\ast},|-z_3 \rangle^{\textstyle\ast}\}$ for particle 3, related to the original basis vectors by

$$|z_{3} \rangle = \cos \frac{\vartheta_3}{2} e^{iz_{3}\varphi_{3}/2} |z_{3} \rangle^{\textstyle \ast} - z_{3} \sin \frac{\vartheta_3}{2} e^{iz_{3}\varphi_{3}/2} |-z_{3} \rangle^{\textstyle\ast},$$

$$|-z_{3} \rangle = z_{3} \sin \frac{\vartheta_3}{2} e^{-iz_{3}\varphi_{3}/2} |z_{3} \rangle^{\textstyle\ast}
+ \cos \frac{\vartheta_3}{2} e^{-iz_{3}\varphi_{3}/2} |-z_{3} \rangle^{\textstyle\ast}.$$

Using Eqs. (3a) and (3b) we can rewrite the three-particle state (2) as $$|\Psi \rangle = p^{1/2}_{+}|\Psi^{+}_{12} \rangle |z_{3} \rangle^{\textstyle\ast}
+ p^{1/2}_{-}|\Psi^{-}_{12} \rangle |-z_{3} \rangle^{\textstyle\ast},$$ where

$$\begin{aligned}
|\Psi^{+}_{12} \rangle &=& p^{-1/2}_{+} \left( c_1 \cos \frac{\vartheta_{3}}{2}
|z_1 \rangle |z_2 \rangle \right. \nonumber   \\
&& + \left. c_2 z_3 \sin \frac{\vartheta_{3}}{2} e^{-iz_{3}\varphi_{3}}
|-z_1 \rangle |-z_2 \rangle \right),\end{aligned}$$

$$\begin{aligned}
|\Psi^{-}_{12} \rangle &=& p^{-1/2}_{-} \left( -c_1 z_3 \sin \frac{\vartheta_{3}}{2}
|z_1 \rangle |z_2 \rangle \right. \nonumber   \\
&& + \left. c_2 \cos \frac{\vartheta_{3}}{2} e^{-iz_{3}\varphi_{3}}
|-z_1 \rangle |-z_2 \rangle \right),\end{aligned}$$

and

$$\begin{aligned}
p_+ &=& c^{2}_{1} \cos^2 \left( \frac{\vartheta_{3}}{2} \right)
+ c^{2}_{2} \sin^2 \left( \frac{\vartheta_{3}}{2} \right),  \\
p_- &=& c^{2}_{1} \sin^2 \left( \frac{\vartheta_{3}}{2} \right)
+ c^{2}_{2} \cos^2 \left( \frac{\vartheta_{3}}{2} \right).\end{aligned}$$

Since the basis vectors $|\pm z_3 \rangle$ represent definite spin $\pm z_3$ along the $z$ axis for particle 3, the basis vectors $|\pm z_3 \rangle^{\textstyle\ast}$ represent definite spin $\pm z_3$ for particle 3 along the direction characterized by the angles $\vartheta_3$ and $\varphi_3$. In 