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	\begin{document}   
 
	\begin{center}   
	{\bf Hypercharge and Weak Isospin \\}   
	{\small John\ C.\ Baez, May 12 2003\\}   
	\vspace{0.3cm}   
        \end{center}


In the Standard Model, the weak and electromagnetic forces are two
aspects of something called the `electroweak force', which is
described by the group $\SU(2) \times \U(1)$.  Curiously, it turns out
that the familiar concept of `electric charge' is less fundamental
than the concepts of `weak isospin' and `hypercharge'.  The weak
isospin of a particle describes how it transforms under $\SU(2)$,
while its hypercharge describes how it transforms under $\U(1)$.  
The electric charge is computed in a funny way from these two!

In the following problems, you will examine how this works.  
The full symmetry group of the Standard Model is
\[      \ISpin(3,1) \times \SU(3) \times \SU(2) \times \U(1), \]
but we will focus on the electroweak force, so we'll ignore
$\ISpin(3,1)$ and $\SU(3)$ and only think of particles as
irreps of $\SU(2) \times \U(1)$.   You can read these irreps
off the elementary particle chart at the end of this handout.

We begin by introducing bases for $\su(2)$ and $\u(1)$:

\begin{itemize}
\item The Lie algebra $\su(2)$ consists of traceless skew-adjoint
$2 \times 2$ complex matrices, so it has a basis consisting of 
the matrices $\frac{i}{2}\sigma_j$, where 
\[  \sigma_1 = \left( \begin{array}{cc}
        0 & 1  \\
        1 & 0
         \end{array} \right) 
\]
\[ \sigma_2 = \left( \begin{array}{cc}
        0 & -i  \\
        i & 0
         \end{array} \right)
\]
\[ \sigma_3 = \left( \begin{array}{cc}
        1 & 0  \\
        0 & -1
         \end{array} \right). 
\]

\item The Lie algebra $\u(1)$ consists of skew-adjoint $1 \times 1$
complex matrices, which are the same as imaginary numbers.  For these
we choose the basis vector $\frac{i}{3}$.
\end{itemize}

In quantum theory, observables correspond to self-adjoint
operators.  Given any unitary representation $\rho$ of $\SU(2)$
on some Hilbert space, the above basis of $\su(2)$
gets mapped by $d\rho$ to skew-adjoint operators on that Hilbert space.
Dividing these by $i$ we get self-adjoint operators called
the three components of {\bf weak isospin}: $I_1,I_2$ and $I_3$.
In short: 
\[        I_j = d\rho(\textstyle{\frac{1}{2}}\sigma_j)  \]
where we have cancelled some factors of $i$ in a slightly underhanded
manner.

Similarly, given any unitary representation $\rho$ of $\U(1)$
on some Hilbert space, the above basis of $\U(1)$
gets mapped by $d\rho$ to a skew-adjoint operator.
Dividing this by $i$ we get a self-adjoint operator called
{\bf hypercharge}, $Y$.  In short: 
\[        Y = d\rho(\frac{1}{3}) . \]

\vskip 1em
{\bf Digression on History and Terminology}
\vskip 1em

{\it The following stuff is not strictly necessary for doing the
homework.  But if you want to understand the terms `isospin'
and `hypercharge' and that weird factor of $\frac{1}{3}$, 
read this!}

The group $\SU(2)$ first showed up in physics because it is the double
cover of the rotation group $\SO(3)$.  The matrices $\frac{i}{2}\sigma_j$
serve as a convenient basis for the Lie algebra $\su(2)$ because they
satisfy the commutation relations
\[        [\textstyle{\frac{i}{2}}\sigma_1, 
\textstyle{\frac{i}{2}} \sigma_2] = 
\textstyle{\frac{i}{2}} \sigma_3 \qquad {\rm and
\; cyclic \; permutations\; thereof, }  \]
which show that $\su(2)$ is isomorphic to $\R^3$ with its usual
cross product:
\[        {\bf i} \times {\bf j} = {\bf k} \qquad {\rm and
\; cyclic \; permutations\; thereof. }  \]
While the matrices $\textstyle{\frac{i}{2}}\sigma_j$ are skew-adjoint,
the matrices $\textstyle{\frac{1}{2}}\sigma_j$ are self-adjoint, so
they correspond to observables of any quantum system with $\SU(2)$
symmetry.  When $\SU(2)$ is used to describe the rotation symmetries
of some system, the
observables corresponding to the matrices $\frac{1}{2}\sigma_j$ are
called the components of its {\bf angular momentum}.  The
angular momentum intrinsic to an elementary particle is called its
{\bf spin}.  The matrices $\sigma_j$ are called {\bf Pauli matrices}
because they were introduced by Pauli in his explanation of the spin
of the electron.

Later, Heisenberg proposed $\SU(2)$ as a symmetry group for the strong
nuclear force.  His idea was that this group would explain a symmetry
between protons and neutrons: both these particles would really be two
states of a single particle called the {\bf nucleon}, which would
transform under the spin-$\frac{1}{2}$ representation of $\SU(2)$.
The proton would be the spin-up state:
\[  p = \left( \begin{array}{c}
        1  \\
        0  
         \end{array}\right) \in \C^2  , \]
while the neutron would be the spin-down state:
\[ n =  \left( \begin{array}{c}
        0  \\
        1  
         \end{array} \right) \in \C^2.  \] 
However, the `spin' in question here has nothing to do with angular
momentum --- we're just reusing that word because the same group $\SU(2)$ was
used to describe rotation symmetries.  To avoid confusion, Heisenberg
needed another name for this new sort of `spin'.  Since different
isotopes of an element differ in how many neutrons they have in their
nucleus, he coined the term `isotopic spin'.  Later this got shortened
to `isospin'.  So, we call the observables corresponding to
$\frac{1}{2}\sigma_j$ in this new context the three components of a
particle's {\bf isospin}.

Still later, Glashow, Weinberg and Salam used $\SU(2)$ as a symmetry 
group for the weak nuclear force, and called the observables
corresponding to $\frac{1}{2}\sigma_j$ the three components of 
a particle's {\bf weak isospin}, $I_j$.  
There is a close relation between weak isospin and Heisenberg's 
original isospin: in particular, the isospin of a nucleon is
the sum of the weak isospins of the three quarks it is
made of.  However, weak isospin is now considered to be important
for the weak rather than the strong nuclear force.

\centerline{\it In short: the same math keeps getting recycled for different physics!}

Similarly, $\U(1)$ started out being used as a symmetry group for
the electromagnetic force.  For each integer 
$q$ there is a unitary irrep $\rho$ of $\U(1)$ on $\C$ 
called the {\bf charge-$q$ irrep}.  This is given by
\[        \rho(e^{i\theta}) \psi = e^{iq\theta} \psi  \]
for any unit complex number $e^{i\theta} \in \U(1)$ and any 
vector $\psi \in \C$.  Differentiating with respect to $\theta$ 
and setting $\theta = 0$, we get an irrep $d\rho$ of 
the Lie algebra $\u(1)$ on $\C$ with
\[        d\rho(i) \psi = iq \psi . \]
The operator $d\rho(i)$ is skew-adjoint, but we can divide it by $i$
to get a self-adjoint operator
\[        Q = i^{-1} d\rho(i),  \]
or $d\rho(1)$ if we cancel some factors of $i$ in a slightly 
underhanded way.  It's easy to see that
\[        Q \psi = q \psi . \]

The observable corresponding to $Q$ is called {\bf electric charge},
and the above equation says that any state of a particle
described by the charge-$q$ irrep of $\U(1)$ has electric
charge equal to $q$.
Using the group $\U(1)$ this way gives a nice `explanation' of the
fact that the electric charge is quantized: the charge of any
particle is an integer times some smallest charge.  However, it
doesn't say what this smallest charge actually is!

For a long time people thought that the electron had the smallest
possible charge, so they said the electron has charge 1.  
Actually they said it has charge $-1$: an unfortunate
convention which we can blame on Benjamin Franklin, because he was
mixed up about which way the electricity flowed in a current.  But
what do you expect from someone who {\it flies a kite with a key
hanging on it during a thunderstorm, to attract lightning bolts?}
Dumb!  But lucky: the next two people to try that experiment were killed.

Much later, people discovered that quarks have electric charges
smaller than that of the electron.  Measured in units of the electron
charge, quark charges are integral multiples of $\frac{1}{3}$.
Mathematically it would be nicest to redefine our units of charge so
the smallest possible charge is still $1$, but people are too
conservative to do this, so now the smallest charge is taken to be
$\frac{1}{3}$.

Still later, people reused the group $\U(1)$ as a symmetry group for
the electroweak force, and used the term `hypercharge' for the
observable corresponding to this new $\U(1)$.  Since hypercharge is
closely related to charge, physicists also measure hypercharge in
integral multiples of $\frac{1}{3}$.

Here's how we accomodate this foolish factor of $\frac{1}{3}$.
For each number $y$ with $3y \in \Z$, there is a unitary 
irrep $\rho$ of $\U(1)$ on $\C$ called the {\bf hypercharge-$y$
irrep}.  This is given by
\[        \rho(e^{i\theta}) \psi = e^{3iy\theta} \psi.  \]
Differentiating with respect to $\theta$ as before, 
we get an irrep $d\rho$ of $\u(1)$ on $\C$ with
\[        d\rho(\frac{i}{3}) \psi = iy \psi . \]
The operator $d\rho(\frac{i}{3})$ is skew-adjoint, but dividing
it by $i$ we get a self-adjoint operator
\[        Y = i^{-1} d\rho(\frac{i}{3}),  \]
or $d\rho(\frac{1}{3})$ for short.  It's easy to see that
\[        Y \psi = y \psi  .\]
We call the observable corresponding to $Y$ {\bf hypercharge}, and
the above equation says that any state of a particle described by
the hypercharge-$y$ irrep of $\U(1)$ has hypercharge $y$.  

\vskip 1em
{\bf Back to Business}
\vskip 1em

Any particle in the Standard Model corresponds to some unitary irrep
of $\SU(2) \times \U(1)$.  This is a unitary rep of both 
$\SU(2)$ and of $\U(1)$, so we get self-adjoint operators 
$I_1,I_2,I_3$ and $Y$ on this irrep, corresponding to weak isospin
and hypercharge.  The observable {\bf electric charge} is
related to these by the mystical formula
\[      Q = I_3 + \frac{Y}{2} . \]
Now let's use this to work out the electric charges
of all the elementary particles!

I'll do an example: consider the left-handed
electron neutrino $\nu_e^L$.  As indicated in
the chart at the end of this handout, 
this is the first member of the standard basis of the irrep 
$\C^2 \tensor \C_{-1}$ of $\SU(2) \times \U(1)$:
\[ \nu_e^L = \left( \begin{array}{c}
        1  \\
        0  
         \end{array}\right)  , \quad
  e^L =  \left( \begin{array}{c}
        0  \\
        1  
         \end{array} \right) .
\]
Note that
\[ I_3 \nu_e^L = \left(\begin{array}{cc}
       \frac{1}{2} & 0  \\
        0 & -\frac{1}{2} 
         \end{array} \right)
\left( \begin{array}{c}
        1  \\
        0  
         \end{array} \right)   =
\textstyle{\frac{1}{2}} \left( \begin{array}{c}
        1  \\
        0  
\end{array} \right) = \textstyle{\frac{1}{2}} \nu_e^L  .\]
Since the eigenvalue is $\frac{1}{2}$, 
a physicist reading this equation will say
`the left-handed electron neutrino has $I_3 = \frac{1}{2}$'. 

In the chart at the end of this handout, the hypercharge-$y$
irrep of $\U(1)$ is denoted $\C_y$.  
As explained in the Digression, the hypercharge operator $Y$ acts
as multiplication by the number $y$ on any vector in this representation.
Since the left-handed electron lives in the 
hypercharge-$(-1)$ rep, it follows that
\[ Y \nu_e^L = - \nu_e^L .\]

Now that we know $I_3$ and $Y$ for the left-handed electron
neutrino, we can use the magic formula to work out its electric charge:
\[      Q \nu_e^L = (I_3 + \frac{Y}{2}) \nu_e^L = 0 . \]
Since the eigenvalue is 0, 
the left-handed electron neutrino has electric
charge 0.  And indeed, this particle is neutral!  

\centerline{\it Finally, here's where you come in....}

1. Use this idea to fill out as much of the following chart
as you can.  If you know enough representation theory you can
do it all!  It may help to reread the list of
conventions in the previous homework on elementary particles. 

\begin{center}
\begin{tabular}{|l|l|l|l|}                    \hline   
type of particle $\qquad \qquad\qquad$ eigenvalue of: & $Y$  
  &  $I_3$ & $Q$ \\ \hline
{\bf GAUGE BOSONS}          &          &        &        \\
$g_{rg},g_{rb},     
  g_{gr},g_{gb},
  g_{br},g_{bg},
  g_{rr}-g_{bb}, g_{bb}-g_{gg}$  &       &           &          \\  
$W_1$  &       &           &          \\  
$W_2$  &       &           &          \\  
$W_3$  &       &           &          \\  
$W_0$  &       &           &          \\  \hline
{\bf HIGGS BOSON}         &           &           &          \\ 
$H^+$   &       &           &          \\  
$H^0$   &       &           &          \\  \hline
{\bf FIRST GENERATION} 
{\bf FERMIONS}         
          &           &           &          \\  
{\sl Leptons:}          &           &           &          \\ 
$\nu_e^L$     &           &           &          \\ 
$e^L$         &           &           &          \\ 
$\nu_e^R$     &           &           &          \\ 
$e^R$         &           &           &          \\ 
{\sl Quarks:}           &           &           &          \\ 
$u_r^L,u_g^L,u_b^L$    &           &           &          \\
$d_r^L,d_g^L,d_b^L$    &           &           &          \\
$u_r^R,u_g^R,u_b^R$    &           &           &          \\
$d_r^R,d_g^R,d_b^R$    &           &           &          \\
\hline
{\bf SECOND GENERATION FERMIONS}         
          &           &           &          \\ 
{\sl Leptons:}          &           &           &          \\ 
$\nu_\mu^L$     &           &           &          \\ 
$\mu^L$         &           &           &          \\ 
$\nu_\mu^R$     &           &           &          \\ 
$\mu^R$         &           &           &          \\ 
{\sl Quarks:}            &           &           &          \\ 
$c_r^L,c_g^L,c_b^L$    &           &           &          \\
$s_r^L,s_g^L,s_b^L$    &           &           &          \\
$c_r^R,c_g^R,c_b^R$    &           &           &          \\
$s_r^R,s_g^R,s_b^R$    &           &           &          \\
\hline
{\bf THIRD GENERATION FERMIONS}         
           &           &           &          \\  
{\sl Leptons:}          &           &           &          \\ 
$\nu_\tau^L$     &           &           &          \\ 
$\tau^L$         &           &           &          \\ 
$\nu_\tau^R$     &           &           &          \\ 
$\tau^R$         &           &           &          \\ 
{\sl Quarks:}            &           &           &          \\ 
$t_r^L,t_g^L,t_b^L$    &           &           &          \\
$b_r^L,b_g^L,b_b^L$    &           &           &          \\
$t_r^R,t_g^R,t_b^R$    &           &           &          \\
$b_r^R,b_g^R,b_b^R$    &           &           &          \\
\end{tabular}
\end{center}

2.  In problem 1 of the previous homework you may have noticed
that for leptons and quarks,
the average of the hypercharge of the right-handed ones is equal
to the hypercharge of the left-handed one.  Use your new-found
knowledge to say more about the significance of this fact.

3.  What is the sum of the hypercharges of all the fermions in a given 
generation?  To do this right you have to sum over all 16 basis 
vectors of the fermion rep, e.g. 
$\nu_e^L$, $e^L$, $\nu_e^R$, $e^R$, $u_r^L$,$u_g^L$,$u_b^L$  
$d_r^L$,$d_g^L$,$d_b^L$,  
$u_r^R$,$u_g^R$,$u_b^R$,  
$d_r^R$,$d_g^R$,$d_b^R$.    

4.  What is the sum of the eigenvalues of $I_3$
over all the fermions in a given generation?  

5.  What is the sum of the electric charges of all the fermions in 
a given generation?  

The answers to questions 3-5 are very important in {\bf grand unified
theories}.  These are theories where $\su(3) \oplus \su(2) \oplus \u(1)$ 
is embedded as a Lie subalgebra of some {\bf simple} Lie algebra like
$\su(5)$ or $\so(10)$: i.e., a Lie algebra with no nontrivial ideals.  
The fermion rep can only extend to a rep of a simple Lie algebra
if the answers to questions 3-5 take a certain special form!


\vskip 2em
\centerline{\bf ELEMENTARY PARTICLES IN THE STANDARD MODEL}
\vskip 2em

\begin{tabular}{|l|l|l|l|l|}                    \hline   
type of particle & $\ISpin(3,1)$ irrep &  $\SU(3)$ irrep &  $\SU(2)$ irrep
& $\U(1)$ irrep \\ 
{\bf GAUGE BOSONS}     &           &           &          &          \\
$\bullet$ gluons ($\SU(3)$ force carriers):           
&           &           &           &          \\
$(g_{rg},g_{rb},     
  g_{gr},g_{gb},
  g_{br},g_{bg},
  g_{rr}-g_{bb}, g_{bb}-g_{gg})$
& massless spin-1& $\su(3)$ & $\R$  & $\R$     \\
$\bullet$ $\SU(2)$ force carriers:                     
&           &           &           &          \\
$(W_1,W_2,W_3)$   
& massless spin-1 &  $\R$    &  $\su(2)$     &   $\R$ \\
$\bullet$ $\U(1)$ force carrier:                      
&           &           &           &          \\
$(W_0)$      
 & massless spin-1 &  $\R$    &  $\R$     &   $\u(1)$  \\  \hline
{\bf HIGGS BOSON}     &           &           &           &          \\ 
$\bullet$ Higgs:           
&           &           &           &          \\
$(H^+,H^0)$    & massless spin-0  & $\C$     &   $\C^2$  
               &  $\C_1$ \\     
and its antiparticle!   
      &           &           &           &          \\   \hline
{\bf FIRST GENERATION}         
{\bf FERMIONS}         
&           &           &           &          \\  
{\sl Leptons:}      &           &           &           &          \\ 
$\bullet$ left-handed electron neutrino and electron:
                     &           &           &           &          \\
$(\nu_e^L,e^L)$ &   left-handed massless spin-\hf        & $\C$           & $\C^2$      & $\C_{-1}$  
 \\ 
$\bullet$ right-handed electron neutrino:                       &           &           &           &          \\
$(\nu_e^R)$  &   right-handed massless spin-\hf        & $\C$           & $\C$      & $\C_{0}$   \\ 
$\bullet$ right-handed electron:                       &           &           &           &          \\
$(e^R)$  &   right-handed massless spin-\hf        & $\C$           & $\C$      & $\C_{-2}$   \\ 
{\sl and their antiparticles!}  
  &           &           &           &          \\   \hline
{\sl Quarks:}
  &           &           &           &          \\
$\bullet$ left-handed up and down quarks:
                     &           &           &           &          \\
$(u_r^L,u_g^L,u_b^L,d_r^L,d_g^L,d_b^L)$ &  left-handed massless spin-\hf        & $\C^3$           & $\C^2$      & $\C_{\third}$  
 \\ 
$\bullet$ right-handed up quark:                       &           &           &           &          \\
$(u_r^R,u_g^R,u_b^R)$  &  right-handed massless spin-\hf        & $\C^3$           & $\C$      & $\C_{\fourthirds}$   \\ 
$\bullet$ right-handed down quark                       &           &           &           &          \\
$(d_r^R,d_g^R,d_b^R)$  &  right-handed massless spin-\hf        & $\C^3$           & $\C$      & $\C_{\minustwothirds}$   \\ 
{\sl and their antiparticles!}  
  &           &           &           &          \\   \hline
{\bf SECOND GENERATION}         
{\bf FERMIONS}         
  &           &           &           &          \\   \hline
{\sl Leptons:}  
  &           &           &           &          \\   \hline
$\bullet$ left-handed mu neutrino and muon:
                     &           &           &           &          \\
$(\nu_\mu^L,\mu^L)$ &   left-handed massless spin-\hf        & $\C$           & $\C^2$      & $\C_{-1}$  
 \\ 
$\bullet$ right-handed mu neutrino:                       &           &           &           &          \\
$(\nu_\mu^R)$  &   right-handed massless spin-\hf        & $\C$           & $\C$      & $\C_{0}$   \\ 
$\bullet$ right-handed muon:                       &           &           &           &          \\
$(\mu^R)$  &   right-handed massless spin-\hf        & $\C$           & $\C$      & $\C_{-2}$   \\ 
{\sl and their antiparticles!}  
  &           &           &           &          \\   \hline
{\sl Quarks:}
  &           &           &           &          \\
$\bullet$ left-handed charm and strange quarks:
                     &           &           &           &          \\
$(c_r^L,c_g^L,c_b^L,s_r^L,s_g^L,s_b^L)$ &  left-handed massless spin-\hf        & $\C^3$           & $\C^2$      & $\C_{\third}$  
 \\ 
$\bullet$ right-handed charm quark:                       &           &           &           &          \\
$(c_r^R,c_g^R,c_b^R)$  &  right-handed massless spin-\hf        & $\C^3$           & $\C$      & $\C_{\fourthirds}$   \\ 
$\bullet$ right-handed strange quark                       &           &           &           &          \\
$(s_r^R,s_g^R,s_b^R)$  &  right-handed massless spin-\hf        & $\C^3$           & $\C$      & $\C_{\minustwothirds}$   \\ 
{\sl and their antiparticles!}  
&           &           &           &          \\  \hline
{\bf THIRD GENERATION}         
{\bf FERMIONS}           
&           &           &           &          \\   \hline
{\sl Leptons:}      &           &           &           &          \\ 
$\bullet$ left-handed tau neutrino and tau:
                     &           &           &           &          \\
$(\nu_\tau^L,\tau^L)$ &   left-handed massless spin-\hf        & $\C$           & $\C^2$      & $\C_{-1}$  
 \\ 
$\bullet$ right-handed tau neutrino:                       &           &           &           &          \\
$(\nu_\tau^R)$  &   right-handed massless spin-\hf        & $\C$           & $\C$      & $\C_{0}$   \\ 
$\bullet$ right-handed tau:                       &           &           &           &          \\
$(\tau^R)$  &   right-handed massless spin-\hf        & $\C$           & $\C$      & $\C_{-2}$   \\ 
{\sl and their antiparticles!}  
&           &           &           &          \\  \hline
{\sl Quarks:}
  &           &           &           &          \\
$\bullet$ left-handed top and bottom quarks:
                     &           &           &           &          \\
$(t_r^L,t_g^L,t_b^L,b_r^L,b_g^L,b_b^L)$ &  left-handed massless spin-\hf        & $\C^3$           & $\C^2$      & $\C_{\third}$  
 \\ 
$\bullet$ right-handed top quark:                       &           &           &           &          \\
$(t_r^R,t_g^R,t_b^R)$  &  right-handed massless spin-\hf        & $\C^3$           & $\C$      & $\C_{\fourthirds}$   \\ 
$\bullet$ right-handed bottom quark                       &           &           &           &          \\
$(b_r^R,b_g^R,b_b^R)$  &  right-handed massless spin-\hf        & $\C^3$           & $\C$      & $\C_{\minustwothirds}$   \\ 
{\sl and their antiparticles!}  
&           &           &           &          \\ \hline


\end{tabular}  


\end{document}   
   
   
   
   
