= Solution
Taking the <partial trace> over either photon of the pure pair state gives
$$
\boxed{\rho_A=\rho_B=\frac12I.}
$$
The two <Schmidt coefficients> are both $1/\sqrt2$, so the <entanglement entropy> is one bit for every $\varphi$. A local <phase> gate on Alice maps $|\Psi^0\rangle$ to $|\Psi^\varphi\rangle$, proving local-unitary equivalence. In strict four-vector Bell-basis terminology only $\varphi=0,\pi$ give the standard $|\Psi^+\rangle,|\Psi^-\rangle$; every <phase> nevertheless gives a maximally entangled Bell-type state.
The eigenpolarizations of the requested observable are
$$
|+_\alpha\rangle=\cos\frac\alpha2|H\rangle+\sin\frac\alpha2|V\rangle,\qquad
|-_\alpha\rangle=-\sin\frac\alpha2|H\rangle+\cos\frac\alpha2|V\rangle.
$$
A <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 $\langle Z\otimes Z\rangle=-1$, $\langle X\otimes X\rangle=\cos\varphi$, and $\langle Z\otimes X\rangle=\langle X\otimes Z\rangle=0$. Expanding the two observables therefore gives
$$
\boxed{\langle\sigma_\alpha\otimes\sigma_\beta\rangle=-\cos\alpha\cos\beta+\cos\varphi\sin\alpha\sin\beta.}
$$
The half-angle between the Bloch observable parameter and the physical linear-polarization analyser angle is essential.
\Image[/past-exam-of-the-laser-science-and-quantum-information-processing-course-of-the-university-of-oxford/2018-polarization-analyser-and-bell-signal.png]
{title=<Polarization> measurement and the phase-averaged <Bell inequality> signal}
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