| \Xi_c \bar{K}, I=0 \rangle &=& -\frac{1}{\sqrt{2}} \Big|\Xi_c^+ K^- + \Xi_c^0 \bar{K}^0\Big\rangle\, ,\nonumber\\
| \Xi D, I=0 \rangle &=& -\frac{1}{\sqrt{2}} \Big|\Xi^0D^0 - \Xi^-D^+\Big\rangle\, ,\nonumber\\
| \Xi^*_c \bar{K}, I=0 \rangle &=& -\frac{1}{\sqrt{2}} \Big| \Xi^{* +}_c K^- + \Xi^{* 0}_c \bar{K}^0\Big\rangle\, ,\nonumber\\
| \Xi^* D, I=0 \rangle &=& -\frac{1}{\sqrt{2}} \Big|\Xi^{*0} D^0 + \Xi^{*-} D^+\Big\rangle\, .\end{aligned}$$ With these wave functions and the prescription to calculate the $VPP$ and $VBB$ vertices we can construct the matrix elements of the transition potential between the states of Table \[tab1\]. Some examples are shown in Appendix \[app\].

Following the steps of Appendix \[app\] it becomes easy and systematic to evaluate all the matrix elements and we find $$\label{vij}
V_{ij}= D_{ij}\frac{1}{4f_\pi^2} (p^0+ p^{\prime 0}) \, .$$

Alternatively, we can use another expression which includes relativistic correction in $s$-wave [@Bennhold] $$\label{vijRelat}
V_{ij}= D_{ij}\frac{2\sqrt{s} -M_{B_i} -M_{B_j}}{4f_\pi^2}  \sqrt{\frac{M_{B_i} + E_{B_i}}{2M_{B_i}}} \sqrt{\frac{M_{B_j} + E_{B_j}}{2M_{B_j}}}\, ,$$ where $M_{{B_i},{B_j}}$ and $E_{{B_i},{B_j}}$ stand for the mass and the center-of-mass energy of the baryons, respectively, and the matrix $D_{ij}$ is given in Table \[tabDij\].

\[tabDij\]

           $J=1/2$               $\Xi_c\bar{K}$     $\Xi^{\prime}_c\bar{K}$              $\Xi D$                   $\Omega_c \eta$            $\Xi D^*$          $\Xi_c \bar{K}^*$          $\Xi^{\prime}_c\bar{K}^*$
  --------------------------- ------------------- --------------------------- ------------------------------ ----------------------------- --------------- ------------------------------ -----------------------------
        $\Xi_c\bar{K}$               $-1$                     $0$              $-\frac{1}{\sqrt{2}}\lambda$               $0$                    $0$                    $0$                            $0$
    $\Xi^{\prime}_c\bar{K}$                                  $-1$              $\frac{1}{\sqrt{6}}\lambda$       