Measurement of $CP$ asymmetry in $D^0 \rightarrow K^- K^+$ and $D^0 \rightarrow \pi^- \pi^+$ decays

TL;DR

This paper measures time-integrated $CP$ asymmetries in $D^0$ decays to $K^-K^+$ and $\pi^-\pi^+$ using proton-proton collision data, finding results consistent with no $CP$ violation and improving the precision of these measurements.

Contribution

First measurement of $CP$ asymmetry difference in $D^0$ decays using semileptonic $b$-hadron decays with muon charge tagging at LHC.

Findings

$\Delta A_{CP} = (+0.14 \pm 0.16 ext{(stat)} ext{+} 0.08 ext{(syst)}) ext{ extperthousand}$

$A_{CP}(K^-K^+) = (-0.06 ext{ extpm} 0.15 ext{(stat)} ext{ extpm} 0.10 ext{(syst)}) ext{ extperthousand}$

$A_{CP}(\pi^-\pi^+) = (-0.20 ext{ extpm} 0.19 ext{(stat)} ext{ extpm} 0.10 ext{(syst)}) ext{ extperthousand}$

Abstract

Time-integrated $CP$ asymmetries in $D^0$ decays to the final states $K^- K^+$ and $\pi^- \pi^+$ are measured using proton-proton collisions corresponding to $3\mathrm{\,fb}^{-1}$ of integrated luminosity collected at centre-of-mass energies of $7\mathrm{\,Te\kern -0.1em V}$ and $8\mathrm{\,Te\kern -0.1em V}$. The $D^0$ mesons are produced in semileptonic $b$-hadron decays, where the charge of the accompanying muon is used to determine the initial flavour of the charm meson. The difference in $CP$ asymmetries between the two final states is measured to be \begin{align} \Delta A_{CP} = A_{CP}(K^-K^+)-A_{CP}(\pi^-\pi^+) = (+0.14 \pm 0.16\mathrm{\,(stat)} \pm 0.08\mathrm{\,(syst)})\% \ . \nonumber \end{align} A measurement of $A_{CP}(K^-K^+)$ is obtained assuming negligible $CP$ violation in charm mixing and in Cabibbo-favoured $D$ decays. It is found to be \begin{align} A_{CP}(K^-K^+) = (-0.06 \pm 0.15\mathrm{\,(stat)} \pm 0.10\mathrm{\,(syst)}) \% \ ,\nonumber \end{align} where the correlation coefficient between $\Delta A_{CP}$ and $A_{CP}(K^-K^+)$ is $\rho=0.28$. By combining these results, the $CP$ asymmetry in the $D^0\rightarrow\pi^-\pi^+$ channel is $A_{CP}(\pi^-\pi^+)=(-0.20\pm0.19\mathrm{\,(stat)}\pm0.10\mathrm{\,(syst)})\%$.