|^2 ,\nonumber
\end{eqnarray}
where $\vec{p_f}$ is the momentum of either final particle and $S$ represents a symmetry
factor that we don't have to worry about right now.
For the reaction $A \rightarrow B + C$ we have $\mathcal{M} = g$ (and the symmetry
factor $S = 1$ in this case), so
\begin{eqnarray}
\Gamma   =  \frac{|\vec{p_b}|}{8 \pi m_a^2} g^2 ,\nonumber
\end{eqnarray}
where $\vec{p_b} $ ($-\vec{p_c}$) is determined by energy-momentum conservation.

\subsection{Differential cross section}

The center-of-mass differential cross section for the reaction $1 + 2 \rightarrow 3 + 4$ is given by
\begin{eqnarray}
\frac{d\sigma}{d\Omega}   = \left[ \frac{S}{(8\pi)^2} \right] 
\left[\frac{|\mathcal{M}|^2}{(E_1 + E_2)^2} \right] \frac{|\vec{p_f}|}{|\vec{p_i}|}, \nonumber
\end{eqnarray}
where $\vec{p_i}$ ($\vec{p_f}$) is the momentum of either incoming (outgoing) particle.  
In our example of $A + B \rightarrow A + B$ we have
\begin{eqnarray}
\mathcal{M}   & = &
\left[ \frac{g^2}{t^{\mu} t_{\mu} - m_c^2} \right], \nonumber 
\end{eqnarray}
where $t^{\mu} = p^{\mu}_{a_I} - p^{\mu}_{b_F}$, and represents the momentum transfer
to particle $C$ in Fig. \ref{fig:f3}. 
If I set $m_a = m_b$ we get
\begin{eqnarray}
\mathcal{M}   & = &
\left[ \frac{-g^2}{\vec{t} \cdot \vec{t} + m_c^2} \right]. \nonumber 
\end{eqnarray}
Thus
\begin{eqnarray}
\frac{d\sigma}{d\Omega}   & = &
\left[ \frac{-g^2}{\vec{t} \cdot \vec{t} + m_c^2} \right]^2
\left[\frac{1}{(8\pi)^2 (E_1 + E_2)^2} \right], \nonumber 
\end{eqnarray}
(Note: In the 
the center-of-mass frame $|\vec{p_i}| = |\vec{p_f}|$ for the reaction $A + B \rightarrow A + B$.)

\section{Potential and $\mathcal{L}$}

To determine the form of the potential using our $\mathcal{M}$-amplitude, we will compare our 
cross section in the non-relativistic limit to the cross section we would obtain using the
non-relativistic Born approximation.

For a reaction of the type $A + B \rightarrow A + B$, we can write the non-relativistic 
Born approximation
\begin{eqnarray}
\frac{d\sigma}{d\Omega}   = \frac{m^2_r} {(2 \pi)^2} |\tilde{V}(q)|^2,
\end{eqnarray}
where $m_r$ is the reduced mass ($m_r = m_a m_b/(m_a + m_b)$) and $\tilde{V}(q)$ is the 
potential in momentum space.
Therefore, in the non-relativistic limit, we can make the identification
\begin{eqnarray}
\tilde{V}(q) \rightarrow \frac{1}{4 m_a m_b} \mathcal{M}.
\end{eqnarray}
Thus, we just need to perform a Fourier transform to obtain $V(\vec{r})$ from
$\mathcal{M}$. For our $A + B \rightarrow A + B$ reaction we obtain (remembering that we have set $m_a = m_b$)
\begin{eqnarray}
V(r) & = & 
-\frac{g^2}{4 m^2_a} \int \frac{d^3q}{(2 \pi)^3}\frac{e^{i \vec{q}\cdot \vec{r}}}{\vec{q}\;^2 + m^2}  \nonumber \\
 & = & -\left( \frac{g^2}{16 \pi m^2_a}\right) \frac{e^{-m_c r}}{r}. \label{eq:yukawa}
\end{eqnarray}

The first thing to notice is that the potential is attractive (minus sign in front of the 
right hand side of Eqn. \ref{eq:yukawa}).
In the limit where $m_c \rightarrow 0 $ we would get the potential $V(r) \propto 1/r$. That is,
the potential for the case where $m_c \rightarrow 0$, looks like what we are accustomed to for
gravity and electromagnetism. For the case where $m_c \ne 0$, the inverse radial dependence
is ``damped'' by the exponential factor $e^{-m_c r}$.

You should notice that, for our simple theory, the momentum dependence of $\mathcal{M}$ was determined by the propagator
of the $C$ particle, while the coupling $g$ was determined by the vertex. Thus, both the free part 
(that determines the propagator) and the interaction part (that determines the vertex couplings) are important
in relating the potential to the Lagrange density $\mathcal{L}$.

You should not be too surprised that the potential for the interaction of a scalar field
with a fermion field ($\mathcal{L}_{int} = g\: \overline{\psi}\: \phi\: \psi$, where the over-bar on
$\psi$ is a technical detail we won't worry about right now) yields $V(r) \propto e^{-m_c r}/r$. After all,
the propagator for the $\phi$-field is the same as we have seen in our $A B C$ theory. 
This becomes more interesting if we were to make a strong-force model where baryons (three-quark composite particles
like protons and neutrons) interact with scalar mesons. In fact, such theories have been proposed, where one
particularly famous  
model came from Hideki Yukawa. The potential $V(r) \propto e^{-m_c r}/r$ is commonly called the 
``Yukawa potential'' and represents a major achievement in the early theoretical understanding of the 
nucleon-nucleon force. 
 
Unfortunately, the nucleon-nucleon force is not quite so nice. For one thing, the most common mesons are not
scalar particles but, instead, pseudo-scalar. This means that the parity of these particles change
sign under a ``space-inversion'' ($\vec{r} \rightarrow -\vec{r}$).

\subsection{One-pion exchange potential for proton-proton scattering}


In relativistic quantum mechanics, anti-particles are possible, and this creates a 
new degree of freedom that is not seen in the non-relativistic theory.  
In non-relativistic quantum mechanics, a fermion can be represented by a two component spinor, whereas
in a relativistic treatment, a four-component spinor is found that describes both the spin and 
particle/anti-particle nature of the field excitation.
As a reminder of relativistic quantum mechanics, we write the Dirac equation as:
\begin{eqnarray}
(i \gamma^{\mu} \partial_{\mu} - m) \psi = 0,
\end{eqnarray}
where the $4 \times 4$ gamma matrices $\gamma^{\mu}$ can be written (``Bjorken and Drell'' convention):
\begin{eqnarray}
\gamma^0  = 
\begin{pmatrix}
1 & 0 \\
0 & -1 \\
\end{pmatrix} ,
\gamma^i  =
\begin{pmatrix}
0 & \sigma^i \\
- \sigma^i & 0 \\
\end{pmatrix} ,
\gamma^5  =
\begin{pmatrix}
0 & 1 \\
1 & 0 \\
\end{pmatrix} \nonumber ,
\end{eqnarray}
and $\sigma$ are the Pauli spin matrices. 

Since $\psi^{\dagger} \psi$ is not a relativistic invariant, it is convenient to work with
$\overline{\psi} \psi$, where $\overline{\psi} \equiv \psi^{\dagger} \gamma^0$. 
It can be shown that $\overline{\psi} \psi$ is a relativistic invariant, and that the ``bilinears''
transform as follows:
\newcounter{bean}
\begin{list}{\arabic{bean}.}
{\usecounter{bean}
\setlength{\rightmargin}
{\leftmargin}}
\item $\overline{\psi} \psi = $ scalar,
\item $\overline{\psi} \gamma^5 \psi = $ pseudo-scalar,
\item $\overline{\psi} \gamma^{\mu}\psi = $ vector,
\item $\overline{\psi} \gamma^{\mu} \gamma^5 \psi = $ pseudo-vector,
\item $\overline{\psi} \sigma^{\mu \nu} \psi = $ antisymmetric tensor,
\end{list}
where $\sigma^{\mu \nu} \equiv i/2 [\gamma^{\mu},\gamma^{\nu}]$.

The free Lagrange density for fermions is written
\begin{eqnarray}
\mathcal{L}_0 = \overline{\psi} (i \partial^{\mu} \gamma_{\mu} - m ) \psi.
\end{eqnarray}
To conserve parity in the interaction Lagrange-density for a pseudo-scalar particle interacting
with a fermion spinor field (in this case we will only consider proton fields),
one needs to use a $\gamma^5$ factor so that,
%\begin{eqnarray}
%\mathcal{L}_{int} = -g \overline{\psi} \gamma^5 \vec{\tau} \psi \cdot \vec{\phi}
%\end{eqnarray}
\begin{eqnarray}
\mathcal{L}_{int} = -g \overline{\psi} \gamma^5  \psi \phi.
\end{eqnarray}
The full Lagrange density can now be written
\begin{eqnarray}
\mathcal{L} = \overline{\psi} [i \partial^{\mu} \gamma_{\mu} - m_p ] \psi + 
 \frac{1}{2}[(\partial \phi)^2 - m^2_{\pi} \phi^2] 
 -g \overline{\psi} \gamma^5  \psi \phi, \nonumber
\end{eqnarray}
and is invariant under a parity transformation.

Without going into the technical details 
(a full derivation can be found in section 9.9 of 
``Relativistic Quantum Mechanics and Field Theory'' by Franz Gross) 
the resulting non-relativistic
potential becomes:
\begin{eqnarray}
V(r)  =  \frac{g^2 m^2_{\pi}}{48 \pi m^2_p} 
\left[ \frac{e^{-m_{\pi} r}}{r} \right] \times \left\{  
\vec{\sigma_1} \cdot \vec{\sigma_2} + \right. \nonumber \\ 
\left. (3 \vec{\sigma_1} \cdot \hat{r} \vec{\sigma_2} \cdot \hat{r} - \vec{\sigma_1} \cdot \vec{\sigma_2}) 
\left( 1 + \frac{3}{m_{\pi} r} + \frac{3}{(m_{\pi} r)^2}\right)
\right\} \label{eq:ope} . 
\end{eqnarray}
%where 
%$\tau$ is the isospin operator (Pauli spin matrix operating on the isospin space of the nucleon), and 
What a difference a $\gamma^5$ makes!
At first glance, this seems to be a disaster, but if we look closer, it turns
out not to be all that bad. For one, we see the common factor of $e^{-m_c r}/r$. Also, we can see terms:
\begin{itemize}
\item $(\vec{\sigma_1} \cdot \vec{\sigma_2})$: central spin dependence,
\item $(3 \vec{\sigma_1} \cdot \hat{r} \vec{\sigma_2} \cdot \hat{r} - \vec{\sigma_1} \cdot \vec{\sigma_2})$: tensor,
\end{itemize}
that are needed for a description of nucleon-nucleon interactions \smiley.
Unfortunately, the central spin dependence is about a factor of ten smaller
than what is found in $N N$ scattering and there is a spin-orbit
part that is completely missing \frownie. 
The model can, however, be improved upon by allowing scalar and vector mesons.

This is a cautionary tale: What may appear to be a simple change to
the vertex can result in large changes in the behavior of the 
reaction. 

\begin{figure}[h]
\includegraphics[scale=0.6]{f4.eps}
\caption{ Spin orbit coupling.
\label{fig:f4}}
\end{figure}

This is a good opportunity to talk about recoil polarimeters.
If our model included vector meson exchange, we would be able to generate another term
to our potential that couples spin and orbital angular momentum \cite{breit}.

To get an intuitive idea as to why we should expect a spin-1 vector meson
to give a spin-orbit interaction term, we will work by analogy. We know that
the photon is a spin-1 massless field, and that the coupling of the photon
field to fermions must produce spin-orbit interactions. (I say this with confidence
because we know that spin-orbit interactions have been measured in atomic 
spectroscopy.) So, if we were to couple a massless spin-1 vector meson to 
a fermion field, we should obtain a spin-orbit coupling. The only difference
between the spin-orbit coupling for a massless spin-1 meson, and a photon, should
come from the different strengths of the coupling constants. In our case,
we are interested in a massive meson field. This means that the propagator
for the massive field is expected to cause a dampening effect in the potential. Otherwise,
we should expect the two fields to act similarly (more on this later on).

We will consider a spin orbit term of the form:
\begin{eqnarray}
V_{SO} = -C \Vec{L} \cdot \vec{S}, \nonumber
\end{eqnarray}
where $C$ is a positive constant.

If we had a target of carbon atoms (which is often the case for a recoil polarimeter),
and incident protons as shown in Fig. \ref{fig:f4}, where the spin of each incident
proton is aligned out of the page, we would obtain a repulsive potential for 
proton 1 and an attractive potential for proton 2. For each case where the proton
spin is directed out of the page, we get a force bending the path to the left!
If the spin were to be pointed into the page, instead, the paths would be bent to the 
right. This is how a recoil polarimeter works: a slab of carbon is placed
in a detector, and the direction and degree of deflection tells us about the 
proton spin polarization.


\section{The QED Lagrangian density}

For a photon field we have the free Lagrange density
\begin{eqnarray}
\mathcal{L}_0 = \frac{-1}{4} F^{\mu \nu} F_{\mu \nu}, \nonumber
\end{eqnarray}
where
\begin{eqnarray}
F^{\mu \nu} & \equiv & \partial^{\mu} A^{\nu} - \partial^{\nu} A^{\mu} \nonumber \\
& = & 
\begin{pmatrix}
0 & -E_x & -E_y & -E_z \\
E_x & 0 & -B_z & B_y \\
E_y & B_z & 0 & -B_x \\
E_z & -B_y & B_x & 0 \\
\end{pmatrix} , \nonumber
\end{eqnarray}
implies that
\begin{eqnarray}
\frac{-1}{4} F^{\mu \nu} F_{\mu \nu} = \frac{1}{2} (\vec{E}^2 - \vec{B}^2).
\end{eqnarray}
This is just the difference between the energy density of the electric and magnetic fields.
So when you see $-F^{\mu \nu} F_{\mu \nu}/4$ you should think
$(\vec{E}^2 - \vec{B}^2)/2$.

Now that the Lagrange density for a free photon field has an intuitive deffinition,
we want to see how the photon field is coupled to a fermion field.
Earlier we saw that the free Lagrangian for a fermion field is written as
\begin{eqnarray}
\mathcal{L}_0 = \overline{\psi} (i \partial^{\mu} \gamma_{\mu} - m ) \psi. \nonumber
\end{eqnarray}
The interaction part for the two fields $\psi$, $A_{\mu}$ will be
\begin{eqnarray}
\mathcal{L}_{int} = - e \overline{\psi} \gamma^{\mu} \psi A_{\mu} . \nonumber
\end{eqnarray}
(To make sense of this $\mathcal{L}_{int}$ term you should notice that we want to
couple a four-vector electromagnetic field $A_{\mu}$ to the fermion field 
forming a scalar [recall that the Lagrange density must be a scalar]. The only way to do this is to
contract the four-vector $A_{\mu}$ with some other four-vector. The only vector bilinear
at our disposal is $\overline{\psi} \gamma^{\mu} \psi$.)
Putting this all together we have
\begin{eqnarray}
\mathcal{L}_{QED} =  \overline{\psi} (i \partial^{\mu} \gamma_{\mu} - m )\psi -
\frac{1}{4} F^{\mu \nu} F_{\mu \nu} - e \overline{\psi} \gamma^{\mu} \psi A_{\mu} . \label{eq:QED1}
\end{eqnarray}
Often you will see this written as
%\begin{eqnarray}
%\mathcal{L}_{QED} =  \overline{\psi} [i \gamma_{\mu} (\partial^{\mu}  + i e A_{\mu}) - m ]\psi -
%\frac{1}{4} F^{\mu \nu} F_{\mu \nu}, \nonumber
%\end{eqnarray}
%or even
\begin{eqnarray}
\mathcal{L}_{QED} =  \overline{\psi} [i \gamma_{\mu} D^{\mu} - m ]\psi -
\frac{1}{4} F^{\mu \nu} F_{\mu \nu}, \nonumber
\end{eqnarray}
where $D^{\mu} \equiv \partial^{\mu}  + i e A_{\mu}$.
These two forms given for $\mathcal{L}_{QED}$ are identical, but look different at first glance.

The interaction vertex has a Feynman diagram as shown in Fig. \ref{fig:f5}, and the vertex factor
can be read directly from the interaction part of $\mathcal{L}_{QED}$
and is $-e \gamma_{\mu}$.
\begin{figure}[t]
\includegraphics[scale=0.5]{f5.eps}
\caption{Vertex for QED.
\label{fig:f5}}
\end{figure}
This is the only vertex possible, as can be seen from the Lagrange density in Eqn. \ref{eq:QED1}. 
Every Feynman diagram you can think of for 
QED is made up of this one vertex. For example we can get a diagram for
photon scattering $\gamma \gamma \rightarrow \gamma \gamma$ as shown in
Fig. \ref{fig:f6} just by
using our one simple vertex over and over.

\begin{figure}[t]
\includegraphics[scale=0.45]{f6.eps}
\caption{A Feynman diagram for $\gamma \gamma \rightarrow \gamma \gamma$.
\label{fig:f6}}
\end{figure}

The propagators for this theory are
\begin{itemize}
\item photons: $-ig^{\mu \nu}/q^2$,
\item fermions: $i/(\gamma^{\mu} q_{\mu} - m)$,
\end{itemize}
and are derivable from the free Lagrange densities.
As would be expected, the non-relativistic potential
is $V \propto 1/r$ for unpolarized fermions, and depends on the photon mass
being zero.

To distinguish between particles and anti-particles within a Feynman diagram,
the time-direction of the arrow is used: Particles will have arrows pointing
in the positive time direction, while anti-particles will point in the
negative time direction. For example, Fig. \ref{fig:f6b} shows $e + \bar{\mu} \rightarrow e + \bar{\mu}$,
where $\bar{\mu}$ represents the $\mu$ anti-particle, and Fig. \ref{fig:f6c}
shows one diagram for the pair annihilation process $e^- + e^+ \rightarrow \gamma + \gamma$.
(The anti-particle for $e$ is denoted $e^+$ instead of as $\bar{e}$ for purely
historical reasons.)

\begin{figure}[h]
\includegraphics[scale=0.5]{f6b.eps}
\caption{A Feynman diagram for $e + \bar{\mu} \rightarrow e + \bar{\mu}$, where
the direction of the arrows determines the particle, anti-paritcle nature.
\label{fig:f6b}}
\end{figure}

\begin{figure}[h]
\includegraphics[scale=0.5]{f6c.eps}
\caption{A Feynman diagram for $e^- + e^+ \rightarrow \gamma + \gamma$.
\label{fig:f6c}}
\end{figure}

For identical particles in the final state, we can not distinguish between the 
Feynman diagram given in Fig. \ref{fig:f6c} and that shown in Fig. \ref{fig:f6d}.
This means that both figures would need to be used in determining the $\mathcal{M}$-amplitude
for the second order $e^- + e^+ \rightarrow \gamma + \gamma$ process.

\begin{figure}[h]
\includegraphics[scale=0.5]{f6d.eps}
\caption{A second Feynman diagram for $e^- + e^+ \rightarrow \gamma + \gamma$.
\label{fig:f6d}}
\end{figure}

One last issue before we move on to the next subsection. 
We need to keep track of the spin of the fermion lines and the polarization
of the photon lines. In Feynman diagrams, you associate a spinor for each
external fermion line and a polarization vector for each external photon line.

Most particle physics books will give you the Feynman rules for QED, and show
you how to calculate the $\mathcal{M}$-amplitude for specific
processes. 

\subsection{Anomalous magnetic moment and effective Lagrange densities}

Since we can always make more complicated Feynman diagrams, it would be good
make modifications to the rules so that, while the Lagrange density
has not changed, the modified rules ``soak up'' some common classes of 
Feynman diagrams. One such modification is associated with something called
the anomalous magnetic moment. This anomalous magnetic moment results
from a class of Feynman diagrams where there are photon lines ``straddling''
a vertex. An example can be found in Fig. \ref{fig:f11} and the sum of
all of these type of corrections is commonly
drawn as in  Fig. \ref{fig:f11b}.

\begin{figure}[h]
\includegraphics[scale=0.5]{f11.eps}
\caption{A Feynman diagram for a vertex correction.
\label{fig:f11}}
\end{figure}

\begin{figure}[h]
\includegraphics[scale=0.5]{f11b.eps}
\caption{A Feynman diagram for all possible vertex corrections.
\label{fig:f11b}}
\end{figure}

We can make a simple change to our vertex function that accounts for the 
anomalous magnetic moment by changing our vertex function to be
\begin{eqnarray}
\gamma^{\mu} F_1(q^2) + \frac{i \sigma ^{\mu \nu}q_{\nu}}{2 m_N} F_2(q^2), \nonumber
\end{eqnarray}
where $F_1$ and $F_2$ are functions of momentum-transfer that are determined for
the particular fermion fields involved, and $m_N$ is the proton mass.
While the fundamental Lagrange density has not changed, we could write down
an effective interaction Lagrange density such that
\begin{eqnarray}
\mathcal{L}_{int\; \rm{Effective}} = 
- F_1(q^2) \overline{\psi} \gamma^{\mu} \psi A_{\mu} + 
F_2(q^2) \overline{\psi} \left( \frac{ \sigma^{\mu \nu} }{2 m_N} \right) \psi F_{\mu \nu}, \nonumber
\end{eqnarray}
where it is understood that the class of Feynman diagrams as shown in Fig. \ref{fig:f11} are
already taken care of.

It might seem a bit strange to replace $i \sigma^{\mu \nu} q_{\nu} A_{\mu}$ with
$\sigma^{\mu \nu} F_{\mu \nu}/2$. To make sense of this substitution, you need to know
that the $A^{\mu}$ field can be written as
\begin{eqnarray}
A^{\mu} = \sum_{s} \int \frac{d^4q \:  \epsilon^{\mu}_s}{(2 \pi)^3 \sqrt{E}}
\left( a_{s \vec{q}} + a^{\dagger}_{s \vec{-q}} \right) \: e^{-i x \cdot q},
\nonumber
\end{eqnarray} 
where $\epsilon^{\mu}$ is the polarization vector, 
and implies that
\begin{eqnarray}
\partial^{\nu} A^{\mu} & = & i \sum_{s} \int \frac{d^4q \: q^{\nu}  \epsilon^{\mu}_s}{(2 \pi)^3 \sqrt{E}}
\left( a_{s \vec{q}} \: + a^{\dagger}_{s \vec{-q}} \right) \: e^{-i x \cdot q} .
\nonumber \\
& = & -i q^{\nu} A^{\mu} \nonumber .
\end{eqnarray}
You also need to recognize that since $\sigma^{\mu \nu}$ is antisymmetric under $\mu \rightarrow \nu$:
\begin{eqnarray}
\sigma^{\mu \nu} q_{\nu} A_{\mu} & = & - \sigma^{\mu \nu} q_{\mu} A_{\nu} \Rightarrow \nonumber \\ 
 & = &\frac{1}{2} ( \sigma^{\mu \nu} q_{\nu} A_{\mu} -
\sigma^{\mu \nu} q_{\mu} A_{\nu} ) \nonumber \\
& = & \frac{-i}{2}\sigma^{\mu \nu} F_{\mu \nu} \nonumber .
\end{eqnarray}
Thus, $i \sigma^{\mu \nu} q_{\nu} A_{\mu} = \sigma^{\mu \nu} F_{\mu \nu}/2$.


What we will want to know next is how massive
spin-1 mesons (e.g. $\rho$, $\omega$, and $\phi$) can be incorporated into a 
model for nucleon interactions.

\section{Massive spin-1 vector field}

As a reminder, we can write the free Lagrange density for a photon field as
\begin{eqnarray}
\mathcal{L}_0 = \frac{-1}{4} F^{\mu \nu} F_{\mu \nu}, \nonumber
\end{eqnarray}
where
\begin{eqnarray}
F^{\mu \nu} & \equiv & \partial^{\mu} A^{\nu} - \partial^{\nu} A^{\mu}. \nonumber 
\end{eqnarray}

The most general Lagrange density we could write for a 
gauge field and fermion field is
\begin{eqnarray}
\mathcal{L} =  \overline{\psi} [i \gamma_{\mu} D^{\mu} - m ]\psi -
\frac{1}{4} F^{\mu \nu} F_{\mu \nu}
- c \epsilon_{\alpha \beta \mu \nu} F^{\alpha \beta} F^{\mu \nu}.
\label{eq:vGeneral}
\end{eqnarray}
(see section 15.1 of ``An Introduction to Quantum Field Theory'', by M.E. Peskin and D.V. Schroeder.)
The term $\epsilon_{\alpha \beta \mu \nu} F^{\alpha \beta} F^{\mu \nu}$ violates parity and 
is thrown out. We will come back to this equation (Eqn. \ref{eq:vGeneral}) later on.

For a spin-1 massive vector field we set
\begin{eqnarray}
\mathcal{L}_0 = \frac{-1}{4} V^{\mu \nu} V_{\mu \nu} - \frac{1}{2} m^2_V V^{\mu} V_{\mu}, \nonumber
\end{eqnarray}
where
\begin{eqnarray}
V^{\mu \nu} & \equiv & \partial^{\mu} V^{\nu} - \partial^{\nu} V^{\mu}, \nonumber
\end{eqnarray}
$V^{\mu}$ represents the massive vector field, and $m_V$ is the mass of the particle excitation
of the vector field $V^{\mu}$.
The interaction term for our massive vector field coupled to a fermion field is given as
\begin{eqnarray}
& & \mathcal{L}_{int}  =  - \lambda \overline{\psi} \gamma^{\mu} \psi V_{\mu} , \nonumber \\
& & \mathcal{L}_{int\; \rm{Effective}}  =  \nonumber \\
& & - G_1(q^2) \overline{\psi} \gamma^{\mu} \psi V_{\mu} + 
G_2(q^2) \overline{\psi} \left( \frac{i \sigma ^{\mu \nu}q_{\nu}}{2 m_N} \right) 
\psi V_{\mu \nu}, \nonumber
\end{eqnarray}
where $G_1$, $G_2$ are analogs of the form factor $F_1$, $F_2$.
Immediately, we see that if we take $m_V = 0$, then the only difference between the Lagrange 
density for our meson vector-field and the photon field, is the coupling constants.  
(For the effective Lagrangian we also see different form factors.)
Also, the transformation properties of $F^{\mu \nu}$ will
be identical to the transformation properties of $V^{\mu \nu}$, and thus there are 
quantities within $V^{\mu \nu}$ that are a direct analog to the $\vec{E}$ and $\vec{B}$ fields
of electromagnetism. 

Because of the mass term in the free Lagrange density, the propagator for massive vector field
will be different than that for the massless photon field ($-ig^{\mu \nu}/q^2$), and is 
\begin{eqnarray}
\frac{-i(g^{\mu \nu} - q^{\mu} q^{\nu}/m^2_V)}{q^2 - m^2_V} \label{eq:vProp} .
\end{eqnarray}
The first thing you should notice is that the propagator given in Eqn. \ref{eq:vProp} does
not behave nicely in the limit that $m_V \rightarrow 0$. What we would like to see, is 
the massive propagator, in the limit $m_V \rightarrow 0$, to become the photon propagator 
$-ig^{\mu \nu}/q^2$. The reason that the two propagators look so different has to
do with the fact that a massless vector field has only two polarization degrees of freedom. 

A technical detail:
For an arbitrary spin-1 particle we can have three polarization states: One longitudinal
state and two transverse states. For a massless particle, the velocity must be the
speed of light. This causes the length contraction in the direction of motion to completely
``wipe-out'' any vector quantity in that direction. This is why a photon is spin-1 with
only two polarizations. The photon field has ``lost'' a degree of freedom. Because of this
lost degree of freedom, we have a field that is under-determined. This 
manifests itself in the photon field having a non-determined ``gauge''. We have 
the freedom to choose an electromagnetic gauge transformation such that  
\begin{eqnarray}
A^{\mu} \rightarrow A^{\mu} + \partial^{\mu} f(\vec{x},t),
\end{eqnarray}
where $f$ is any function of position and time.
This choice of gauge transformation manifests itself in the photon propagator such that
the propagator can be written as
\begin{eqnarray}
\frac{-i}{q^2} \left[g^{\mu \nu} - (1 - \zeta) \frac{q^{\mu} q^{\nu}}{q^2} \right],
\end{eqnarray}
where $\zeta$ can be any number and represents a choice of gauge. In this document
$\zeta = 1$ is chosen so that the photon propagator becomes $-ig^{\mu \nu}/q^2$.
I still have not shown how the massive vector-field propagator becomes the
massless vector propagator, but have instead shown that determining the
massless vector propagator is a tricky affair (that we won't go into any further).

%In this document, we are primarily concerned with the meson vector-fields $\rho$, and $\omega$.
%However, it is a small detour for us to briefly discuss the $W^{\pm}$, $Z^0$ electro-weak
%mediators.

%\subsection{A touch of Weak theory}
%I am just going to state a few results and move on.

%The $W^{\pm}$, $Z^0$ are massive spin-1 vector fields, with masses $M_w \approx 80$ GeV,
%and $m_Z \approx 90$ GeV. Remembering that a proton has $m_p = 0.938$ GeV, we 
%can see that the $W^{\pm}$, $Z^0$ are very heavy. This will result in the
%weak force as being short range.

%The weak interaction has a property that is very unusual: It is neither vector, nor
%psuedo-vector! It is a mixture of both! The weak vertex is
%\begin{eqnarray}
%-i g (\gamma^{\mu} - \gamma^{\mu} \gamma^5).
%\end{eqnarray}
%Parity is not conserved for weak interactions!

%\begin{figure}[h]
%\includegraphics[scale=0.5]{f9.eps}
%\caption{A Feynman diagram for the $W$.
%\label{fig:f9}}
%\end{figure}


%A possible vertex for the Weak interaction can be found in Fig \ref{fig:f9}.
%From this vertex we can draw the Feynman diagram for the reaction 
%$\mu \rightarrow \nu_{\mu} + e + \overline{\nu_e}$ as seen 
%in Fig \ref{fig:f10}.

%\begin{figure}[h]
%\includegraphics[scale=0.5]{f10.eps}
%\caption{A Feynman diagram for the $\mu \rightarrow \nu_{\mu} + e + \overline{\nu_e}$.
%\label{fig:f10}}
%\end{figure}

%In addition to changing leptons into neutrinos, the Weak interaction can change the 
%$u$ quark into $d$, and an $u$ quark into a $s$ and every flavor changing combination possible.
%The particular mediator ($W^{\pm}$, $Z^0$) involved will be detrmined so that the reaction makes 
%sense when considering charge conservation. The strength of each 
%flavor changing reaction is not predicted by the standard model, is determined 
%through empirical measurement and is related to something called the CKM
%(Cabibbo–Kobayashi–Maskawa) matrix.

%The Weak interaction is very interesting and you might want to 
%study it further. I only took the opportunity to touch upon the Weak
%interaction because we are looking at massive vector fields.

\section{Effective Lagrangian theory for $\eta$ photoproduction}

I will now briefly describe the interaction parts of an effective
field theory by Benmerrouche, Mukhopadhyay, and Zhang (BMZ). These interaction 
Lagrange densities are also common to the theory of Nakayama and Haberzettl (NH).

I will use the notation of BMZ:
\begin{itemize}
\item $N \rightarrow$ Proton field.
\item $\eta \rightarrow$ $\eta$ field.
\item $V \rightarrow$ Vector meson field ($\rho$ or $\omega$).
\item $m_N \rightarrow$ Proton mass.
\item $A^{\mu} \rightarrow$ Photon field.
\item $R \rightarrow$ Nucleon resonance field.
\end{itemize}

\subsection{Interaction Lagrange density for $\gamma N N$}
We have already worked out the effective Lagrange density for a photon-field
interacting with a fermion field. As previously discussed:
\begin{eqnarray}
\mathcal{L}_{\gamma N N} =
-e \overline{N} \gamma_{\mu} N A^{\mu} + \frac{\kappa_t}{2 m_N}\overline{N} \sigma_{\mu \nu} N F^{\mu \nu}. \nonumber
\end{eqnarray}
In this case $F_1 = e$, and $F_2 = \kappa_t$.

\subsection{Interaction Lagrange density for $V N N$}
We have also shown that for a vector-field coupled to a fermi-field, we expect 
(by direct analogy to the photon case):
\begin{eqnarray}
\mathcal{L}_{V N N} =
-g_v \overline{N} \gamma_{\mu} N A^{\mu} + \frac{g_t}{2 m_N} \overline{N} \sigma_{\mu \nu} N F^{\mu \nu}. \nonumber
\end{eqnarray}
In this case $G_1 = g_v$, and $G_2 = g_t$.

\subsection{Interaction Lagrange density for $\eta N N$}
The Lagrange density for $\eta N N$ has an additional 
pseudo-vector coupling term we have not discussed thus far. This new pseudo-vector
coupling can be seen in the second term of the Lagrange density 
\begin{eqnarray}
\mathcal{L}_{\eta N N} =
g_{\eta} 
\left[ 
-i \zeta \overline{N} \gamma_5 N \eta + 
(1 - \zeta) \frac{1}{2} \overline{N} \gamma_{\mu} \gamma_5 N \partial^{\mu} \eta
\right], \nonumber
\end{eqnarray}
where $\zeta$ represents the fraction of the pseudo-scalar to pseudo-vector couplings.
Previously, when discussing the anomalous magnetic moment, we found that the 
effective Lagrange density can be described using field-derivative couplings.
It turns out that the general form for our Lagrange density for $\eta N N$ can
include an $\eta$-field derivative. In the case of $\pi N N$, the pseudo-vector
piece dominates to the extent that the pseudo-scalar piece is often left out
entirely ($\zeta \rightarrow 0$) for pion-nucleon-nucleon interactions.


\subsection{Interaction Lagrange density for $\eta N R$}
These terms are as expected, once the parity of the resonance state $R$ is taken account of:
\begin{eqnarray}
\mathcal{L}^{PS}_{\eta N R} & = & -i g_{\eta N R} \overline{N} \Gamma R \eta \nonumber \\
\mathcal{L}^{PV}_{\eta N R} & = & 
-i \frac{f_{\eta N R}}{m_{\eta}} \overline{N} \Gamma_{\mu} R \partial^{\mu} \eta  , \nonumber\\ 
\mathcal{L}_{\gamma N R} & = &
\frac{e \kappa}{2 (m_R + m_N)} \overline{R}\: \Gamma_{\mu \nu} N F^{\mu \nu} \nonumber,
\end{eqnarray}
where 
\begin{itemize}
\item Odd parity $\Rightarrow$ $\Gamma = 1$, $\Gamma_{\mu} = \gamma_{\mu}$, 
$\Gamma_{\mu \nu} = \gamma_5 \sigma_{\mu \nu}$
\item Even parity $\Rightarrow$ $\Gamma = \gamma_5$, $\Gamma_{\mu} = \gamma_{\mu} \gamma_5$, 
$\Gamma_{\mu \nu} = \sigma_{\mu \nu}$,
\item Super script $PV \Rightarrow$ pseudo-vector coupling, 
\item Super script $PS \Rightarrow$ pseudo-scalar coupling.
\end{itemize}
We can readily see that the only difference between the odd- and even-resonance cases is
a factor of $\gamma_5$ to keep the overall parity to be even.

\subsection{Interaction Lagrange density for $\eta \gamma V$}

We want to include a term that couples the $\eta$-field to the $\gamma$-field and vector meson $V$-field.
As usual, the fact that the $\eta$-field is pseudo-scalar means that we have to be careful about
parity. In our discussion regarding the most general form for the Lagrange density for a Dirac field coupled
to a vector gauge field seen in Eqn. \ref{eq:vGeneral}, I stated that the term
$\epsilon_{\alpha \beta \mu \nu} F^{\alpha \beta} F^{\mu \nu}$ was odd under a parity transformation.

To easily see the parity properties of $F^{\mu \nu}$, it is useful to look at the individual components.
The electromagnetic $A^{\mu}$ field transforms like a regular vector, and since
\begin{eqnarray}
\vec{E} & = & - \vec{\nabla} A^0 - \frac{\partial \vec{A}}{\partial t} ,  \nonumber \\
\vec{B} & = & \vec{\nabla} \times \vec{A} \nonumber ,
\end{eqnarray}
then $\vec{E}$ is a vector and $\vec{B}$ is a pseudo-vector. Another way to think about the
magnetic transformation properties is to imagine an infinite positive line charge
moving in the $+z$ direction. The magnetic field for this moving line charge will have a 
magnetic field pointed in the $+\hat{\phi}$-direction (taking cylindrical coordinates).
If we were to invert space ($\vec{r} \rightarrow -\vec{r}$), then the direction of
the $\vec{B}$-field remains in the $+\hat{\phi}$-direction. 
Since $F^{\mu \nu}$ can be made from $E$ and $B$ components, then
$F^{\mu \nu}$ is of mixed parity. The combination $-F^{\mu \nu} F_{\mu \nu}/4 = 
(\vec{E}^2 - \vec{B}^2)/2$, however, is even under a parity transformation. 

Since the strong and electromagnetic interactions conserve parity, then a term
like $\epsilon_{\alpha \beta \mu \nu} F^{\alpha \beta} F^{\mu \nu}$
must be excluded in the QED Lagrangian. Now, however,
what was a fatal flaw for including this type of term for QED, is an asset when
considering an interaction term for coupling $\eta$ to $V$ and $\gamma$.
It is now reasonable to take the form of $\mathcal{L}_{V \eta \gamma}$ to be:
\begin{eqnarray}
\mathcal{L}_{V \eta \gamma} =
\frac{e \lambda_V}{4 m_{\eta}} \epsilon_{\mu \nu \alpha \beta} F^{\mu \nu} V^{\alpha \beta} \eta. \nonumber
\end{eqnarray}
As discussed previously, $V^{\mu \nu}$ has the same transform properties as $F^{\mu \nu}$. 
This means that there will be quantities within $V^{\mu \nu}$ that act as direct analogs
to the electromagnetic vectors $\vec{E}$, and $\vec{B}$. Let me write these analogs
as $\vec{\underline{E}}$, and $\vec{\underline{B}}$ so that
\begin{eqnarray}
V^{\mu \nu} & \equiv & \partial^{\mu} V^{\nu} - \partial^{\nu} V^{\mu} \nonumber \\
& = &
\begin{pmatrix}
0 & -\underline{E}_x & -\underline{E}_y & -\underline{E}_z \\
\underline{E}_x & 0 & -\underline{B}_z & \underline{B}_y \\
\underline{E}_y & \underline{B}_z & 0 & -\underline{B}_x \\
\underline{E}_z & -\underline{B}_y & \underline{B}_x & 0 \\
\end{pmatrix} , \nonumber
\end{eqnarray}
and
\begin{eqnarray}
\frac{1}{4}\epsilon_{\mu \nu \alpha \beta} F^{\mu \nu} V^{\alpha \beta} =
 \vec{E} \cdot \vec{\underline{B}} + \vec{\underline{E}} \cdot \vec{B}.
\end{eqnarray}
We see that the electromagnetic $\vec{E}$-field couples to the vector-analog $\vec{\underline{B}}$-field,
and the electromagnetic $\vec{B}$-field couples to the vector-analog $\vec{\underline{E}}$-field. 
Moreover, since $\vec{E}$ and $\vec{\underline{E}}$ are vector, and $\vec{B}$ and $\vec{\underline{B}}$
are pseudo-vector, then 
$\epsilon_{\mu \nu \alpha \beta} F^{\mu \nu} V^{\alpha \beta} =  \vec{E} \cdot \vec{\underline{B}} + \vec{\underline{E}} \cdot \vec{B}$ 
will be a pseudo-scalar as we desire for coupling to the pseudo-scalar $\eta$ meson.

\subsection{Contact term $N N \gamma \eta$}
To maintain the photon-field gauge invariance, a contact term is required. The sole purpose 
of this $N N \gamma \eta$ interaction is to subtract out bad (gauge violating) behavior.
You can think of it as an Ad Hoc addition to the model to keep it from falling apart. 

\subsection{Resonances with spin $3/2$ ($R_{3/2}$)}
The NH model includes resonances with spin 3/2. This extra degree of freedom for 
the resonance mediator adds additional complications. The Lagrange density for 
this spin 3/2 case we have
\begin{eqnarray}
\mathcal{L}_{N R_{3/2} \eta} =
\frac{g_{N R_{3/2} \eta}}{m_{\eta}} \overline{R}^{\mu} \Theta_{\mu \nu}(z)
\Gamma^{(\pm)} N \partial \eta, \nonumber
\end{eqnarray}
where $\Gamma{(+)} = \gamma_5$, $\Gamma{(+)} = 1$, and 
$\Theta_{\mu \nu}(z) = g_{\mu \nu} - (z - 1/2)\gamma_{mu} \gamma_{\nu}$.

We also have a modification to the $\gamma R_{3/2} N$ coupling:
\begin{eqnarray}
\mathcal{L}_{\gamma R_{3/2} N} =
& i & e  \frac{g_{1 \gamma R_{3/2} N}}{m_{R_{3/2}}} \overline{R}_{\mu} \gamma_5 \gamma_{\nu}
N F^{\mu \nu} + \nonumber \\
& & e \frac{g_{2 \gamma R_{3/2} N}}{m_{R_{3/2}}} \partial_{\nu} \overline{R}_{\mu} \gamma_5
N F^{\mu \nu}.
\nonumber
\end{eqnarray}
Since I am not comfortable talking about fields with spin $\ge 3/2$, I'll have to
end the discussion of $R_{3/2}$ here \frownie. 

It has taken some time, but now we have shown all of the Lagrange densities
associated with the BMZ and NH models for $\eta$ and $\eta^{\prime}$ photoproduction $\smiley$
Now we want to see how one of these models compares to actual data.


\section{Results of the NH model applied to ASU $\eta^{\prime}$ photoproduction data}

\begin{figure}[h]
\includegraphics[scale=0.5]{NH1.eps}
\caption{Feynman diagrams for the NH model. The variable $q$ is associated with
the $\eta^{\prime}$, $k$ the incident photon, $p$ the target proton, $p'$ the 
recoil proton, $R$ the resonance, and $\rho/\omega$ the vector mesons.
\label{fig:NH1}}
\end{figure}
\begin{figure}[h]
\includegraphics[scale=0.5]{NH2.eps}
\caption{NH model differential cross sections for five different sets of assumed resonance contributions
(I through V).
\label{fig:NH2}}
\end{figure}
\begin{figure}[h]
\includegraphics[scale=0.4]{NH4.eps}
\caption{NH model beam asymmetries for the five different sets of assumed resonance contributions
(I through V).
\label{fig:NH4}}
\end{figure}


The NH effective Lagrangian model has the Feynman diagrams as shown in Fig. \ref{fig:NH1} 


When all of the production mechanisms shown in Fig. \ref{fig:NH1} are
added together for five different sets of assumed resonance contributions,
NH obtain the fits to the ASU differential cross sections shown in Fig. \ref{fig:NH2}.

We can see that each of these five sets (I through V) compare well with the ASU results.
This means that we will need another type of observable to help deconvolute the
spectrum. One such observable is the beam asymmetry. A plot of the beam asymmetry
from the NH model for the five sets of resonances can be found in Fig. \ref{fig:NH4}.


%\begin{figure}[h]
%\includegraphics[scale=0.5]{NH3.eps}
%\caption{NH model for five different sets of assumed resonance contributions
%(I through V).
%\label{fig:NH3}}
%\end{figure}

It is easy to see from Fig. \ref{fig:NH4} that the beam asymmetries should help
greatly in determining the best possible set of included resonances for 
$\eta^{\prime}$ photoproduction. A former member of the ASU group, Patrick Collins, is
currently working on the beam asymmetry measurements for this reaction.




%\subsection{QCD}

%\subsection{The weak interaction}
%Unfortunately the electroweak Lagrange density is very complex and we just can't go
%into the details in this document. However, it is usefull to look at some of the
%possible vertices and obtain some physics straight from the Feynman diagrams.

\section{Conclusion}
We have discussed interaction Lagrange densities for photoproduction of $\eta$ 
and $\eta^{\prime}$ (from the proton) commonly found in the literature regarding
these reactions. We have also seen how the results of the NH model compare
to ASU $\eta^{\prime}$ photoproduction data.

\section{Acknowledgments}
I thank Barry Ritchie and William Kaufmann for their comments.

This work was supported at Arizona State University
by the National Science Foundation award PHY-0653630.

\begin{thebibliography}{0}
\bibitem{breit}  
G. Breit, Phys.\ Rev.\  \textbf{120}, 287 (1960)
\end{thebibliography}

\clearpage



{\bf APPENDIX: FINDING THE $Z$ AMPLITUDE}
\linebreak[4]

There are two common methods for finding the $\mathcal{M}$-amplitude: (1)
Canonical method using creation and annihilation operators; (2)
Path integrals. Each of these approaches has strengths and weaknesses.
One strength of the path integral approach is that the
relation of $\mathcal{M}$-amplitude to the Lagrange density is more
clearly seen than in the canonical formalism.


Most field theory text include a description of both formalisms. It is probably more common 
for the canonical treatment to be emphasized initially 
(e.g. ``An Introduction to Quantum Field Theory'' by Michael E. Peskin, and Daniel V. Schroeder;
``Quantum Field Theory'', by Michio Kaku;
``Relativistic Quantum Mechanics and Field Theory'', by Franz Gross;
``The Quantum Theory of Fields, Vol 1'', by Stephen Weinberg \ldots). 
Most likely, this is 
because the canonical approach looks more like the quantum mechanics you have encountered
as undergraduates, with operators acting on quantum-mechanical bra's and ket's. However, some texts
will start by emphasizing the path integral, with one such example being ``Quantum Field Theory in a Nutshell'', 
by A. Zee.  
In what follows, I have chosen to discuss the path integral approach of determining $Z$.

It is not the intent of this appendix to show how the path integral is put together from first
principles. For a complete treatment of the path integral formalism you might want
to read Zee. I will, instead, state a few results, so as to more clearly show why I
constructed the $Z$ amplitudes in the manner that I have shown in the document.

I start by simply stating the the time development of a state out-of and into the vacuum 
can be written as 
\begin{eqnarray}
Z \equiv \langle 0 | e^{-i H T} | 0 \rangle =
\int D\phi e^{i S},
\end{eqnarray}
where $H$ is the Hamiltonian, $\int D\phi \Rightarrow $ integration over all possible paths,
and the action $S = \int dt L = \int d^4x \mathcal{L}$.

Since we are constructing states that must come from the vacuum
at some space-time points and then returning to the vacuum at 
other space-time points (we have constructed the $Z$-amplitude so that the initial and final states
are the vacuum), we need a mechanism that acts as sources and sinks for our ``external'' particles
within any given Feynman diagram. To accomplish this, we add terms to the Lagrange density:
$J \phi$, where $\phi$ represents the field of our external line and $J$ represents sources/sinks.
Fortunately, we can take the sources and sinks to be space-time delta functions $J = \delta^{(4)}(x - x_1)$,
where $x_1$ represents the space-time point that a field $\phi$ is ``pulled out'' of the vacuum (source), 
or ``put into'' the vacuum (sink).

The amplitude $Z$ can now be seen as a function of the sources and sinks, and by noticing that
$\mathcal{L} = \mathcal{L}_0 + \mathcal{L}_{int} + J\phi$, we can write
\begin{eqnarray}
Z = \int D\phi e^{i \int (d^4x \mathcal{L}_0 + \mathcal{L}_{int} + J\phi)} \nonumber.
\end{eqnarray}
We are almost done! We are simply going to expand $Z$ in terms of $J$, and then
expand $Z$ in terms of the interaction to get:
\begin{eqnarray}
Z & = & \sum_{j = 1}^{\infty} \sum_{k = 1}^{\infty} Z(j,k) 
 =  \int D\phi e^{i \int d^4x \mathcal{L}_0} \times \nonumber \\
& & \sum_{j = 1}^{\infty} \frac{1}{j!} \left[i \int d^4x J\phi \right]^j
\sum_{k = 1}^{\infty} \frac{1}{k!} \left[i \int d^4x \mathcal{L}_{int} \right]^k \nonumber .
\end{eqnarray} 

Let us see what we get when we take $J = \delta^{(4)}(x - a) + \delta^{(4)}(x - b) + \delta^{(4)}(x - c)$,
and $\mathcal{L}_{int} = -g \phi^3/(3!)$ to the order $j = 3$, and $k = 1$:
\begin{eqnarray}
Z(3,1) & = & -i\frac{g}{3!} \int d^4x \int D\phi e^{i \int d^4x \mathcal{L}_0} \phi(x) \phi(x) \phi(x) \times \nonumber \\
& & \frac{1}{3!} \left[i \int d^4x J\phi \right]^3 \nonumber .
\end{eqnarray}
The first thing that we notice is that the terms in $J$ are going to give us a mess! 

We can fix the terms in $J$ by noticing that combination like $[\delta^{(4)}(x - a)]^3$, and
$[\delta^{(4)}(x - a)]^2 \delta^{(4)}(x - b)$ will describe a process where one or more
of our external lines share the same space-time point of origin or ``death''. We are not interested
in processes that have external lines that share a common beginning or end, so we only consider
the term $3!\: \delta^{(4)}(x - a) \delta^{(4)}(x - b) \delta^{(4)}(x - c)$. Now we have
\begin{eqnarray}
Z(3,1) & = &  -\frac{(i)^4}{3!} \int d^4x \int D\phi e^{i \int d^4x \mathcal{L}_0} \times \nonumber \\
& & \phi(x) \phi(x) \phi(x) 
\phi(a) \phi(b) \phi(c) \label{eq:a1} .
\end{eqnarray}

To evaluate the integral in Eqn. \ref{eq:a1} we need an identity that I will state without proof:
\begin{eqnarray}
& & \frac{(i)^3}{Z(0,0)} \int D\phi e^{i \int d^4x \mathcal{L}_0} \times \nonumber 
 \phi(x) \phi(x) \phi(x)
\phi(a) \phi(b) \phi(c) \nonumber \\
& = &
D(a - x)D(b - x)D(c - x) +  D(a - x)D(c - x)D(b - x)  \nonumber \\
& + & D(b - x)D(c - x)D(a - x) + D(b - x)D(c - x)D(a - x)  \nonumber \\
& + & D(c - x)D(b - x)D(a - x) + D(c - x)D(a - x)D(b - x)  \nonumber \\
& + & D(x - x)D(a - b)D(c - x) + \cdots \label{eq:a2}  ,
\end{eqnarray}
where I have connected field pairs in every possible way to construct the 
propagators. 

The first $3!$ (six) terms of Eqn. \ref{eq:a2} can be represented by the single Feynman diagram
shown in Fig. \ref{fig:f7}. Whereas, a term like $D(x - x)D(a - b)D(c - x)$ will yield a ``disconnected''
diagram as shown in Fig. \ref{fig:f8}.

\begin{figure}[t]
\includegraphics[scale=0.4]{f7.eps}
\caption{A Feynman diagram for the first $3!$ (six) terms in Eqn. \ref{eq:a2}.
\label{fig:f7}}
\end{figure}

\begin{figure}[t]
\includegraphics[scale=0.4]{f8.eps}
\caption{A Feynman diagram for $D(x - x)D(a - b)D(c - x)$.
\label{fig:f8}}
\end{figure}

We are not interested in the disconnected diagrams, so we can now write
\begin{eqnarray}
Z & = &\frac{Z(3,1)}{Z(0,0)} = \nonumber \\
 & & -i g \int d^4x D(a - x)D(b - x)D(c - x) \nonumber .
\end{eqnarray}

\end{document}
