 Solution

For classical events, $A\setminus C\subset (A\setminus B)\cup (B\setminus C)$: an outcome in $A$ but not in $C$ either fails $B$ or belongs to $B\setminus C$. Monotonicity and the union bound prove the Fréchet-Bell inequalityMalus' law gives transmission ${\cos }^2\gamma$ for a photon polarized by the first filter, so its conditional rejection probability at the second is $\boxed{{\sin }^2\gamma }$. Conditional and joint probabilities must be distinguished: $P(A\setminus B)=P(A)P(B^c\mid A)$.

For equal unbiased marginals $P(A)=P(B)=1/2$, the quantum two-setting rejection probabilities are $\frac{1}{2}{\sin }^2({\theta }_A-{\theta }_B)$. A classical common assignment of all three outcomes would therefore require ${\sin }^2({\theta }_A-{\theta }_B)+{\sin }^2({\theta }_B-{\theta }_C)\ge {\sin }^2({\theta }_A-{\theta }_C)$. Choose directions $0,\gamma ,2\gamma$. ThenFor example $\gamma =\pi /6$ gives $1/2<3/4$. More generally ${\sin }^{2}a+{\sin }^{2}b-{\sin }^2(a+b)=-2\sin a\sin b\cos (a+b)$. In a triangle whose side directions are these filters, the internal angle supplementary to $a+b$ is obtuse exactly in this violation range. This specifies the angle convention behind the triangle description: with polarization axes regarded as unoriented lines, a claim that every sufficiently large physical axis separation violates the inequality would not be correct. The contradiction concerns a single classical joint model for all settings, not the validity of ordinary probability or a comparison of arbitrarily renormalized sequential experiments.

Use $|0⟩=|{\alpha }_0⟩$ and $|1⟩=|{\alpha }_1⟩$. Projecting the first photon with $|0⟩⟨0|$ gives the unnormalized product vector $c_0|00⟩$, with success probability $|c_0{|}^2$. Projection of the second onto $|\theta ⟩=\cos \theta |0⟩+\sin \theta |1⟩$ givesConditioned on success, normalize these vectors; a zero amplitude means that outcome never occurs. For the maximally entangled equal-amplitude pair, rotation of both real polarization bases gives the same correlations for arbitrary settings. Comparing the appropriately normalized coincidence/rejection frequencies at three setting pairs tests the inequality above. A general unequal-amplitude state need not reproduce the unbiased rotationally invariant probabilities, so the maximally entangled choice supplies the clean explicit example.

For swapping, label the four systems in order ${\xi }_1,{\eta }_1,{\xi }_2,{\eta }_2$ and let $|\Phi {⟩}_{\eta }=2^{-1/2}{\sum }_{j=0}^1|{\eta }_j{\eta }_j⟩$. Its overlap with $|{\eta }_r{\eta }_s⟩$ is ${\delta }_{rs}/\sqrt{2}$. Applying its rank-one projector to Alice's two systems givesAfter regrouping the remote and local systems, this factorizes into $[{\sum }_r{b}_r^2|{\xi }_r{\xi }_r⟩/\sqrt{2}]\otimes |\Phi {⟩}_{\eta }$. The selected outcome has probability $p=\frac{1}{2}{\sum }_r|b_r{|}^4$. Therefore the normalized conditional remote state isIt has Schmidt rank at least two whenever the original pair was entangled, since squaring a nonzero coefficient does not make it zero. Thus the two photons that never interacted become entangled conditional on Alice's joint measurement outcome. This is entanglement swapping. Alice is projecting onto a specified Bell state, not performing a universal test of whether an arbitrary unknown state is entangled. Averaging all her outcomes leaves the remote reduced density operator unchanged, so no faster-than-light signal is available without her classical outcome label.