view of Eq. (4) it thus follows that, provided $\vartheta_3 \neq n\pi$, $n=0,\pm1,\pm2,\ldots ,$ a spin measurement on particle 3 along (an otherwise arbitrary) direction $\vec{e}_3 (\vartheta_{3},\varphi_{3})$ leaves the particles 1 and 2 in an entangled state. Specifically, if the measurement result for particle 3 is found to be $z_3$, the normalized two-particle state for particles 1 and 2 becomes $|\Psi^{+}_{12}\rangle$, whereas for a measurement result equal to $-z_3$ it becomes $|\Psi^{-}_{12}\rangle$. However, from expression (4), we can see that the probability of obtaining the result $z_3$ ($-z_3$) in a spin measurement on particle 3 along the axis $\vec{e}_3 (\vartheta_{3},\varphi_{3})$ is $p_+$ ($p_-$). Of course we have $p_+ + p_- =1$. It will further be noted that the states $|\Psi^{+}_{12}\rangle$ and $|\Psi^{-}_{12}\rangle$ are orthogonal to each other provided that $|c_1|=|c_2|=1/{\sqrt{2}}$.

If we analyze the results of observers 1 and 2 only when observer 3 obtains the result $z_3$ ($-z_3$) (i.e., if we restrict ourselves to either one of subensembles $+$ or $-$) then the adequate state describing the spin of particles 1 and 2 will be the pure state $|\Psi^{+}_{12}\rangle$ ($|\Psi^{-}_{12}\rangle$). However, if we attemp to analyze the results of observers 1 and 2 irrespective of what result happens to 