<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" display="inline"><mml:mi>U</mml:mi></mml:math>-spin puzzle in <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" display="inline"><mml:mi>B</mml:mi></mml:math> decays

We impose $U$-spin symmetry [$SU(2{)}_{\text{Uspin}}$] on the Hamiltonian for $B$ decays. As expected, we find the equality of amplitudes related by the exchange $d\ensuremath{\leftrightarrow}s$. We also find that the amplitudes for the $\mathrm{\ensuremath{\Delta}}S=0$ processes ${B}^{0}\ensuremath{\rightarrow}{\ensuremath{\pi}}^{+}{\ensuremath{\pi}}^{\ensuremath{-}}$, ${B}_{s}^{0}\ensuremath{\rightarrow}{\ensuremath{\pi}}^{+}{K}^{\ensuremath{-}}$, and ${B}^{0}\ensuremath{\rightarrow}{K}^{+}{K}^{\ensuremath{-}}$ form a $U$-spin triangle relation. The amplitudes for ${B}_{s}^{0}\ensuremath{\rightarrow}{K}^{+}{K}^{\ensuremath{-}}$, ${B}^{0}\ensuremath{\rightarrow}{\ensuremath{\pi}}^{\ensuremath{-}}{K}^{+}$, and ${B}_{s}^{0}\ensuremath{\rightarrow}{\ensuremath{\pi}}^{+}{\ensuremath{\pi}}^{\ensuremath{-}}$ form a similar $\mathrm{\ensuremath{\Delta}}S=1$ triangle relation. And these two triangles are related to one another by $d\ensuremath{\leftrightarrow}s$. We perform fits to the observables for these six decays. If perfect $U$ spin is assumed, then the fit is very poor. If $U$-spin-breaking contributions are added, then we find many scenarios that can explain the data. However, in all cases, 100% $U$-spin breaking is required, considerably larger than the naive expectation of $\ensuremath{\sim}20%$. This is the $U$-spin puzzle; it may be strongly hinting at the presence of new physics.

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