%         Created on  11-JANUARY-1996 by Phillip Gutierrez
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\rhead{P. Gutierrez}
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\cfoot{Lecture 21-\thepage}
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\def\reff{r_{\text{eff}}}
\def\ko{$K^0$}
\def\kob{$\bar{K}^0$}
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\begin{document}

\begin{center}
\LARGE\bf Lecture 21
          Physics 4213/5213
\end{center}
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\section{Introduction}
The combined symmetry of $CP$ has been introduced
since, this was found to be a 
good symmetry for the neutrino, while the individual symmetries $C$ and $P$ were found to 
not be good symmetries. Therefore, the assumption is made that $CP$ is a better 
symmetry for weak interactions than the two individually and that it is a conserved 
quantity. This assumption will be used to study the $K^0$--$\bar{K}^0$  system.
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\section{The Neutral Kaon}
The neutral kaons, $K^0$ and \kob, both decay into two pions and three pions through the weak 
interaction. Since they decay to the same states, a valid question to ask, 
is how can they be distinguished. Both these particles contain a strange 
quark---\ko($\bar{s}d$) and \kob($\bar{d}s$)---being particle a,ni-particle they
have opposite strangeness, implying that 
they should be produced in different reactions. These reactions correspond to:
\begin{align}
  \pi^-+p &\rightarrow \Lambda + K^0 \notag \\
  \pi^-+p &\rightarrow \bar{\Lambda} + \bar{K}^0 +n+n
\end{align}
The two neutrons in the second reaction are needed to conserve baryon number. Because 
these processes involve the strong interactions, strangeness must be conserved---the 
strong interaction is blind to quark flavor and therefore cannot change it.

\begin{figure}[h]
\begin{center}
  \epsfig{file=kpi.eps}
  \caption{Transition from $K^0$ to $\bar{K}^0$ through virtual pions.\label{fig:kpi}}
\end{center}
\end{figure}

Given that they are distinguishable and decay into the same states, they might be able 
to mix. That is a $K^0$  can transform itself into a $\bar{K}^0$  through the exchange of 
virtual pions
(see fig \ref{fig:kpi}).
The mixing must come about through the weak interaction, since there is a change of 
strangeness ($\Delta S=2$). Therefore, in the quark model and using the exchange of the $W$, 
the Feynman diagram that correspond to this process is (see fig \ref{fig:wkk}).
%
\begin{figure}
\begin{center}
  \epsfig{file=k0_k0b.eps}
  \caption{Kaon oscillations as view from the quark model and the theory of weak interactions.
    \label{fig:wkk}}
\end{center}
\end{figure}
\section{$CP$ of the Kaon---The Quark Model}
The parity and charge conjugation quantum numbers of the two kaons can be determined from
the quark model. The kaons are the lightest neutral mesons containing a strange quark. This
implies that they are spin zero mesons, with the quarks having zero relative orbital angular
momentum. The parity is then given by $(-1)^{\ell+1}=-1$ for both of the kaons.

As far as charge conjugation goes, neither of the two are eigen-states. The reason for this
is that the flavor of the meson (this refers to the strangeness in this case) changes under
the charge conjugation operator: 
\begin{align}
  C|K^0\rangle &= |\bar{K}^0\rangle \notag \\
  C|\bar{K}^0\rangle &= |K^0\rangle
\end{align}
Therefore 
neither of the two kaons is an eigen-state of $CP$.
\section{$CP$ of the Pions}
Start with the decays $K^0 \rightarrow \pi^0+\pi^0$ and $\bar{K}^0 \rightarrow \pi^0+\pi^0$; here
nothing will be assumed about the spin of the kaon. Since the system is composed of
identical bosons, the system must be symmetric under particle exchange. This implies that
the parity of the system must be even since parity and particle exchange are equivalent
operations; the intrinsic parity of each pion is $-1$, therefore the combined intrinsic
parity of the two pions is $+1$. 
Since the system is even under parity, the angular momentum must be even. If the angular
momentum has an even value greater than or equal to two, then the decay $K^0\rightarrow
\pi^0+\gamma$ should occur; note this could never be in a state of zero total spin, since
by taking the $z$-axis as the photon direction $s_z=\pm1$. This
decay has never been seen, therefore it ia assumed that 
the spin of the $K^0$ and similarly for the $\bar{K}^0$ must
be zero.

The charge conjugation of the two $\pi^0$'s is given by the intrinsic charge conjugation
of the two
pions and is therefore $+1$. This leads to $CP=+1$ for the two pion system. The parity
of the two charged pions is given by $(-1)^\ell$ as is the charge conjugation (since the
charge conjugation inverts the signs of the particles, this being equivalent to rotating
the system by $\pi$). This
leads to $CP=(-1)^{2 \ell}=+1$.

Next, determine $CP$ for the decays $K^0 \rightarrow \pi^0+\pi^0+\pi^0$ and
$\bar{K}^0 \rightarrow \pi^0+\pi^0+\pi^0$. The charge conjugation is again just given
by the intrinsic charge conjugation of the three pions ($+1$). The parity is given by
the intrinsic parity of the three pions $(-1)^3=-1$ and the spatial part is derived by
knowing that the $K^0$ has spin zero. In this case, the relative angular momentum about
the common center of mass of
two of $\pi^0$ is given by $\ell$. The relative orbital angular momentum about the
common center of mass of the $2\pi^0$ system and the other $\pi^0$ is $L$. Since the
total angular momentum is in the range $L=\ell \rightarrow |L-\ell|$, implying $\ell=L$.
Therefore the parity is given
by $(-1)(+1)=-1$, which finally gives $CP=-1$. Again similar arguments can be used to determine 
$CP$ for the decay modes $K^0\rightarrow \pi^++\pi^-+\pi^0$ and
$\bar{K}^0\rightarrow \pi^++\pi^-+\pi^0$, except that here the fact that the 
relative orbital angular momentum between the various pions is zero must be used (this being
an experimental fact).
\begin{center}
\begin{tabular}{|l|r|r|r|} \hline
  Pion State & P & C & CP \\ \hline \hline
  $\pi^+\pi^-$ & $(-1)^\ell$& $(-1)^\ell$ & 1\\
  $\pi^0 \pi^0$ & 1 &1& 1\\
  $\pi^+\pi^-\pi^0$ & $-1$ & $(-1)^0$& $-1$\\
  $\pi^0\pi^0\pi^0$& $-1$&1&$-1$\\ \hline
\end{tabular}
\end{center}
\section{Eigen-states of $CP$}
Earlier in this lecture, it was pointed out that the two kaons can mix, that is oscillate
between the two. It has also been pointed out that the kaons are not states with a definite
$CP$ but do have a definite strangeness. Further, based on the $CP$ properties of the neutrino
it is assumed that even thought $C$ and $P$ are not conserved, the combined operation $CP$
might be conserved. Also, in weak interactions strangeness is not conserved, strangeness
is a property of strong interactions. Therefore, from all this information it might be
inferred that the weak interaction states are states with a definite $CP$ while the 
strong interaction states are states with definite strangeness. The $CP$ states can be written
as:
\begin{align}
  |K_1\rangle &= \frac{1}{\sqrt{2}} \left[|K^0\rangle - |\bar{K}^0\rangle\right] \Rightarrow CP=+1
  \notag \\
  |K_2\rangle &= \frac{1}{\sqrt{2}} \left[|K^0\rangle + |\bar{K}^0\rangle\right] \Rightarrow CP=-1
\end{align}
where the eigen-value of $CP$ on the kaons is $-1$ since the parity is odd ($-1$). Given these
states, the decay modes are:
\begin{alignat}{2}
K_1 & \rightarrow \pi^+\pi^- & \qquad K_1 & \rightarrow \pi^0\pi^0 \notag \\
K_2 & \rightarrow \pi^+\pi^-\pi^0 & \qquad K_2 & \rightarrow \pi^0\pi^0\pi^0
\end{alignat}

Given that the $K_1$ decays to two pions and the $K_2$ decays to three pions, it might
be expected that the $K_1$ has a shorter lifetime, it being easier to create two rather
than three pions. In fact this is what is observed. The lifetime for the $K_1$ is
\nunit{0.9\times10^{-10}}{s}, while that for the $K_2$ has a lifetime of
\nunit{0.5\times10^{-7}}{s}; the $K_1$ is usually referred to as the $K_S$ ($K$-short) and the
$K_2$ as the $K_L$ ($K$-long).
\section{$CP$ Violations}
In 1964 Christenson, {\bf Cronin}, {\bf Fitch} and Turley, discovered that on 
occasion the $K_L$  decays to two pion---a definite violation of  $CP$:
\begin{equation}
\frac{\text{Amplitude}\left(K_L\rightarrow \pi^+\pi^-\right)}{\text{Amplitude}
  \left(K_S\rightarrow \pi^+\pi^-\right)} = 2\times10^{-3}
\end{equation}
This violation can have two sources, one is that the $K_L$  is really not an 
eigenstate of $CP$ as is assumed and is really:
\begin{equation}
  |K_L\rangle =\left(\frac{1}{\sqrt{1+\epsilon^2}}\right) \left[|K_2\rangle +
    \epsilon |K_1\rangle\right]
\end{equation}
where $\epsilon$  is a small parameter. The second is that $CP$  is not 
conserved---direct $CP$  violation $K_2$ decaying to two pions.
Which of the two is valid is at present not 
known. Experiments in this area are being conducted but these are not simple 
experiments. Currently the errors on the measurements are not small enough to 
distinguish between the two cases.

As a final comment on $CP$ violation, a second mode is also seen. This
mode is as follows:
\begin{align}
  K_L &\rightarrow \pi^+ + e^- + \bar{\nu}_e \notag \\
      &\rightarrow \pi^- + e^+ \nu_e.
\end{align}
Note that the two equations are $CP$ conjugates of each other, which implies
that the decay rates should be the same. In fact the decay rates are different
and allow one to distinguish matter from anti-matter:
\begin{equation}
  \frac{\Gamma(\pi^+ + e^-+\bar{\nu}_e) -\Gamma(\pi^-+e^+ +\nu_e)}
     {\Gamma(\pi^+ + e^-+\bar{\nu}_e) +\Gamma(\pi^-+e^+ +\nu_e)}=
    3.3\times10^{-3}.
\end{equation}
 This has been stated
as a necessary condition for the universe to have a matter anti-matter asymmetry.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\vfill
%\clearpage
%\section{Outline}
%\begin{enumerate}
%  \item Parity of the photon
%  \begin{enumerate}
%    \item Parity of a single free photon
%    \item Parity of a radiated photon
%  \end{enumerate}
%\end{enumerate}
%\begin{thebibliography}{9}
%  \bibitem{r1} R
%\end{thebibliography}
\end{document}