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\begin{document}
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\noindent
\centerline{{\bf TRIUMF Experiment 497}} \\ \\
{\bf Measurement of the Flavour Conserving Hadronic
Weak Interaction} \\ \\
A.R. Berdoz$^{\rm 6}$,
\underline{J. Birchall$^{\rm 1}$},
J.D. Bowman$^{\rm 2}$,
\underline{J.R Campbell$^{\rm 1}$},
\underline{C.A Davis$^{\rm 4}$},\\
\underline{A.A. Green$^{\rm 1}$},
P.W. Green$^{\rm 3}$,
\underline{A.A Hamian$^{\rm 1}$},
D.C. Healey$^{\rm 4}$,
R. Helmer$^{\rm 4}$,
W.Kellner$^{\rm 4}$, \\
Y.Kuznetsov$^{\rm 5}$,
\underline{L.Lee$^{\rm 1}$},
C.D.P Levy$^{\rm 4}$. \\ \\
$^{\rm 1}$ University of Manitoba \\
$^{\rm 2}$ Los Alamos National Laboratory \\
$^{\rm 3}$ University of ALberta \\
$^{\rm 4}$ TRIUMF \\
$^{\rm 5}$ Institute of Nuclear Research \\
$^{\rm 6}$ Carnegie Mellon University \\ \\ \\
{\bf Introduction and Background} \\ \\
The outgoing parity violation experiment at
TRIUMF$^{\rm 1}$, in which the University
of Manitoba group plays a leading role, will
determine the parity-violation longitudinal
analyzing power
$A_{z}=(\sigma^{+}-\sigma^{-})/
(\sigma^{+}+\sigma^{-})$
in p-p elastic scattering at 221MeV ,
where $\sigma^{+}$ and $\sigma^{-}$
are the scattering cross sections for positive and
negative helicity. The goal of E497 is to measure
$A_{2}$ with the precision of
$\pm 0.2 \times 10^{-7}$. \\ \\
At a given energy, the sign and magnitude of
$A_{z}$ are determined by the weak interaction,
but the variation of $A_{z}$ with angle
is determined by the strong interaction.
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\begin{itemize}
\item Morning
\item Noon
\item Afternoon
\begin{enumerate}
\item Coffee
\item Sandwich
\item Biscuit
\end{enumerate}
\item \LaTeX\ permits nice documents
\item Night
\end{itemize}
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\section{Integrals of complex time and the partition function}
The partition function of a particle can be written as
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\begin{equation}
Z=Tr e^{-\beta \hat{H}} = \int dx \langle
x|e^{-\beta \hat{H}}|x \rangle
\label{imaginary time}
\end{equation}
%=========================================
Someone can then think that an addition on all the diagonal
elements of the matrix of the operator of imaginary time.
%==================
\begin{equation}
U \left(x_{f} \tau_{f},x_{i} \tau_{i} \right) =
\langle x_{f}
| e^{-\left(\tau_{f}-\tau_{i}\right)\frac{\hat{H}}{\hbar}}
|x_{i} \rangle
\label{giou}
\end{equation}
%==================
estimated in the interval $\tau_{f}-\tau_{i}=\beta \hbar$.
All the steps of calculation of the integral for real times, can
be repeated in the case of imaginary time [4].\\
For the hamiltonian $H_{s}$ we have:
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\begin{eqnarray}
U\left(x_{f} \tau_{f},x_{i} \tau_{t} \right) &=&
\lim_{M \rightarrow \infty} \int
\prod_{\kappa=1}^{M-1} d^3 x_{\kappa}
\prod_{\kappa=1}^{M} \langle x_{\kappa}|
e^{-\frac{\epsilon}{\hbar} H \left( {\bf \hat{P}} ,{\bf \hat{x}} \right)}
|x_{\kappa -1} \rangle \nonumber \\
%-------------------------------------
&=& \lim_{M \rightarrow \infty} \int
\prod_{\kappa=1}^{M-1} d^3 x_{\kappa}
\prod_{\kappa=1}^{M} d^3 P_{\kappa}
\langle x_{\kappa}| P_{\kappa} \rangle
\langle P_{\kappa} | :
e^{-\frac{\epsilon}{\hbar} H \left( {\bf \hat{P}} ,{\bf \hat{x}} \right)}
:+ \vartheta \epsilon^{2} |x_{\kappa-1} \rangle \nonumber \\
%-----------------------------------------
&=& \lim_{M \rightarrow \infty} \int
\prod_{\kappa=1}^{M-1} d^3 x_{\kappa}
\prod_{\kappa=1}^{M}
\frac{d^3 P_{\kappa}}{\left( 2 \pi \hbar \right)^3}
e^{\sum^{M}_{\kappa=1} \bigg[
\frac{i P_{\kappa}}{\hbar} \left( x_{\kappa}-x_{\kappa-1} \right)
-\frac{\epsilon}{\hbar} \left( \frac {P^{2}_{\kappa}}{2m} +
V \left( x_{\kappa-1} \right) \right) \bigg] } \nonumber \\
%-----------------------------------------------
&=& \lim_{M \rightarrow \infty} \int
\prod_{\kappa=1}^{M-1} d^3 x_{\kappa}
\left( \frac{m}{2 \pi \epsilon \hbar} \right)^{\frac{3M}{2}}
e^{-\frac {\epsilon}{\hbar} \sum_{\kappa=1}^{M}
\Bigg [ \frac{m}{2} \frac{ \left( x_{\kappa}-x_{\kappa-1} \right)^2}
{{\epsilon}^2}
+V \left( x_{\kappa -1} \right) \Bigg] } \nonumber \\
%---------------------------------------------------
&=& \int_{\left( x_{i} , \tau_{i} \right)}^{\left( x_{f} , \tau_{f} \right)}
D \bigg[ x \left( \tau \right) \bigg]
e^{- \frac{1}{\hbar} \int_{\tau_{i}}^{\tau_{f}} d \tau
\left( \frac{m}{2} \left( \frac{dx \left( \tau \right)}{d \tau} \right)^2
+V \left( x \left( \tau \right) \right) \right) } \nonumber \\
%------------------------------------------------------
&=& \int_{\left( x_{i} , \tau_{i} \right)}^{\left( x_{f} , \tau_{f} \right)}
D \bigg[ x \left( \tau \right) \bigg]
e^{- \frac{1}{\hbar} \int_{\tau_{i}}^{\tau_{f}} d \tau
L \bigg[ x \left( \tau \right) \Bigg] }
%-----------------------------------------------------
\label{makrinari}
\end{eqnarray}
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Where $ \epsilon= \frac{1}{M} \left( \tau_{f} -\tau_{i} \right)$.
So the integral at imaginary times is the sum through all the
roads that start from $x_{i}$ at time $\tau_{i}$ and finish at time
$x_{f}$ at time $\tau_{f}$ of the exponation at which cause of the
change of sign of the kinetic energy we end up in the lagragian
at the Eucledian Space.
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This is a table that I would like to create:
\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline
{\bf {Monday}} & {\bf{Tuesday}} & {\bf{Wednesday}}
& {\bf{Thursday}} \\ \hline
10 & 10 & 12 & 14 \\ \hline
11 & 9 & 25 & 903 \\ \hline
4 & 3 & 593 & 1003 \\ \hline
1005 & 9876 & 6431 & 5351 \\ \hline
\end{tabular}
\end{center}
Is this table a nice one? I hope I was successful.
We will see about this.
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And now I am going to create three matrices. Look at this:
\begin{equation}
\lambda_{1}=
\left( \begin{array}{ccc}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 0
\end{array} \right) , \hspace{.5in}
\lambda_{2}=
\left( \begin{array}{ccc}
0 & -i & 0 \\
i & 0 & 0 \\
0 & 0 & 0
\end{array} \right) , \hspace{.5in}
\lambda_{3}=
\left( \begin{array}{ccc}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 0
\end{array} \right)
\label{matrices}
\end{equation}
I hope that now it is going to be right. We will see about it.
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\section{The fermionic table}
\subsection{The eigenvalues of the fermionic table}
\underline{This is for spaces}.
\begin{equation}
ABCD=3D^{q}_{t} PR_{trd} FRT
\label{spaces1}
\end{equation}
\begin{equation}
A \, B \: C \; D=3 \, D^{q}_{t} \, P \:
R_{t \, r \: d} F \; R \; T
\label{spaces2}
\end{equation}
Matrix Q is equal to :
\begin{equation}
Q=\sum_{\mu=0,1,2,3}
\left[ \left( 1-\gamma_{\mu} \right) U_{\mu} (n)
\delta_{n,n'-\hat{\mu}}
+\left( 1+\gamma_{\mu} \right) U_{\mu}^{+}
\left( n' \right) \delta_{n,n'+\hat{\mu}} \right]
\label{Q}
\end{equation}
is usually called {\bf hopping matrix}.
We set operator ${\bf \Xi}$ so that
\begin{equation}
{\bf \Xi} (t,x,y,z) = \left\{
\begin{array}{rl}
-1 & \mbox{For odd sites of the lattice} \\
1 & \mbox{For even sites of the lattice}
\end{array}
\right.
\label{xi}
\end{equation}
For the operator ${\bf \Xi}$ it can be said that
\begin{equation}
{\bf \Xi}^{2}={\bf I} \; and \; {\bf \Xi}^{+}={\bf \Xi}
\label{idiot}
\end{equation}
If operator Q is acted on both right and
left side we have:
\begin{equation}
{\bf \Xi} \, Q \, {\bf \Xi}=-Q
\label{result}
\end{equation}
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