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Expand a normalized product state asThus $\alpha =ac$, $\beta =ad$, $\gamma =bc$, $\delta =bd$, givingConversely the nonzero coefficient matrix $C=\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right)$ has determinant zero precisely when it has rank one, so the condition also suffices for a pure two-qubit state to factor.

For any fixed pair of local measurement settings with operators $E_i,F_j$, the Born rule factorizes:Hence a product state has independent local results in every pair of local bases. This assumes independent settings and local measurements; shared classical randomization of preparations or a subsequent joint interaction can generate correlations. A separable mixture is not necessarily a single product state.

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.