 Solution

Taking the partial trace over either photon of the pure pair state givesThe two Schmidt coefficients are both $1/\sqrt{2}$, so the entanglement entropy is one bit for every $\phi$. A local phase gate on Alice maps $|{\Psi }^0⟩$ to $|{\Psi }^{\phi }⟩$, proving local-unitary equivalence. In strict four-vector Bell-basis terminology only $\phi =0,\pi$ give the standard $|{\Psi }^{+}⟩,|{\Psi }^{-}⟩$; every phase nevertheless gives a maximally entangled Bell-type state.

The eigenpolarizations of the requested observable areA polarizing beam splitter with its transmitted polarization axis at angle $\alpha /2$ to horizontal separates these two orthogonal states. Assign outcome $+1$ to its first detector and $-1$ to the other. Equivalently, use a fixed horizontal/vertical splitter preceded by a half-wave plate whose fast axis is at $\alpha /4$; the diagram shows this equivalent implementation. Bob uses $\beta /2$ for his splitter axis, or $\beta /4$ for the preceding half-wave plate.

The pair correlations are $⟨Z\otimes Z⟩=-1$, $⟨X\otimes X⟩=\cos \phi$, and $⟨Z\otimes X⟩=⟨X\otimes Z⟩=0$. Expanding the two observables therefore givesThe half-angle between the Bloch observable parameter and the physical linear-polarization analyser angle is essential.