be measured by observer 3 in the corresponding measurement along direction $\vec{e}_3$ (i.e., if we consider the total ensemble formed out of subensembles $+$ and $-$), then the appropriate state accounting for particles 1 and 2 will consist of a mixture of both pure states $|\Psi^{+}_{12}\rangle$ and $|\Psi^{-}_{12}\rangle$ with respective weights $p_+$ and $p_-$; that is, $$\rho_{12} = p_+ |\Psi^{+}_{12}\rangle \langle \Psi^{+}_{12}|
+ p_- |\Psi^{-}_{12}\rangle \langle \Psi^{-}_{12}| .$$ It is a rather simple matter to show that the density matrix (7) can also be decomposed in the form $$\rho_{12} = c^{2}_{1} |\phi^{+}_{12}\rangle \langle \phi^{+}_{12}|
+ c^{2}_{2} |\phi^{-}_{12}\rangle \langle \phi^{-}_{12}| ,$$ with $|\phi^{+}_{12}\rangle = |z_1\rangle |z_2\rangle$, and $|\phi^{-}_{12}\rangle = |-z_1\rangle |-z_2\rangle$. It should be noted at this point that both expressions (7) and (8) for $\rho_{12}$ can also be obtained by taking the partial trace of the density operator $|\Psi\rangle \langle\Psi|$ over the states corresponding to particle 3. So, on choosing the basis states for particle 3 to be $\{|z_3 \rangle^{\textstyle\ast},|-z_3 \rangle^{\textstyle\ast}\}$, we can put $$\rho_{12} = { }^{\textstyle\ast} \langle z_3 |\Psi\rangle\langle\Psi|z_3 \rangle^{\textstyle\ast}
+ { }^{\textstyle\ast} \langle -z_3 |\Psi\rangle\langle\Psi|-z_3 \rangle^{\textstyle\ast} .$$ Replacing now $|\Psi\rangle$ by the state in Eq. (4) we obtain quickly the density matrix (7). Likewise, by tracing over 