The $D_s$-meson leading-twist distribution amplitude within the QCD sum rules and its application to the $B_s \to D_s$ transition form factor

TL;DR

This paper calculates the $D_s$ meson distribution amplitude using QCD sum rules, then applies it to determine the $B_s o D_s$ transition form factor, branching ratio, and CKM matrix element, aligning with lattice QCD results.

Contribution

It introduces a precise calculation of the $D_s$ meson distribution amplitude moments and applies them to improve the $B_s o D_s$ transition form factor estimation.

Findings

The moments of the $D_s$ meson distribution amplitude are accurately computed.

The $B_s o D_s$ transition form factor agrees with lattice QCD predictions.

The branching ratio and CKM element are extracted with quantified uncertainties.

Abstract

We make a detailed study on the $D_s$ meson leading-twist LCDA $\phi_{2;D_s}$ by using the QCD sum rules within the framework of the background field theory. To improve the precision, its moments $\langle \xi^n\rangle _{2;D_s}$ are calculated up to dimension-six condensates. At the scale $\mu = 2{\rm GeV}$, we obtain: $\langle \xi^1\rangle _{2;D_s}= -0.261^{+0.020}_{-0.020}$, $\langle \xi^2\rangle _{2;D_s} = 0.184^{+0.012}_{-0.012}$, $\langle \xi^3\rangle _{2;D_s} = -0.111 ^{+0.007}_{-0.012}$ and $\langle \xi^4\rangle _{2;D_s} = 0.075^{+0.005}_{-0.005}$. Using those moments, the $\phi_{2;D_s}$ is then constructed by using the light-cone harmonic oscillator model. As an application, we calculate the transition form factor $f^{B_s\to D_s}_+(q^2)$ within the light-cone sum rules (LCSR) approach by using a right-handed chiral current, in which the terms involving $\phi_{2;D_s}$ dominates the LCSR. It is noted that the extrapolated $f^{B_s\to D_s}_+(q^2)$ agrees with the Lattice QCD prediction. After extrapolating the transition form factor to the physically allowable $q^2$-region, we calculate the branching ratio and the CKM matrix element, which give $\mathcal{B}(\bar B_s^0 \to D_s^+ \ell\nu_\ell) = (2.03^{+0.35}_{-0.49}) \times 10^{-2}$ and $|V_{cb}|=(40.00_{-4.08}^{+4.93})\times 10^{-3}$.