Past exam of the particle physics course of the University of Oxford / 2015 / 3

Direct formation in $e^{+}{e}^{-}$ through one virtual photon selects the vector meson quantum numbers $J^{PC}=1^{--}$. The nonzero $K_S^0{K}_L^0$ mode and absence of $K_S^0{K}_S^0$ are consistent with this assignment: the two spin-zero kaons require odd orbital angular momentum for negative parity, whereas identical $K_S^0$ bosons cannot have an odd relative wave. This argument neglects the tiny CP violation in neutral kaons.

Assuming the $\rho \pi$ decays conserve isospin, their equal rates indicate $I=0$. The Clebsch-Gordan coefficients giveAn $I=1,I_3=0$ state lacks ${\rho }^0{\pi }^0$, while $I=2,I_3=0$ has squared coefficients $1:4:1$ rather than $1:1:1$. The branching fractions alone determine neither $J$ nor these relative phases; the single-photon formation and approximate strong interaction isospin conservation are essential additional inputs.

For the resonant contribution only, subtract nonresonant Bhabha scattering and its interference before interpreting a fitted area. The integrated narrow vector resonance area follows from the Breit-Wigner resonance cross section, with $J=1$, $s_{e^{-}}=s_{e^{+}}=1/2$, and $p\simeq M/2$, becomesAcross a narrow scattering resonance, treat the prefactor and partial decay widths as constant and extend the integral over $E-M$ to the real line. Since ${\int }_{-\infty }^{\infty }du/(u^2+a^2)=\pi /a$,A normalized beam-energy response convolves the line shape without changing this full area. It can completely change the observed peak height and apparent width. If a uniform acceptance $ϵ$ is not already unfolded, use ${\mathcal{I}}_{ee}={\mathcal{I}}_{ee}^{\mathrm{obs}}/ϵ$. The distinction between $B_{ee}^2$ for this exclusive final state and $B_{ee}$ for an inclusive final state is essential.

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The recoverable numerical conversion is nevertheless definite. With $M\simeq 3.1\mathrm{GeV}$, $B_{ee}=0.060$, $ϵ=0.50$, and $1{\mathrm{GeV}}^{-2}=389400\mathrm{nb}$, b givesThis coefficient assumes the plotted exclusive electronic sample has not already been corrected for acceptance. A restored area, with continuum subtraction and acceptance convention checked, would complete the estimate. A tabulated modern J/psi meson width would be external information rather than a reading of these data.

The narrow scattering resonance indicates suppressed strong interaction decay. In the quark model, $c\bar{c}$ is below the open-charm quark $D\bar{D}$ threshold, so its constituents cannot simply emerge in separate charmed mesons. The OZI rule suppresses annihilation into light hadrons; for a $1^{--}$ color-singlet state the leading purely gluonic channel needs three gluons. The observed leptonic channels remain competitive. Its mass and $I=0$ fit charmonium; the small width alone would not uniquely identify the new quark flavour without the wider spectrum and threshold evidence.

For a neutral $c\bar{c}$ bound state, meson parity and charge conjugation give $P=(-1{)}^{L+1}$ and $C=(-1{)}^{L+S}$. In spectroscopic notation $n{}^{2S+1}{L}_J$, the integer $n$ labels radial excitation, $L$ is orbital angular momentum, $S$ the total constituent spin, and $J$ their total angular momentum. The surviving level diagram contains eight states:

The spin-dependent splittings distinguish the singlet and triplet levels and the three triplet P levels. The radial excitation changes neither parity nor charge conjugation.

The four allowed two-real-photon states are ${\eta }_c$, ${\eta }_c(2S)$, ${\chi }_{c0}$ and ${\chi }_{c2}$. Two photons have $C=+1$, excluding $J/\psi$, $\psi (2S)$ and $h_c$; the Landau-Yang theorem excludes ${\chi }_{c1}$ because $J=1$. Both scalar and pseudoscalar two-photon amplitudes exist, so $0^{++}$ and $0^{-+}$ are allowed. The J/psi meson was discovered first because its $1^{--}$ assignment permits direct, resonantly enhanced single-photon formation in $e^{+}{e}^{-}$ and a clean dilepton peak in hadron production. The lighter ${\eta }_c$ cannot be formed by one photon; it needs less favorable production channels and has a broader two-gluon decay.

A spin-zero ${\eta }_c$ at rest has no preferred axis, so $d\Gamma /d{\Omega }_p$ is isotropic after summing final spins. A J/psi meson formed in unpolarized high-energy $e^{+}{e}^{-}$ has transverse $m=\pm 1$ alignment. Its $p\bar{p}$ distribution can be writtenwhere $a,b$ contain the two independent helicity amplitudes. It is generally anisotropic; the approximate massless, helicity-conserving limit has $b\to 0$. The tabulated branching fraction does not fix $a/b$. An ensemble equally populated in all three spin projections would instead be isotropic: alignment of the production ensemble matters.