= Solution
Use the <quantum fidelity>[fidelity] between the original <pure state> and the <quantum measurement>[measurement] output averaged over Calvin's results. Put $p=|\alpha|^2$, so $1-p=|\beta|^2$. The output is $\rho_Z=\operatorname{diag}(p,1-p)$, and
$$
\boxed{F=\langle\psi|\rho_Z|\psi\rangle
=p^2+(1-p)^2=\frac{1+r_z^2}{2}.}
$$
Here $r_z=2p-1$ is the input Bloch component. If a particular result is specified instead, the conditional fidelities are $F_0=p$ and $F_1=1-p$; their outcome-weighted mean is precisely the displayed $F$. This distinction fixes what is meant by the <quantum fidelity>[fidelity] of the <quantum measurement>[measurement] procedure.
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