 Solution

The equal mixture of $|{\Phi }^{+}⟩$ and $|{\Phi }^{-}⟩$ cancels their off-diagonal coherence, leaving the separable quantum stateA trivial LHV model chooses a shared random bit $j$, uniformly zero or one, and prepares both local qubits in $|j⟩$. Each party uses the local Born rule for its own chosen spin direction, with independent local random seeds if a deterministic description is desired. This reproduces the entire joint measurement distribution, not merely the $z$-axis correlation.

Since the state has an explicit local hidden-variable theory for these local measurements, it cannot violate the CHSH inequality, by the derivation in part (b). Classical correlation of the shared preparation bit is sufficient; the cancellation of coherence removes the entanglement.