 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 isAfter 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 $K_a$,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.