= 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 inequality>
$$
P(A\setminus C)\leq P(A\setminus B)+P(B\setminus C).
$$
<Malus' 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 $\tfrac12\sin^2(\theta_A-\theta_B)$. A classical common assignment of all three outcomes would therefore require $\sin^2(\theta_A-\theta_B)+\sin^2(\theta_B-\theta_C)\geq\sin^2(\theta_A-\theta_C)$. Choose directions $0,\gamma,2\gamma$. Then
$$
\boxed{2\sin^2\gamma<\sin^2(2\gamma)\quad\text{when }0<\gamma<\pi/4.}
$$
For example $\gamma=\pi/6$ gives $1/2<3/4$. More generally $\sin^2a+\sin^2b-\sin^2(a+b)=-2\sin a\sin b\cos(a+b)$. 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 $|0\rangle=|\alpha_0\rangle$ and $|1\rangle=|\alpha_1\rangle$. Projecting the first photon with $|0\rangle\langle0|$ gives the unnormalized product vector $c_0|00\rangle$, with success <probability> $|c_0|^2$. <Projection> of the second onto $|\theta\rangle=\cos\theta|0\rangle+\sin\theta|1\rangle$ gives
$$
\boxed{c_0\cos\theta\,|0\rangle|\theta\rangle,\qquad P(\text{both transmissions})=|c_0|^2\cos^2\theta.}
$$
Conditioned 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 $|\Phi\rangle_\eta=2^{-1/2}\sum_{j=0}^1|\eta_j\eta_j\rangle$. Its overlap with $|\eta_r\eta_s\rangle$ is $\delta_{rs}/\sqrt2$. Applying its rank-one <projector> to Alice's two systems gives
$$
\boxed{\frac12\sum_{r,k}b_r^2|\xi_r\rangle|\eta_k\rangle|\xi_r\rangle|\eta_k\rangle.}
$$
After regrouping the remote and local systems, this factorizes into $[\sum_rb_r^2|\xi_r\xi_r\rangle/\sqrt2]\otimes|\Phi\rangle_\eta$. The selected outcome has <probability> $p=\tfrac12\sum_r|b_r|^4$. Therefore the normalized conditional remote state is
$$
\boxed{|\chi\rangle_\xi=\frac{\sum_rb_r^2|\xi_r\xi_r\rangle}{\sqrt{\sum_r|b_r|^4}}.}
$$
It 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> \b[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.
Back to article page
