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arXiv 0705.3915hep-phhep-ex
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$B\toϕK^*$偏振拼图的测试方法

Testing Explanations of the $B\toϕK^*$ Polarization Puzzle

Alakabha Datta, Andrei V. Gritsan, David London, Makiko Nagashima, Alejandro Szynkman

摘要:

$B\toϕK^*$($\btos$)是三个独立的衰变,一个对应于末态矢量介子的每个极化(一个纵向,两个横向)。观察到横向衰变的分数$\fT$和纵向衰变的分数$\fL$大致相等:$\fTfL \simeq 1$,与天真的期望$\fT \ll \fL$相反。如果需要一个单一的解释所有的极化之谜,两种可能性仍然在标准模型:企鹅湮灭和重新散射。在本文中,我们研究的预测这两种解释$\btod$衰变的$\fTfL$。在$B \to ρρ$衰变中,只有$\bd \to ρ^0ρ^0$可能表现出大的$\fTfL$。在与U自旋有关的B衰变中,我们发现了两种有希望的可能性:(i)$B^+ \to K^{*0} ρ^+$($\btos$)和$B^+ \to \Kbar^{*0} K^{*+}$($\btod$);(ii)$\bs \to K^{*0} \Kbar^{*0}$($\btos$)和$\bd \to \Kbar^{*0} K^{*0}$($\btod$)。测量这些衰变对中的$\fTfL$将使我们能够测试企鹅湮灭和重新散射。最后,通过对$\bd \to \Kbar^{*0} K^{*0}$进行随时间变化的角度分析,可以区分企鹅湮灭和再散射。

英文摘要:

$B\toϕK^*$ ($\btos$) is three separate decays, one for each polarization of the final-state vector mesons (one longitudinal, two transverse). It is observed that the fraction of transverse decays, $\fT$, and the fraction of longitudinal decays, $\fL$, are roughly equal: $\fTfL \simeq 1$, in opposition to the naive expectation that $\fT \ll \fL$. If one requires a single explanation of all polarization puzzles, two possibilities remain within the standard model: penguin annihilation and rescattering. In this paper we examine the predictions of these two explanations for $\fTfL$ in $\btod$ decays. In $B \to ρρ$ decays, only $\bd \to ρ^0ρ^0$ can possibly exhibit a large $\fTfL$. In B decays related by U-spin, we find two promising possibilities: (i) $B^+ \to K^{*0} ρ^+$ ($\btos$) and $B^+ \to \Kbar^{*0} K^{*+}$ ($\btod$) and (ii) $\bs \to K^{*0} \Kbar^{*0}$ ($\btos$) and $\bd \to \Kbar^{*0} K^{*0}$ ($\btod$). The measurement of $\fTfL$ in these pairs of decays will allow us to test penguin annihilation and rescattering. Finally, it is possible to distinguish penguin annihilation from rescattering by performing a time-dependent angular analysis of $\bd \to \Kbar^{*0} K^{*0}$.

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

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