Fibonacci fusion measurement branches and zero outcomes

The coefficients have squared moduli $3\mathrm{/}5$ and $2\mathrm{/}5$, whose sum is one. The channel projectors are $P_1=\mathrm{\mid }1⟩⟨1\mathrm{\mid }$ and $P_{\tau }=\mathrm{\mid }\tau ⟩⟨\tau \mathrm{\mid }$, so they send $\mathrm{\mid }\psi ⟩$ to the corresponding coefficient times its basis vector. Thus the probabilities are $3\mathrm{/}5$ and $2\mathrm{/}5$, and after division by the coefficient magnitudes the conditional states are $\mathrm{\mid }1⟩$ and $\mathrm{\mid }\tau ⟩$ by [F1]–[F2].

If the coefficient of one channel is zero, its projected vector is zero and the corresponding probability is zero. The branch-validity theorem says that this case has no normalized conditional state. If the surviving channel is $a$ with coefficient $\alpha$, normalization gives $\mathrm{\mid }\alpha \mathrm{\mid }=1$, so its conditional vector is $\alpha \mathrm{\mid }a⟩$, representing the same pure state as $\mathrm{\mid }a⟩$. This checks both zero-coefficient boundary cases without fixing a global phase.