 Solution

Assume the proposed device distinguishes the four displayed eigenstates, as an ideal rank-one projective measurement. This assumption matters: if all four had the same eigenvalue, the identity observable would admit that eigenbasis but its Lüders rule measurement would do nothing and could not signal. The printed eigenvectors alone do not exclude this degeneracy.

For the complete resolving measurement, let $c=\cos \theta$, $s=\sin \theta$, and ${\Pi }_j$ be the four rank-one projectors. If the nonlocal outcome is ignored, the nonselective projective measurement channel is ${\mathcal{D}}_{\theta }(\rho )={\sum }_j{\Pi }_j\rho {\Pi }_j$. Each listed state's local reduced density matrix is diagonal in $Z_A$, with $Z_A$ expectation $\cos 2\theta$ for either plus state and $-\cos 2\theta$ for either minus state. ConsequentlyCompute $C_{\theta }$ in its even and odd two-dimensional blocks: both have diagonal entries $\cos 2\theta ,-\cos 2\theta$ and off-diagonal entries $\sin 2\theta$. ThusandTo signal, prepare Alice in $|+⟩$ and Bob initially in $|+⟩$. Bob encodes a bit by either doing nothing or applying a local Pauli Z gate, which changes his state to $|-⟩$. Alice's input reduced density matrix is identical in both cases, but after the hypothetical instantaneous measurement her local $Z_A$ expectation isThe two probabilities for Alice's outcome $0$ are $(1\pm \sin 2\theta \cos 2\theta )/2$, so their difference is $\sin 2\theta \cos 2\theta$. It is strictly positive for $0<\theta <\pi /4$. Repeated trials let Alice infer Bob's bit while their operations are still spacelike, violating quantum no-signalling and relativistic causality. The relativistic causality constraint on an ideal nonlocal measurement therefore permits onlywithin the specified interval. The argument requires no rapid communication of the hypothetical nonlocal outcome: Alice reads her own changed local statistics. It rules out the full ideal instrument, not merely the later classical comparison of locally obtained records.