Past exam of the sub-atomic physics course of the University of Oxford / 2016 / 2

The light pseudoscalar meson nonet has $J^P=0^{-}$ and flavor structure $3\otimes \bar{3}=8\oplus 1$. Its members are the three pions, four kaons, and the two neutral isospin-zero states eta meson and eta prime meson. For mesons the baryon number is zero, so the Gell-Mann--Nishijima formula reduces toHere $S$ is strangeness, with an $s$ quark carrying $-1$ and an antiquark $\bar{s}$ carrying $+1$. Thus the upper row is $K^0=d\bar{s}$ at $(I_3,S)=(-1/2,1)$ and $K^{+}=u\bar{s}$ at $(1/2,1)$. The lower row is $K^{-}=s\bar{u}$ at $(-1/2,-1)$ and ${\bar{K}}^0=s\bar{d}$ at $(1/2,-1)$. The charged pions lie at $(\pm 1,0)$, while ${\pi }^0,\eta ,{\eta }^{\prime }$ share $(0,0)$ as flavor-weight coordinates.

The neutral pion direction is $(u\bar{u}-d\bar{d})/\sqrt{2}$. The octet and singlet isoscalar directions mix to make the physical eta and eta-prime states; counting three central states is essential to obtaining nine rather than seven particles. Horizontal lines in the diagram have constant $S$, while lines $I_3+S/2=Q$ mark constant charge. Meson exchange describes part of the residual nuclear force; the underlying quark-level strong interaction is mediated by gluons, not by treating pions as elementary gauge bosons.

The charged meson decays proceed through a virtual $W^{+}$:The first quark vertex has CKM factor $V_{ud}$, while the second has $V_{us}$. In the simple Cabibbo description these are approximately $\cos {\theta }_C$ and $\sin {\theta }_C$, so the strange amplitude carries a suppression $\tan {\theta }_C$ relative to the nonstrange one, or ${\tan }^2{\theta }_C\simeq 0.053$ in their corresponding squared weak-coupling factors.

For a charged pseudoscalar and a negligibly massive neutrino, the leading leptonic energy width has the formin natural units. The helicity factor $m_{\mu }^2$ is common to these two muon channels, while the threshold factor is much less favorable for the pion. Using the usual masses, its value is about $0.182$ for the pion and $0.910$ for the charged kaon: the pion's threshold factor alone is only about one fifth as large. Including the explicit $m_P$ factor gives a kinematic ratio about $0.0564$. The pseudoscalar decay constants $f_P$ also matter, so Lorentz-invariant phase space and CKM factors alone do not determine the full lifetime ratio.^[ref]

For $K^{+}\to {\pi }^{+}{\pi }^0$, one tree topology has the spectator $u$ unchanged and $\bar{s}\to \bar{u}+W^{+\ast }$ followed by $W^{+\ast }\to u\bar{d}$. The spectator $u$ can combine with $\bar{d}$ to form ${\pi }^{+}$, and the $u\bar{u}$ component supplies ${\pi }^0$. This carries the strangeness-changing weak factor $V_{us}^{\ast }{V}_{ud}$. Hadronization, other quark rearrangements and strong final-state interactions determine the hadronic amplitude; the diagram is not a complete precision width calculation. The small pion–muon energy release suppresses the pion leptonic channel kinematically, whereas the strange weak vertex suppresses the kaon channels through Cabibbo suppression.

Use $c=1$ and let $E$ denote the total center-of-momentum energy, not one beam's energy. Opposite equal momenta give $E=2E_{\mathrm{beam}}=498$ MeV, so the kaon is at its stipulated resonance mass $m_K=498$ MeV. Write the unspecified selected exit branching fractions as $b_{\rho }$ and $b_K$. The supplied line shape givesThe distinction between inclusive and exclusive resonance branching factors matters here. For inclusive resonance formation, sum over all decays, so $b_{\rho }=b_K=1$. At $E=m_K$ the small kaon width cancels from the peak value:If instead the measured cross section is specifically resonant elastic ${\pi }^{+}{\pi }^{-}\to {\pi }^{+}{\pi }^{-}$, its exit branching fractions equal the entrance ones: $b_{\rho }=1$, $b_K=0.69$. ThenThe source says production without specifying an observed decay channel, so both conventions are made explicit rather than silently squaring a branching fraction for an inclusive measurement. The quoted kaon width and two-pion branching fraction identify the short-lived propagation state $K_S$; a pure flavor state $K^0$ is a superposition of short- and long-lived states. This exercise uses the stipulated $K_S$ resonance model and equal omitted proportionality factors.

Raising each beam energy to 250 MeV makes $E=500$ MeV. The kaon denominator is now dominated byrather than by $E^2{\Gamma }_K^2$. Substituting into a givesThe enormous change is a tiny-width effect, not a comparably abrupt change in the broad rho contribution. The center-of-momentum energy has moved about $2/(7.4\times 10^{-12})\simeq 2.7\times 10^{11}$ kaon widths from the peak. A width of $7.4\times 10^{-12}$ MeV is only $7.4\times 10^{-6}$ eV, requiring an energy-selection scale of order $10^{-14}$ relative to the several-hundred-MeV beam energy.

Real beams have vastly broader energy distributions. Averaging over such a distribution suppresses the useful narrow-resonance contribution: the finite peak value does not imply a large energy-integrated formation rate, whose resonant area shrinks with the width at fixed branching fractions. The weak inverse-decay entrance coupling and the broad strong-interaction rho background therefore make this a poor way to study kaon resonant formation. The supplied broad-resonance formula is used algebraically as requested; a practical calculation would convolve an appropriate narrow line shape with the actual beam distribution.

The beam energy is a total pion energy, so the momentum follows the relativistic relation $p^2{c}^2=E_{\pi }^2-m_{\pi }^2{c}^4$. Taking the standard charged-pion rest energy $m_{\pi }{c}^2=139.57$ MeV,^[ref]For motion perpendicular to a uniform magnetic field, the Lorentz force supplies the transverse momentum change and gives $r=p/(qB)$. With the stipulated $qB=1$ eV s m${}^{-2}$,The unit cancellation leaves length after dividing by $c$ in metres per second. Equivalently, the usual magnetic-rigidity conversion is $p[\mathrm{GeV}/c]\simeq 0.2998B[\mathrm{T}]r[\mathrm{m}]$ for unit charge. The sign of the pion charge chooses the bending direction, while the radius uses its magnitude. Treating the pion as massless would give about $0.831$ m and is not a good approximation at this modest total energy. The production target does not determine the downstream orbit once the pion momentum and magnetic field are specified.