 Solution

Assume $|\alpha {|}^2+|\beta {|}^2=1$. Let Alice's first recorded quantum measurement outcome be $a\in \{0,1\}$. Immediately afterward she assigns the conditional pure quantum state $|a⟩$, with probabilities $p_0=|\alpha {|}^2$, $p_1=|\beta {|}^2$. Bob, who has no record, assigns the density operatorThis is a nonselective quantum measurement, which removes the initial computational basis coherence. It is not generally the original pure quantum state.

After the Hadamard gate, Alice assigns $H|a⟩=|+⟩$ for $a=0$, or $|-⟩$ for $a=1$. Bob assignsBoth $|+⟩$ and $|-⟩$ give either final computational basis outcome with probability $1/2$. Thus, after Alice records the second outcome $b$, she assigns $|b⟩$, whereas Bob assignsAlice's conditional quantum states are pure; Bob's are averages over her unobserved records. They agree on all probabilities when supplied the same information. Their first descriptions can coincide if the initial quantum state is a computational basis state; after the second quantum measurement, Bob's maximally mixed state differs from each conditional pure quantum state.