\input phyzzx

\def\Re{\mathop{\rm Re}\nolimits}

%\hsize=6.8in	\hoffset=-0.2in	%\vsize=8.5in	\overfullrule=0pt
\noindent PHY--396~K.\quad
Problem set \#11.\quad
Due December 3, 1998.
\par\smallskip\hrule\kern1pt\hrule\vbox{}\medskip

\pointbegin
As discussed in problem 1 of set\#9, the muon decays into an electron,
a muon neutrino an an electron antineutrino with amplitude
$$
\bra{e^-,\bar\nu_e,\nu_\mu}{\cal M}\ket{\mu^-}
=\ {G_F\over\sqrt{2}}\,
\left[\bar u(\nu_\mu){1-\gamma^5\over2}\gamma^\alpha u(\mu^-)\right]\times
\left[\bar v(\bar\nu_e) {1-\gamma^5\over2}\gamma_\alpha u(e^-)\right] .
\eqn\MuonAmplitude
$$
Show that
$$
\half\sum_{\rm all\atop spins}
	\left|\bra{e^-,\bar\nu_e,\nu_\mu}{\cal M}\ket{\mu^-}\right|^2\
=\ 64G^2_F (p_\mu\cdot p_{\bar\nu})\,(p_e\cdot p_\nu) .
\eqn\MuonAverage
$$

\point
Consider so-called Bhabha scattering $e^-e^+\to e^-e^+$.
In QED, there are two tree-level Feynman diagrams contributing to this
process; their contributions must be added {\sl before} squaring the
amplitude and adding/averaging over spins,
$$
\left|{\cal M}_1+{\cal M}_2\right|^2\
=\ |{\cal M}_1|^2\ +\ |{\cal M}_2|^2\ +\ 2\Re\bigl({\cal M}_1^*{\cal M}_2\bigr)\
\neq\ |{\cal M}_1|^2\ +\ |{\cal M}_2|^2.
\eqno\eq
$$
Calculate the partial cross-section $d\sigma/d\Omega$ for the
Bhabha scattering.
For simplicity, assume $E\gg m_e$ and neglect the electron's mass throughout
your calculation.
%\subpar
You may find it convenient to use Mandelstam's Lorentz-invariant kinematic
variables $s$, $t$ and $u$, see eq.~(5.69) of the Peskin\&Schroeder textbook
for details.
Notice $s+t+u=4m_e^2\approx0$ in the $m_e\approx0$ approximation.
\subpar
The answer to this problem is simple:
$$
\left({d\sigma\over d\Omega}\right)_{\rm c.m.}\
=\ {\alpha^2\over 2s}\,\left[ \left({t\over s}\right)^2\,
    +\,\left({s\over t}\right)^2\,+\,\left({u\over s}+{u\over t}\right)^2\right]
\eqno\eq
$$
but the intermediate steps are quite complicated, so beware.


\bye
