for the proton [@close]. Then, we can see that one gets the same result as using Eq. , and this is also the case for all transitions. Therefore we use this method to obtain the $VBB$ vertex in the charm sector. The extension of the vertex $\mathcal{L}_{VPP}$ to the charm sector is easier. We take the same structure as in Eq.  but now $P$ and $V$ are $$\label{Pmatrix}
P =
\left(
\begin{array}{cccc}
\frac{1}{\sqrt{2}} \pi^0 + \frac{1}{\sqrt{3}} \eta  + \frac{1}{\sqrt{6}} \eta^{\prime} & \pi^+ & K^+ & \bar{D}^0 \\
\pi^- &  - \frac{1}{\sqrt{2}} \pi^0 +\frac{1}{\sqrt{3}} \eta + \frac{1}{\sqrt{6}} \eta^{\prime} & K^0 & D^- \\
K^- & \bar{K}^0 & -\frac{1}{\sqrt{3}} \eta +\sqrt{\frac{2}{3}} \eta^{\prime} & D_s^- \\
D^0 & D^+ & D_s^+ & \eta_c\\
\end{array}
\right)\, ,$$ where we include the mixing between $\eta$ and $\eta^{\prime}$ [@bramon], and

$$\label{Vmatrix}
V =
\left(
\begin{array}{cccc}
\frac{1}{\sqrt{2}} \rho^0 + \frac{1}{\sqrt{2}} \omega & \rho^+ & K^{* +} & \bar{D}^{* 0} \\
\rho^- & -\frac{1}{\sqrt{2}} \rho^0 + \frac{1}{\sqrt{2}} \omega & K^{* 0} & \bar{D}^{* -} \\
K^{* -} & \bar{K}^{* 0} & \phi & D_s^{* -} \\
D^{* 0} & D^{* +} & D_s^{* +} & J/\psi\\
\end{array}
\right)\, .$$

It has been shown in Ref. [@luisbd] (see section 