III of Ref. [@miyahara] (the same assignment is also used in Ref. [@Pavao:2017cpt]). Thus $$\begin{aligned}
    \label{xiwave}
    \Xi^0 \equiv \frac{1}{\sqrt{2}}(\phi_{MS}\,\chi_{MS}+\phi_{MA}\,\chi_{MA}),\end{aligned}$$ with $$\begin{aligned}
    \phi_{MS} = \frac{1}{\sqrt{6}} [s(us+su) - 2uss]\, ,\end{aligned}$$ $$\begin{aligned}
    \phi_{MA} = -\frac{1}{\sqrt{2}}[s(us - su)]\, ,\end{aligned}$$ and $\chi_{MS}$ is given in Eq. , while $\chi_{MA}$ is given by $$\label{ma}
    \chi_{MA}=
    \left\{
      \begin{array}{rl}
        \hbox{$\displaystyle \frac{1}{\sqrt{2}}\uparrow (\uparrow\downarrow - \downarrow\uparrow)$}, & \hbox{for $S_z=1/2$}, \\
        \hbox{$\displaystyle \frac{1}{\sqrt{2}}\downarrow (\uparrow\downarrow - \downarrow\uparrow)$}, & \hbox{for $S_z=-1/2$};
      \end{array}
    \right.$$

7.  $\Xi^-$: as in Eq.  with $$\begin{aligned}
    \phi_{MS} = -\frac{1}{\sqrt{6}} [s(ds + sd) - 2dss] \, ,\end{aligned}$$ $$\begin{aligned}
    \phi_{MA} = \frac{1}{\sqrt{2}}[s(ds - sd)]\, .\end{aligned}$$ For the baryon states of spin $J^P={3/2}^+$ we have

8.  $\Xi^{*+}_c$: $c\frac{1}{\sqrt{2}}(us+su)$, and the symmetric spin wave function, $\chi_S = \, \uparrow\uparrow\uparrow$, ... ;

9.  $\Xi^{*0}_c$: $c\frac{1}{\sqrt{2}}(ds+sd)$, and $\chi_S$;

10. $\Omega^*_c$: $css$, and $\chi_S$;

11. $\Xi^{*0}$: $\frac{1}{\sqrt{3}}(sus + ssu +uss)$, and $\chi_S$;

12. $\Xi^{*-}$: $\frac{1}{\sqrt{3}}(sds + ssd +dss)$, and $\chi_S$;

We have to construct states with $I=0$ to match the $\Omega_c$. For that recall that our isospin multiplets are: $$\begin{aligned}
\bar{K} = {\ensuremath{\begin{pmatrix}\bar{K}^0\\-K^-\end{pmatrix}}};\,\,D = {\ensuremath{\begin{pmatrix}D^+\\-D^0\end{pmatrix}}};\,\,
 \Xi = {\ensuremath{\begin{pmatrix}\Xi^0\\-\Xi^-\end{pmatrix}}};\,\, \Xi^\ast = {\ensuremath{\begin{pmatrix}\Xi^{\ast 0}\\ \Xi{^{\ast -}}\end{pmatrix}}}; \nonumber\\
 \vspace{15pt}
 \Xi_c = {\ensuremath{\begin{pmatrix}\;\Xi_c^+\\ \;\Xi_c^0\end{pmatrix}}};\,\,
 \Xi^{\prime}_c = {\ensuremath{\begin{pmatrix}\;\Xi^{\prime +}_c\\ \; \Xi^{\prime 0}_c\end{pmatrix}}};\,\,\,\,\,
\,\,  \Xi^{*}_c = {\ensuremath{\begin{pmatrix}\Xi^{* +}_c\\\Xi^{*0}_c\end{pmatrix}}}\, ;\end{aligned}$$ and thus $$\begin{aligned}
