${{\boldsymbol \chi}_{{c0}}{(3860)}}$ $I^G(J^{PC})$ = $0^+(0^{+ +})$
The assignment $\mathit J{}^{P} = 0+$ is preferred over ${}^{}2{}^{+}$ by 2.5 sigma. Observed by
CHILIKIN 2017 using full amplitude analysis of the process ${{\mathit e}^{+}}$ ${{\mathit e}^{-}}$ $\rightarrow$ ${{\mathit J / \psi}}{{\mathit D}}{{\overline{\mathit D}}}$ , where ${{\mathit D}}$ = ${{\mathit D}^{0}}$, ${{\mathit D}^{+}}$.