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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}
\hrule 

\vskip .2in

%% begin of body 

\leftline{\large\bf 3. Fermions and Scattering} \medskip 
\begin{itemize} 
\item[\bf 1)] The Dirac action is given by 
\begin{equation} 
S[\phi]= \frac{1}{2}\int d^4 x\, \bar \psi 
\left(i\dslash - 
m\right) \psi(x)\,. 
\label{eq:0}\end{equation} 
The field operator of a fermionic field operator is given by 
\begin{equation} 
\psi(x)=\int \frac{d^3 p}{(2 \pi)^3}\frac{1}{2 p^0}\sum_{s=\pm 1/2}
\left\{ 
e^{i p\,x } v_s(p) b^\dagger (\vec k)+e^{-i p\,x } u_s(p) a(\vec p)\right\}\,, 
\end{equation}
with $p^0=\sqrt{\vec p^2+m^2}$, and annihilation/creation operators 
of electrons and fermions $a,a^\dagger$ and $b,b^\dagger$ respectively with  
\begin{equation}
\{a_s(\vec p)\,,\,a_r^\dagger (\vec p')\}=
2 p^0 (2 \pi)^3 \delta_{sr}\delta(\vec p-\vec p')\,,\qquad 
\{b_s(\vec p)\,,\,b_r^\dagger (\vec p')\}=
2 p^0 (2 \pi)^3 \delta_{sr}\delta(\vec p-\vec p')
  \label{eq:1}\end{equation} 
and 
 \begin{equation}
 \sum_{s=\pm 1/2} u_s(p) \bar u_s(p)=\slh{p} +m\,,\qquad    \sum_{s=\pm 1/2} 
v_s(p) \bar v_s(p)=\pslash -m\,. 
\label{eq:ssum}\end{equation} 

\begin{itemize}
\item[\bf a)] Show with (\ref{eq:1}) that $\psi(x)$ and $\bar\psi(x)$ 
satisfy the anti-commutation relations 
\begin{equation}
\hspace{-1cm}\{  \psi(\vec x,t)\,,\,  \bar\psi(\vec y,t)\}=\gamma^0
\delta(\vec x-\vec y)\,, 
\qquad \{  \psi(\vec x,t)\,,\,  \psi(\vec y,t)\}=
\{\bar\psi(\vec x,t)\,,\,  \bar\psi(\vec y,t)\}=0\,.
\end{equation} 
\vspace{.2cm}
\item [\bf b)] Show that 
\begin{equation}\label{eq:expect}
  \langle 0|\,\psi(x)\,|e^-(\vec p,s\rangle = e^{-i px} u_s(p)\,, 
  \qquad  \langle 0|\,\bar\psi(x)\,|e^+(\vec p,s\rangle = e^{-i px} \bar 
  v_s(p)\,,\end{equation} 
with $|e^-(\vec p,s\rangle =a^\dagger_s(\vec p)|0\rangle$, and 
$|e^+(\vec p,s\rangle =b^\dagger_s(\vec p)|0\rangle$. 
\end{itemize}
\vspace{.2cm}
\item[\bf 2)] Consider the scattering of particles in the presence of an
  interaction Lagrangian ${\cal L}'(x)=e \int d^4 x \, A_\mu(x) 
:\bar\psi \gamma^\mu\psi(x):$, with the $S$-matrix 
\begin{equation} \label{eq:4}
S = T\,e^{i e \int d^4 x\, a_\mu :\bar\psi\gamma^\mu\psi(x):}.
\end{equation} 

\begin{itemize}
\item[\bf a)] Expand the $S$-matrix up to the second order, and
  perform the time-ordering.
  \\[-1ex]






\item[\bf b)] Reduce the matrix element 
\begin{equation} 
\langle e^+(p_4) e^-(p_3)|:\psi(x)\gamma^\mu \psi(x):\,
:\psi(x')\gamma^\mu \psi(x'): 
| e^+(p_2) e^-(p_1)\rangle\,  
\end{equation}
to products of the matrix elements \eq{eq:expect}. This matrix element
is relevant for $e+ e-$-scattering (Bhabba-scattering). Compare the 
result to that for $e+e-\to \mu+\mu^-$-scattering. \\[-1ex]

\item[\bf c)] Show that 
\begin{eqnarray}\nonumber 
&&\hspace{-1cm} 
\sum_{s,s',r,r'} |\bar u_s(p_3)\gamma^\mu v_{s'}(p_4)\, \bar v_r(p_2)
\gamma_\mu u_r'(p_1)|^2\\ 
&&=\Tr\, \left[\gamma^\mu (\slh{p}_4-m)\gamma_\rho (\slh{p}_3+m) \right] 
\Tr\, \left[\gamma^\mu (\pslash_1+m)\gamma_\rho (\pslash_2-m) \right] \,.
\label{eq:spinsumsav}\end{eqnarray}
The spin sums in \eq{eq:spinsumsav} are relevant for  
$e^+ e^-\to e^+ e^- $ and $e^+ e^-\to  \mu+\mu^-$-scattering matrix elements. 
%\end{description}
\end{itemize}
\end{itemize}

\eject 

 
\begin{itemize} 
\item[\bf 3)]
Consider the scattering process a($p_a$) + b($p_b$) $\rightarrow$ c($p_c$)
+ d($p_d$), where the 4-momenta $p_i$ are that in the lab-system. The
 Mandelstam-variables $s,\ t,\ u$ are Lorentz-invariant variables,
\begin{equation} 
s=( p_a + p_b)^2 \; , \;\;\;\;  t=( p_a - p_c)^2 \;,  \;\;\;\;   
u=( p_a - p_d)^2\,, 
\end{equation} 
with 
\begin{equation} 
s=( p_a + p_b)^2 =( p_a^* + p_b^*)^2\,,
\end{equation}
where the $p_i^*$ are the momenta of the particles in 
the centre of mass (CM) system: $\vec{p}_a^* =- \vec{p}_b^* $.

\begin{itemize} 
\item[\bf a)]
The particle b is at rest in the lab system. Show that 
\begin{equation}
|\vec{p}_a| = \frac{1}{2m_b} w(s, m_a^2, m_b^2)\,,
\end{equation}
with  
\begin{equation}
w(x,y,z) = (x^2 + y^2 + z^2 -2xy -2xz -2yz)^{\frac{1}{2}}  
\end{equation}
Show also that the energy $E_a^*$ ($E_a^*$) of the particle  a
(b) in the  CMS is given by 
\begin{equation}
E_a^* = \frac{s+m_b^2-m_a^2}{2\sqrt{2}} \;\; \mbox{und} \;\; 
E_b^* = \frac{s+m_a^2-m_b^2}{2\sqrt{2}}
\end{equation} 
Equivalent relations hold for $E_c^*$ and $E_d^*$.
\vspace{0.3cm}

\item[\bf b)] Show that 
\begin{equation}
|\vec{p}_b^*|= |\vec{p}_a^*| = \frac{1}{2\sqrt{s}} w(s, m_a^2, m_b^2)\,\qquad 
|\vec{p}_c^*|= |\vec{p}_d^*| = \frac{1}{2\sqrt{s}} w(s, m_c^2, m_d^2)\,,
\end{equation} 
and write $t$ as a function of $s$ and the scattering angle 
$\theta^*$ in the CMS. \\[-1ex] 
\item[\bf c)] The flow F for the above configuration is (b is at rest,
  volume $V=1$): 
\begin{equation}
F = |\vec{v}|2E_a 2 m_b \,. 
\end{equation}
Express the flow as a
  function of the momentum $|\vec{p}_a^*|$ in CMS.\\[-1ex]
\item[\bf d)]
%\vspace{0.5cm}
%\leftline{\bf Phasenraumintegration} 
%\vspace{0.5cm}
Perform the integration of the 2-particle phase space for the above 
scattering process in the CMS. Show that 
\begin{equation}
\int{d\Phi_2} = \frac{1}{16\pi^2} \frac{|\vec{p}_c^*|}{\sqrt{s}}
\int{d\Omega_C}
\,,
\end{equation} 
by using the relation 
\begin{equation}
\int\delta [ f(\omega) ] g(\omega) d\omega = \left( g \left|
\frac{df}{d\omega} \right| ^ {-1} \right)_{f=0}  \,.
\end{equation} 
\end{itemize}
\end{itemize} 





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