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Let me introduce the problem:

In a two electron fixed nucleus problem the "body" is the atom, whose electrons are located relative to the nucleus by the coordinates $r_1$ and $r_2$, and the angle $\Theta_{12}$ is the angle between the vectors $r_1$ and $r_2$. The orientation of the body relative to the space-fixed frame is specified by the Euler angles $\alpha$, $\beta$, $\gamma$ and $\theta_1$, $\phi_1$, $\theta_2$, $\phi_2$ are the spherical angles for the two electrons in the space-fixed frame.

After some of algebra one can find the relations between $\alpha$, $\beta$, $\gamma$, $\Theta_{12}$ and $\theta_1$, $\theta_2$, $\phi_1$, $\phi_2$:

$$\sin\Theta_{12}\cos\beta=\sin\theta_1\sin\theta_2\sin(\phi_2-\phi_1)$$ $$\sin\Theta_{12}\sin\beta \sin\alpha= \sin\theta_1\sin\phi_1\cos\theta_2- \cos\theta_1\sin\theta_2\sin\phi_2$$ $$\sin\Theta_{12}\sin\beta \cos\alpha= \sin\theta_1\cos\phi_1\cos\theta_2- \cos\theta_1\sin\theta_2\cos\phi_2$$ $$2\sin\frac{\Theta_{12}}{2}\cos\gamma= \sin\theta_2\cos(\phi_2-\alpha)-\sin\theta_1\cos(\phi_1-\alpha)$$ $2\sin\frac{\Theta_{12}}{2}\sin\gamma= (\cos\theta_2-\cos\theta_1)\sin\beta + \sin\theta_2 \sin(\phi_2-\alpha)-\sin\theta_1\sin(\phi_1-\alpha)\cos\beta$ $$\cos\Theta_{12}= \cos\theta_1\cos\theta_2+\sin\theta_1\sin\theta_2\cos(\phi_1-\phi_2)$$

I know the components of the total angular momentum operator are readily expressed in terms of the particles' spherical angles. Thus, for example,


If above we showed that $\theta_1$, $\theta_2$, $\phi_1$, $\phi_2$ are implicit functions of the for angles $\alpha$, $\beta$, $\gamma$ and $\Theta_{12}$. I applied the chain and product rules of partial diferentiation to find $L_x$ in terms of $\alpha$, $\beta$, $\gamma$ and $\Theta_{12}$:

$$-\frac{i}{\hbar}L_x=A_\alpha\frac{\partial}{\partial\alpha}+A_\beta\frac{\partial}{\partial\beta}+A_\gamma\frac{\partial}{\partial\gamma}+A_{\Theta_{12}}\frac{\partial}{\partial\Theta_{12}}$$ where

$$A_{\chi}=\sin\phi_1\frac{\partial\chi}{\partial\theta_1}+\sin\phi_2\frac{\partial\chi}{\partial\theta_2}+\cot\theta_1\cos\phi_1\frac{\partial\chi}{\partial\phi_1}+\cot\theta_2\cos\phi_2\frac{\partial\chi}{\partial\phi_2}$$ and $\chi$ can be anyone of the angles $\alpha$, $\beta$, $\gamma$ and $\Theta_{12}$.

I've found in the literature $$A_{\Theta_{12}}=0$$ $$A_{\alpha}=\sin\alpha \cot\beta$$ $$A_{\beta}=-\cos\alpha$$ $$A_{\gamma}=-\sin\alpha/\sin\beta$$

I've been following the derivation of these equations (starting at section 4) from this article Symmetric Euler-Angle Decomposition of the Two-Electron Fixed-Nucleus Problem, however i don't know how to obtain these coefficients. ¿Could you give me some hints in order to proceed?

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