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\markright{{\small\bf Standard Model}  
\hfill {\small\it WS 07/08, Sheet \theBlt,}}

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\begin{document}

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\vspace*{-25mm}\parbox{60mm}
{{\sf Universit\"at Heidelberg}\\ 
\small Jan M. Pawlowski, Ulrich Uwer}
\hfill \parbox{48mm}{\rightline{\it
\begin{tabular}{l}
WS 07/08\\ 
%\today\\
Sheet \theBlt\end{tabular}} }
      

                

\vskip .5cm

\leftline{{\bf Standard Model}\hfill 
- 1 - } 
\vspace{.05cm}
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\leftline{\large\bf 1. Poincar\'e-Transformations} \medskip The
inhomogenous Lorentz-transformations of a Minkowski space vector $x$ 
are given by
\begin{equation}
\label{eq:1}
    \mathcal{P} \ni g(\Lambda , a ) : \; x \mapsto x' = \Lambda x + a \qquad {\rm mit} \qquad  
 \Lambda^T \eta \Lambda = \eta\;,\qquad a\in \RR^4\; \; 
\end{equation}
This defines Poincar\'e-group $\mathcal{P}$.
\begin{description}
\item[a)]
Show, that the multipication in $\mathcal{P}$ can be represented by 
\begin{equation}
    \mathcal{P} \ni g,g' : \; g' g  \equiv  g(\Lambda' , b )  g(\Lambda , a ) =
    g(\Lambda' \Lambda , \Lambda' a + b ) \; .
\end{equation}
\end{description}
Consider a unitary representation $D:$ $\mathcal{P}\ni g \mapsto 
U_g \equiv
U(\Lambda, a)$ 
\begin{description}
\item[b)]
Show that $U$ can be decomposed as follows 
\begin{equation}
       U(\Lambda, a)  = T(a)   U(\Lambda) \; , \;\; 
            U(\Lambda) \equiv U(\Lambda,0) \;\; \mbox{und}\;\;  
            T(a) \equiv  U(\ID,a) =  \exp\{ -i a^\mu P_\mu \}\; ,
\end{equation}
where $T(a)$ is a translation, generated by the momentum operator
$P^\mu$. Show that the translations are an invariant subgroup of
$\mathcal{P}$, 
\begin{equation}
   U(\Lambda,b)\, T(a) \, U^{-1}\!(\Lambda,b) = T(\Lambda a) \; .
\end{equation}
\end{description}

The proper orthochronous Lorentz-group
$\mathcal{L}_{+}^{\scriptscriptstyle{
\uparrow}}$ 
is represented by the $(4\times 4)$ matrices $\Lambda $, which 
leave the Minkowski space metric $\eta$ invariant, 
\begin{equation} 
  \Lambda^T \eta \Lambda = \eta \; , \quad \eta = \left( \begin{array}{rrrr} 
                            1 & 0 & 0 & 0 \\
                            0 & -1 & 0 & 0 \\
                            0 & 0 & -1 & 0 \\
                            0 & 0 &  0 & -1 \end{array} \right) \qquad 
{\rm und }\quad  \det\Lambda = 1,\quad {\Lambda^0}_0 \ge 1\; ,
\end{equation}
The proper orthochronous Lorentz-group is generated by the exponential 
map of the algebra 
\begin{equation} 
           l_+^{\scriptscriptstyle{
\uparrow}} = \big\{ X \in \RR^{4,4}\; : \;\;  e^{tX} \in \mathcal{L}_+^{\scriptscriptstyle{
\uparrow}}
\, 
           , t \in \RR           \big\} \; .  
\end{equation}

\begin{description}
\item[c)] Show, that  
\begin{equation} 
           X^T \eta = - \eta X \;  \Leftrightarrow \; X \in l_+^{\scriptscriptstyle{
\uparrow}} \; ,
\end{equation}
and hence $\mbox{dim}_{\rm \scriptscriptstyle\mbox{$\RR$}}\,  l_+^{\scriptscriptstyle{
\uparrow}} = 6$. 
A basis of $l_+^{\scriptscriptstyle{
\uparrow}}$ 
ist given by the 3 generators of rotations and boosts respectively,
\[\begin{array}{ccc}
      iJ_1 =   \left( \begin{array}{cccc} 
                            0 & 0 & 0 & 0 \\
                            0 & 0 & 0 & 0 \\
                            0 & 0 & 0 & 1 \\
                            0 & 0 & -1 & 0 \end{array} \right)\;, &
      iJ_2 =   \left( \begin{array}{cccc} 
                            0 & 0 & 0 & 0 \\
                            0 & 0 & 0 & -1 \\
                            0 & 0 & 0 & 0 \\
                            0 & 1 & 0 & 0 \end{array} \right)\;, &
      iJ_3 =   \left( \begin{array}{cccc} 
                            0 & 0 & 0 & 0 \\
                            0 & 0 & 1 & 0 \\
                            0 & -1 & 0 & 0 \\
                            0 & 0 & 0 & 0 \end{array} \right)\;, \\[12pt]
      -iK_1 =   \left( \begin{array}{rrrr} 
                            0 & 1 & 0 & 0 \\
                            1 & 0 & 0 & 0 \\
                            0 & 0 & 0 & 0 \\
                            0 & 0 & 0 & 0 \end{array} \right)\;, &
      -iK_2 =   \left( \begin{array}{rrrr} 
                            0 & 0 & 1 & 0 \\
                            0 & 0 & 0 & 0 \\
                            1 & 0 & 0 & 0 \\
                            0 & 0 & 0 & 0 \end{array} \right)\;, &
      -iK_3 =   \left( \begin{array}{rrrr} 
                            0 & 0 & 0 & 1 \\
                            0 & 0 & 0 & 0 \\
                            0 & 0 & 0 & 0 \\
                            1 & 0 & 0 & 0 \end{array} \right)\; . 
\end{array} 
\]
\item[d)] Show that these generators satisfy the following Lie-algebra 
relations:
\[
\begin{array}{ll}
[ J_i , J_j ] \, = \, i \epsilon_{ijk} J_k \; , \qquad &  
                   [ N_i , N_j ] \, = \, i \epsilon_{ijk} N_k \; ,\\
{ [ J_i , K_j ] }\, = \, i \epsilon_{ijk} K_k \; , \qquad &  
                   [ N^\dagger_i , N^\dagger_j ] \, = 
\, i \epsilon_{ijk} N^\dagger_k \; ,\\
{ [ K_i , K_j ] }\, = \, - i \epsilon_{ijk} J_k \; , \qquad &  
                   [ N_i , N^\dagger_j ] \, = \, 0 \; ,
\end{array}
\]
where we have alos introduced the basis  $N_i = (J_i + i K_i )/2$. 
These relations can be summarised in 
\begin{eqnarray} 
  &M_{0i} = - M_{i0} = K_i \, , \; \; M_{ij} = \epsilon_{ijk} J_k 
\, , \quad 
  \mbox{mit} \\
  &{[ M_{\mu\nu} , M_{\rho\sigma} ] } \, = -i \big( 
       M_{\mu\rho} \eta_{\nu\sigma} -  M_{\mu\sigma} \eta_{\nu\rho} -
      M_{\nu\rho} \eta_{\mu\sigma} +  M_{\nu\sigma} \eta_{\mu\rho} 
\big) \; ,
  \nonumber  
\end{eqnarray}
and we obtain $\displaystyle \Lambda(\omega) = e^{ \frac{i}{2}
\omega^{\mu\nu} M_{\mu\nu}} $. The tensor $ \omega_{\mu\nu} = -
\omega_{\nu\mu} $ contains the 6 real parameters of the transformation. 


\end{description} 
A representation $D: \; g \mapsto U_g$ of a group $G$ as a linear map 
in a vector space $V$ is called irreducible, if there are no 
$D$-invariant sub-spaces of $V$ (except the trivial ones, given by 
the Null vector and $V$ itself. A sub-vector space $W \subset V$ 
is $D$-invariant, if and only if 
$\forall g\in G , |w\rangle \in W \, : \;\;  U_g
|w\rangle  \in W$.    
\begin{description}
\item[e)] Show that the operator $P^2=P_\mu P^\mu$ 
in an irreducible representation of the Poincar\'e group 
on $V$ is provided by a multiple of the unit operator, 
$P^2 = c_1 \,  \hbox{Id}(V) \, , \; c_1 \in \RR$.
\end{description}
$P^2$ is called a Casimir-operator. The fact, that $P^2$ is proportional 
to the unit operator,  is a special case of Schur's (first) lemma. 

\bigskip


The Pauli-Lubanski-vector $W^\mu$ is defined by 
\begin{equation}
\label{eq:4}
        W_{\mu} \, = \, 
        {1\over 2} \,  \epsilon_{\mu\nu\rho\sigma} P^{\nu} M^{\rho\sigma}  \; .
\end{equation}
\begin{description}
\item[f)] Show that it has the following properties:
  \begin{eqnarray}
  \label{eq:5}
      W^\mu P_\mu \, = \, 0  \; , \quad & \quad   [M_{\mu\nu}, W_{\sigma}] \,
  = \, i ( \eta_{\mu\sigma}W_{\nu} - \eta_{\nu\sigma}W_{\mu}) \; , \nonumber \\
      { [W_\mu, P_\nu ] }\, = \, 0   \; , \quad & \quad   [W_\mu, W_\nu ] \, =
  \, i \epsilon_{\mu\nu\rho\sigma} W^\rho P^\sigma \; .  
  \end{eqnarray}
\end{description}
The operator $W^2 = W^\mu W_\mu $ commutates with all elements of the 
 Poincar\'e-algebra (why?). It is the second Casimir operator of the 
Poincar\'e-group, and $W^2 = c_2 \,  \hbox{Id}(V) \, , \; c_2 \in \RR$ 
in irreducible representations.



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