By analogy with the pion and kaon

PHYSICS · Multi-part · Fermi University Physics · Particle Physics & Quantum Fields · Flavour, neutrinos and CP violation · Flavour, neutrinos and CP violation

Symmetry and Flavour, Ch Quark mixing: the Cabibbo angle, the GIM mechanism and the CKM matrix

Problem

By analogy with the pion and kaon, the charged current couples the charm quark to ss with strength ∝cos⁡θC\propto\cos\theta_C (in the Ds+=csˉD_s^+ = c\bar s leptonic decay) and to dd with strength ∝sin⁡θC\propto\sin\theta_C (in the D+=cdˉD^+ = c\bar d leptonic decay), so that Γ(P→μν)=GF28π∣Vqq′∣2fP2mμ2mP(1−mμ2/mP2)2\Gamma(P \to \mu\nu) = \dfrac{G_F^2}{8\pi}|V_{qq'}|^2 f_P^2 m_\mu^2 m_P\left(1 - m_\mu^2/m_P^2\right)^2 applies to both, with ∣Vcd∣|V_{cd}| in place of ∣Vud∣|V_{ud}| and ∣Vcs∣|V_{cs}| in place of ∣Vus∣|V_{us}|. Data: B(D+→μ+ν)=3.74×10−4\mathcal{B}(D^+\to\mu^+\nu) = 3.74\times10^{-4}, τ(D+)=1.033×10−12 s\tau(D^+) = 1.033\times10^{-12}\ \mathrm{s}; B(Ds+→μ+ν)=5.50×10−3\mathcal{B}(D_s^+\to\mu^+\nu) = 5.50\times10^{-3}, τ(Ds+)=5.04×10−13 s\tau(D_s^+) = 5.04\times10^{-13}\ \mathrm{s}; fD=203.7 MeVf_D = 203.7\ \mathrm{MeV}, fDs=249.9 MeVf_{D_s} = 249.9\ \mathrm{MeV}; mD+=1869.66 MeVm_{D^+} = 1869.66\ \mathrm{MeV}, mDs+=1968.35 MeVm_{D_s^+} = 1968.35\ \mathrm{MeV}, mμ=105.658 MeVm_\mu = 105.658\ \mathrm{MeV}. ((a)) Find ∣Vcd/Vcs∣|V_{cd}/V_{cs}| from the two measured leptonic widths. ((b)) Compare it with tan⁡θC\tan\theta_C, using θC=13.0∘\theta_C = 13.0^\circ.

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