= Solution
Let $z_A,z_B\in\{+1,-1\}$ denote the two $Z$ outcomes and $x_C\in\{+1,-1\}$ the $X$ outcome. Measuring $Z_A$ in the <GHZ state> projects $B$ onto the same $Z$ eigenstate, while the later $X_C$ outcome is unbiased. Hence the possible triples, in the order $(z_A,x_C,z_B)$, are
$$
\boxed{(+,+,+),\ (+,-,+),\ (-,+,-),\ (-,-,-)},
$$
each with probability $1/4$.
Choose $(z_A,x_C,z_B)=(+,+,+)$. Immediately before $t=0$, each local readout is $I/2$. The collapse at $(t,x)=(0,0)$ is immediately in $A$'s past light cone, reaches $B$ at $t=1$, and reaches $C$ at $t=2$. Thus
$$
\begin{array}{c|c}
\text{qubit}&\text{state-readout output}\\ \hline
A&I/2\text{ for }t<0,\quad |0\rangle\langle0|\text{ for }t>0,\\
B&I/2\text{ for }0<t<1,\quad |0\rangle\langle0|\text{ for }t>1,\\
C&I/2\text{ for }0<t<2,\quad |0\rangle\langle0|\text{ for }2<t<3,\quad |+\rangle\langle+|\text{ for }t>3.
\end{array}
$$
Here $|+\rangle=(|0\rangle+|1\rangle)/\sqrt2$. The $X_C$ collapse at $(3,2)$ reaches $B$ at $t=4$ and $A$ at $t=5$, but the state is already a product after the first collapse, so it does not change their local states. Likewise, the $Z_B$ collapse at $(5,1)$ reaches $A$ and $C$ at $t=6$ without changing their states. For another allowed outcome, replace $|0\rangle$ by $|1\rangle$ according to $z_A=z_B$ and replace $|+\rangle$ by $|-\rangle$ according to $x_C$.
Solved by gpt-5.6-sol high.
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