       %prepared in LaTeX   
   
   
	\hfuzz = 100pt   
   
        \newcommand{\hf}{$\frac{1}{2}$}
        \newcommand{\third}{\frac{1}{3}}
        \newcommand{\minustwothirds}{-\frac{2}{3}}
        \newcommand{\fourthirds}{\frac{4}{3}}
        \newcommand{\ISpin}{{\rm ISpin}}   
        \newcommand{\End}{{\rm End}}

        \newcommand\F{{\bf F}}   
        \newcommand\G{{\bf G}}   
        \newcommand\HH{{\bf H}}   

        \newcommand\Z{{\mathbb Z}}   
	\newcommand\R{{\mathbb R}}   
	\newcommand\C{{\mathbb C}}   
        \renewcommand\H{{\mathbb H}}   
        \newcommand\K{{\mathbb K}} 
	\renewcommand\O{{\mathbb O}}   
           
        \newcommand\RP{{\mathbb {RP}}}   
        \newcommand\CP{{\mathbb {CP}}}   
        \newcommand\HP{{\mathbb {HP}}}   
        \newcommand\KP{{\mathbb {KP}}}   
        \newcommand\OP{{\mathbb {OP}}}   
        \renewcommand\P{{\mathbb P}}   
   
        \newcommand{\Cliff}{{\rm Cliff}}   
        \newcommand{\J}{{\rm J}} 
        \newcommand{\M}{{\rm M}}
  
	\renewcommand\S{{\rm S}}      
        \newcommand{\OO}{{\rm O}}   
	\newcommand{\SO}{{\rm SO}}   
	\newcommand{\SL}{{\rm SL}}   
        \newcommand{\PSL}{{\rm PSL}}   
	\newcommand{\SU}{{\rm SU}}   
        \newcommand{\Sp}{{\rm Sp}}
        \newcommand{\Spin}{{\rm Spin}}   
        \newcommand{\Pin}{{\rm Pin}}   
	\newcommand{\U}{{\rm U}}   
	\newcommand{\E}{{\rm E}}   

	\newcommand{\so}{{\mathfrak {so}}}   
	\newcommand{\Sl}{{\mathfrak {sl}}}   
	\newcommand{\symp}{{\mathfrak {sp}}}   
	\renewcommand{\u}{{\mathfrak {u}}}   
	\newcommand{\e}{{\mathfrak {e}}}   
	\newcommand{\f}{{\mathfrak {f}}}   
	\newcommand{\g}{{\mathfrak {g}}}   
	\newcommand{\su}{{\mathfrak {su}}}   
        \renewcommand{\a}{{\mathfrak {a}}}   
        \newcommand{\sa}{{\mathfrak {sa}}}   
        \newcommand{\h}{{\mathfrak {h}}}   
        \newcommand{\sh}{{\mathfrak {sh}}}   
        \newcommand{\coll}{{\mathfrak {coll}}}   
        \newcommand{\isom}{{\mathfrak {isom}}}   
        \newcommand{\Isom}{{\rm Isom}} 
        \newcommand{\Der}{{\mathfrak {der}}}   
        \newcommand{\Tri}{{\mathfrak {tri}}}  

   
	\newcommand{\et}{\hspace{-0.08in}{\bf .}\hspace{0.1in}}   
        \newcommand{\DOT}{\hspace{-0.08in}{\bf .}\hspace{0.1in}}   
 	\newcommand{\BOX}{\hbox {$\sqcap$ \kern -1em $\sqcup$}}   
	\newcommand{\qed}{\hskip 3em \hbox{\BOX} \vskip 2ex}   
   
        \renewcommand{\Re}{{\rm Re}}   
        \renewcommand{\Im}{{\rm Im}}   
	\newcommand{\Hom}{{\rm hom}}   
        \newcommand{\Aut}{{\rm Aut}}   
	\newcommand{\Inv}{{\rm Inv}}   
	\newcommand{\tensor}{\otimes}   
        \newcommand{\iso}{\cong}   
        \newcommand{\implies}{\Longrightarrow}   
        \newcommand{\impliedby}{\Longleftarrow}   
   
	\newcommand{\be}{\begin{equation}}   
        \newcommand{\ee}{\end{equation}}   
        \newcommand{\ba}{\begin{eqnarray}}   
        \newcommand{\ea}{\end{eqnarray}}   
        \newcommand{\ban}{\begin{eqnarray*}}   
        \newcommand{\ean}{\end{eqnarray*}}   
  
        \newcommand{\ad}{{\rm ad}}   
	\newcommand{\Cob}{{\rm Cob}}   
	\newcommand{\Vect}{{\rm Vect}}   
	\newcommand{\Hilb}{{\rm Hilb}}   
        \newcommand{\maps}{\colon}   
	\newcommand{\tr}{{\rm tr}}   
	\newcommand{\rank}{{\rm rank}}   
	\newcommand{\id}{{\rm id}}   
	\newcommand{\Fun}{{\rm Fun}}   
	\newcommand{\Ad}{{\rm Ad}}   
	\newcommand{\om}{\omega}   
	\newcommand{\eps}{\epsilon}  
 
	\newtheorem{thm}{Theorem}   
	\newtheorem{prop}{Proposition}   
	\newtheorem{lemma}{Lemma}   
	\newtheorem{corollary}{Corollary}   
	\newtheorem{ex}{Example}   
 
\documentclass{article} 
\usepackage{amsfonts,amssymb}
%\usepackage{latexsym}
\usepackage{diagram}   

	\textwidth 6in   
	\textheight 8.5in	\evensidemargin .25in   
	\oddsidemargin .25in   
	\topmargin .25in   
	\headsep 0in   
	\headheight 0in   
	\footskip .5in   
%If you want single spaced copy, delete the next two lines.   
%	\parskip 1.75\parskip plus 3pt minus 1pt   
%	\renewcommand{\baselinestretch}{1.5}   
	\pagestyle{plain}   
	\pagenumbering{none}   
	\begin{document}   
 
	\begin{center}   
	{\bf The Eightfold Way \\}   
	{\small John\ C.\ Baez, May 27 2003\\}   
	\vspace{0.3cm}   
        \end{center}

It is natural to group the eight lightest mesons into 4 
irreps of isospin $\SU(2)$ as follows:

\vskip 1em
\begin{center}{
\begin{tabular}{|l|c|c|c|}                    \hline   
mesons & $I_3$ & $Y$ & $Q$ \\  \hline
{\bf pions}    &     &        &        \\   
(complexified adjoint rep) &     &        &        \\   
$\pi^+$    &  +1     &   0      &  +1      \\   
$\pi^0$    &  0     &   0      &  0      \\   
$\pi^-$    &  -1    &   0      &  -1     \\   \hline
{\bf kaons}    &     &        &        \\   
(defining rep) &     &        &        \\   
$K^+$    &  +1/2     &   1      &  +1      \\   
$K^0$    &  -1/2     &   1      &  0      \\   \hline
{\bf antikaons}    &     &        &        \\   
(dual of defining rep) &     &        &        \\   
$\overline{K^0}$    &  +1/2     & -1      &  0      \\   
$K^-$               &  -1/2     & -1      &  -1      \\   \hline
{\bf eta}    &     &        &        \\   
(trivial rep) &     &        &        \\   
$\eta$   &  0     & 0      &  0      \\   \hline
\end{tabular}}
\end{center}

\vskip 1em

\noindent
(The dual of the defining rep of $\SU(2)$ is 
isomorphic to the defining rep, but it's always nice to think of
antiparticles as living in the {\it dual} of the rep that the
corresponding particles live in, so above I have said that
the antikaons live in the dual of the defining rep.)

In his theory called
the {\bf Eightfold Way}, Gell-Mann showed that these eight
mesons could be thought of as a basis for the 
the complexified adjoint rep of $\SU(3)$ --- that is,
its rep on the 8-dimensional complex Hilbert space
\[        \su(3) \tensor \C \iso \Sl(3,\C)  .\]
He took seriously the fact that 
\[        \Sl(3,\C) \subset \C[3] \iso \C^3 \tensor (\C^3)^* \]
where $\C^3$ is the defining rep of $\SU(3)$ and $(\C^3)^*$
is its dual.  Thus, he postulated particles called {\bf quarks}
forming the standard basis of $\C^3$:
\[      u = \left( \begin{array}{c}
                   1 \\
                   0 \\
                   0 \\
              \end{array}\right) ,  \qquad
   d = \left( \begin{array}{c}
                   0 \\
                   1 \\
                   0 \\
              \end{array}\right) ,  \qquad
   s = \left( \begin{array}{c}
                   0 \\
                   0 \\
                   1 \\
              \end{array}\right) ,
\]
and {\bf antiquarks} forming the dual basis of $(\C^3)^*$:
\[  \overline{u} = \left( \begin{array}{ccc}
                   1 &  0 &  0 
              \end{array}\right) ,  \qquad
   \overline{d} = \left( \begin{array}{ccc}
                   0 &
                   1 &
                   0 
              \end{array}\right) ,  \qquad
  \overline{s} = \left( \begin{array}{ccc}
                   0 &
                   0 &
                   1 
              \end{array}\right) .
\]
This let him think of the eight mesons as being built from quarks
and antiquarks.  For example, the positive pion corresponds to the matrix
\[    \pi^+ = \left( \begin{array}{ccc} 
                       0 & 1 & 0 \\
                       0 & 0 & 0 \\
                       0 & 0 & 0 \\
               \end{array} \right) \in \Sl(3,\C), \]
but as an element of $\C^3 \tensor (\C^3)^*$ this is just
\[              \left( \begin{array}{c}
                   1 \\
                   0 \\
                   0 \\
              \end{array}\right)  \tensor 
              \left( \begin{array}{ccc}
                   0 &
                   1 &
                   0 
              \end{array}\right) = u \tensor \overline{d}  \]
so he said the positive pion is built from an up quark
and an down antiquark!  After decades of experiment,
we now have lots of evidence that quarks
really exist and that this is true.

\vskip 2em
\centerline{\it Now a little work for you....}

\vskip 2em
\noindent
1. Determine in a similar way how the other mesons are built
from quarks and antiquarks, and fill in this chart:

\[   \begin{array}{cll}

   \pi^+ &=& \left( \begin{array}{ccc} 
                       0 & 1 & 0 \\
                       0 & 0 & 0 \\
                       0 & 0 & 0 \\
               \end{array} \right) = u \tensor \overline{d}  \\

    \pi^0 &=\frac{1}{\sqrt{2}}&\left( \begin{array}{ccc} 
                       1 & 0 & 0 \\
                       0 & -1 & 0 \\
                       0 & 0 & 0 \\
               \end{array} \right) =  \\

    \pi^- &=& \left( \begin{array}{ccc} 
                       0 & 0 & 0 \\
                       1 & 0 & 0 \\
                       0 & 0 & 0 \\
               \end{array} \right) =  \\

    K^+ &=& \left( \begin{array}{ccc} 
                       0 & 0 & 1 \\
                       0 & 0 & 0 \\
                       0 & 0 & 0 \\
               \end{array} \right) =  \\

    K^0 &=& \left( \begin{array}{ccc} 
                       0 & 0 & 0 \\
                       0 & 0 & 1 \\
                       0 & 0 & 0 \\
               \end{array} \right) =  \\

    \overline{K^0} &=& \left( \begin{array}{ccc} 
                       0 & 0 & 0 \\
                       0 & 0 & 0 \\
                       0 & 1 & 0 \\
               \end{array} \right) =  \\

    K^- &=& \left( \begin{array}{ccc} 
                       0 & 0 & 0 \\
                       0 & 0 & 0 \\
                       1 & 0 & 0 \\
               \end{array} \right) =   \\

    \eta &= \frac{1}{\sqrt{6}} &\left( \begin{array}{ccc} 
                       -1 & 0 & 0 \\
                       0 & -1 & 0 \\
                       0 & 0 & 2 \\
               \end{array} \right) =  \\
\end{array}
\]

\vskip 1em
\noindent 
2. In the Eightfold Way, whenever we have particles living
in some rep $\rho$ of $SU(3)$, the third component of weak
isospin corresponds to the operator
\[         I_3 = d\rho \left( \begin{array}{ccc} 
                       1/2 & 0 & 0 \\
                       0 & -1/2 & 0 \\
                       0 & 0 & 0 \\
               \end{array} \right)   \]
while hypercharge corresponds to the operator
\[         Y = d\rho \left( \begin{array}{ccc} 
                       1/3 & 0 & 0 \\
                       0 & 1/3 & 0 \\
                       0 & 0 & -2/3 \\
               \end{array} \right)  . \]
As usual, electric charge corresponds to the operator
\[       Q = I_3 + \frac{Y}{2}  .\]
By calculating how the above operators
act on the standard basis of the defining rep, fill out
the following table of eigenvalues:

\vskip 1em
\begin{center}{
\begin{tabular}{|l|c|c|c|}                    \hline   
quarks & $I_3$ & $Y$ & $Q$ \\  \hline
$u$    &      &         &        \\  \hline   
$d$    &      &         &       \\   \hline
$s$    &      &         &       \\   \hline
\end{tabular}}
\end{center}

\vskip 1em
\noindent
3.  Figure out the same information for the antiquarks:

\vskip 1em
\begin{center}{
\begin{tabular}{|l|c|c|c|}                    \hline   
antiquarks & $I_3$ & $Y$ & $Q$ \\  \hline
$\overline{u}$    &      &         &        \\   \hline
$\overline{d}$    &      &         &       \\    \hline
$\overline{s}$    &      &         &       \\   \hline
\end{tabular}}
\end{center}
\vskip 1em
{\it Hint: Unlike for $\SU(2)$,
the defining representation of $\SU(3)$ is {\it not} isomorphic
to its dual.  So, you should go back to the formula for the
dual of a Lie algebra rep.  If we have a rep of some Lie 
group
\[        \rho \maps G \to \End(V)  \]
then the dual rep
\[        \rho^* \maps G \to \End(V^*)  \]
is defined by
\[         (\rho^*(g)f)(v) = f(\rho(g^{-1})v)  \]
for all $g \in G$, $f \in V^*$ and $v \in V$.   Differentiating
all this, we get a Lie algebra rep
\[        d\rho \maps \g \to \End(V)  \]
whose dual rep
\[        d\rho^* \maps \g \to \End(V^*)  \]
is given by
\[         (d\rho^*(x)f)(v) = -f(d\rho(x)v)  \]
for all $x \in \g$, $f \in V^*$ and $v \in V$.   Taking 
$\rho$ to be the defining rep of $\SU(3)$ on $\C^3$, you 
can use this last formula to work out what $I_3, Y$ and $Q$
do to antiquarks, which form a basis of the dual space
$(\C^3)^*$.   The final answer should be 
intuitively obvious, but it's good to see how it comes
out of the math!}

\vskip 1em
\noindent
4.  By filling out the following chart,
check that you can compute $I_3$, $Y$ or $Q$ for any meson simply
by adding these quantities for the quarks and antiquarks it is 
built from.  For example, the $\pi^+$ is built from a $u$
and a $\overline{d}$.  The $u$ has $I_3 = \frac{1}{2}$ and
the $\overline{d}$ also has $I_3 = \frac{1}{2}$; adding these
we get $I_3 = 1$ for the $\pi^+$.  

\vskip 1em
\begin{center}{
\begin{tabular}{|l|c|c|c|c|}                    \hline   
mesons & quark-antiquark & $I_3$ & $Y$ & $Q$ \\  
       & descriptions &  &  &  \\  \hline
{\bf pions}  &   &     &        &        \\   
$\pi^+$    & $u \tensor \overline{d}$ & $\frac{1}{2} + \frac{1}{2} = +1$  
& $\qquad\qquad$        & $\qquad \qquad$       \\   
$\pi^0$    &  &     &         &        \\   
$\pi^-$    &  &     &         &       \\   \hline
{\bf kaons}    &  &   &        &        \\   
$K^+$          &    &   &         &        \\   
$K^0$    &    &   &       &        \\   \hline
{\bf antikaons}    &   &   &        &        \\   
$\overline{K^0}$   &  &      &       &       \\   
$K^-$            &   &       &       &        \\   \hline
{\bf eta}    &   &  &        &        \\   
$\eta$   &   &   &      &        \\   \hline
\end{tabular}}
\end{center}

\vskip 1em
In terms of the mathematics of representation theory, why does it 
have to work this way?

\vskip 3em
If we had more time, we would now go on to 
explain the baryons --- like the proton and neutron, but also
other particles --- in terms of quarks.  Then we would explain how
the Eightfold Way was eventually incorporated into the Standard Model,
in which quarks are held
together by the `strong force' to form mesons and baryons.  Ironically,
while the strong force is described by a theory with symmetry group $\SU(3)$,
this symmetry group has {\it nothing to do} with Gell-Mann's
$\SU(3)$ symmetry!  Gell-Mann's 
$\SU(3)$ is now seen to be just an {\it approximate}
symmetry coming from the fact that the up, down and strange quarks all 
act roughly the same, though they have different masses and charges.
The $\SU(3)$ symmetry of the strong force
describes how quarks of any sort come in three
`colors'.  

But alas, summer is fast approaching, and there is no time to continue
our adventure into particle physics.  If you want to learn more, try
reading these books, in approximately increasing order of difficulty
and detail:

\begin{itemize}

\item
Robert P. Crease and Charles C. Mann, 
{\sl The Second Creation: Makers of the Revolution in Twentieth-Century},
Physics, MacMillan, New York,
1986.

\item
Emilio Segre, 
{\sl From X-Rays to Quarks: Modern Physicists and Their Discoveries},
W. H. Freeman, San Francisco, 1980.

\item
Abraham Pais, 
{\sl Inward Bound: of Matter and Forces in the Physical World}, 
Clarendon Press, New York, 1986.

\item
T. D. Lee, {\sl Particle Physics and Introduction to Field Theory}, 
Harwood, New York, 1981.  

\item
Kerson Huang, {\sl Quarks, Leptons \& Gauge Fields}, World Scientific, 
Singapore, 1982.	
\end{itemize}


\vfill
\eject
\end{document}

 

