Isospin strikes back

Assuming isospin conservation, the decay of a \(c\bar{c}\) vector meson into the \(\Lambda \bar{\Sigma }^0+\mathrm{c.c.}\) final state is purely electromagnetic. At the leading order, the \(c\bar{c}\) vector meson first converts into a virtual photon that, then produces the \(\Lambda \bar{\Sigma }^0+\mathrm{c.c.}\) final state. Moreover, such a mechanism, i.e., the virtual photon coupling to \(\Lambda \bar{\Sigma }^0+\mathrm{c.c.}\), is the sole intermediate process through which, in Born approximation, the reaction \(e^+e^-\rightarrow \Lambda \bar{\Sigma }^0+\mathrm{c.c.}\) does proceed. It follows that any significant difference between the amplitudes of the processes \(c\bar{c}\rightarrow \Lambda \bar{\Sigma }^0+\mathrm{c.c.}\) and \(e^+e^-\rightarrow \Lambda \bar{\Sigma }^0+\mathrm{c.c.}\) at the \(c\bar{c}\) mass must be ascribed to an isospin-violating contribution in the \(c\bar{c}\) decay. In Ferroli et al. (Eur Phys J C 80: 903, 2020) we studied the decay of the \(\psi (2S)\) vector meson into \(\Lambda \bar{\Sigma }^0+\mathrm{c.c.}\) and, on the light of the large branching fraction

$$\begin{aligned} \textrm{BR}_{18}(\psi (2S)\rightarrow \Lambda \bar{\Sigma }^0+\mathrm{c.c.})=(1.23\pm 0.24)\times 10^{-5}, \end{aligned}$$

published in the 2018 edition of the Review of Particle Physics (Tanabashi et al. in Phys Rev D 98: 030001, 2018), we claimed either the presence of a significant isospin-violating contribution or, with a lesser emphasis, a “not complete reliability of the only available datum”. In any case, we propose a new measurement. Apparently, our second and considered less serious hypothesis was the right one, indeed the branching fraction published in the 2024 edition of the Review of Particle Physics (Navas et al. in Phys Rev D 110: 030001, 2024) is

$$\begin{aligned} \textrm{BR}(\psi (2S)\rightarrow \Lambda \bar{\Sigma }^0+\mathrm{c.c.})=(1.6\pm 0.7)\times 10^{-6}, \end{aligned}$$

more than seven times lower with the error that increased from \(\sim 20\%\) to \(\sim 45\%\).