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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:

%=======================

\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}

%======================================

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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\end{document}

 

 

 

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