= Solution
Alice's outcome is random. Conditional on knowing it, Bob may assign a different <quantum state>, but without that classical record he sees the same <density matrix>. For the displayed pure <two-qubit state>, his reduced <density matrix> before Alice's measurement is
$$
\rho_B=\begin{pmatrix}
|\alpha|^2+|\gamma|^2&\alpha\beta^*+\gamma\delta^*\\
\alpha^*\beta+\gamma^*\delta&|\beta|^2+|\delta|^2
\end{pmatrix}.
$$
After Alice's computational-basis measurement with the outcome unannounced, the weighted mixture $p_0\rho_{B|0}+p_1\rho_{B|1}$ is exactly this same matrix. More generally, for any local trace-preserving operation with measurement operators $K_a$,
$$
\rho'_B=\operatorname{Tr}_A\sum_a(K_a\otimes I)\rho_{AB}(K_a^\dagger\otimes I)
=\operatorname{Tr}_A\rho_{AB}=\rho_B,
\qquad\sum_aK_a^\dagger K_a=I.
$$
The equality follows from the cyclic property of the <partial trace> for operators acting on Alice's subsystem. Bob's every local outcome <probability> therefore remains unchanged. This is the <no-communication theorem>: $\boxed{\text{entanglement alone cannot send information faster than light}}$. Conditional state changes become usable correlations only when the classical outcomes are compared.
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