Past exam of the particle physics course of the University of Oxford / 2019 / 4 / c / Solution

Use $B^0=d\bar{b}$ and ${\bar{B}}^0=\bar{d}b$. A box Feynman diagram with two W bosons and internal up-type quarks changes one neutral flavour into the other. The top quark contribution dominates the dispersive mixing amplitude and carries the squared CKM combination involving $V_{td},V_{tb}$. With a consistent meson phase convention,The illustration in b shows this box and the dominant decay topology. A suitable common final state is $f=J/\psi K_S$, reconstructed for example through $J/\psi \to {\mu }^{+}{\mu }^{-}$ and $K_S\to {\pi }^{+}{\pi }^{-}$. The leading $\bar{b}\to \bar{c}c\bar{s}$ tree transition gives $B^0\to J/\psi K^0$, while the conjugate gives ${\bar{B}}^0\to J/\psi {\bar{K}}^0$. Neutral kaon mixing projects both kaon flavours onto the observed $K_S$.

The final state is approximately a CP eigenstate with ${\eta }_f=-1$: $J/\psi$ and $K_S$ each have CP approximately $+1$, and their relative orbital angular momentum is $L=1$ to couple the vector charmonium to the spin-zero parent. Neglecting the small kaon CP impurity and subleading decay amplitudes,There are two indistinguishable paths, direct decay and mixing followed by decay, so their amplitudes interfere. For $\Delta {\Gamma }_d\simeq 0$ and negligible direct CP violation, the time-dependent neutral-B CP asymmetry isMeasure the production flavour and proper decay time, fit the oscillation coefficient, and extract $\sin 2\beta$. This determines $\beta$ up to the usual trigonometric ambiguities unless further information is used. The cancellation of the dominant decay phase is why this channel isolates the angle involving $V_{td}$ without needing $V_{ub}$.