= Solution
{c}
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> $|a\rangle$, with probabilities $p_0=|\alpha|^2$, $p_1=|\beta|^2$. Bob, who has no record, assigns the <density operator>
$$
\boxed{\rho_{B,1}=|\alpha|^2|0\rangle\langle0|+|\beta|^2|1\rangle\langle1|.}
$$
This 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 $H|a\rangle=|+\rangle$ for $a=0$, or $|-\rangle$ for $a=1$. Bob assigns
$$
H\rho_{B,1}H=|\alpha|^2|+\rangle\langle+|+|\beta|^2|-\rangle\langle-|.
$$
Both $|+\rangle$ and $|-\rangle$ give either final <computational basis> outcome with <probability> $1/2$. Thus, after Alice records the second outcome $b$, she assigns $|b\rangle$, whereas Bob assigns
$$
\boxed{\rho_{B,2}=\tfrac12|0\rangle\langle0|+\tfrac12|1\rangle\langle1|=I/2.}
$$
Alice'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>.
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