Past exam of the laser science and quantum information processing course of the University of Oxford / 2019 / 4 / c

Use the fidelity between the original pure state and the measurement output averaged over Calvin's results. Put $p=|\alpha {|}^2$, so $1-p=|\beta {|}^2$. The output is ${\rho }_Z=\mathrm{diag}(p,1-p)$, andHere $r_z=2p-1$ 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 fidelity of the measurement procedure.

Since $F=1-2p(1-p)$ with $0\le p\le 1$, its maximum occurs at $p=0$ or $p=1$:up to a global phase. These are the eigenstates of the measured observable, so their ideal measurement does not disturb them.

Complete the square: $F=2(p-1/2{)}^2+1/2$. ThereforeEvery equatorial Bloch-sphere state attains the minimum, not just $|+⟩$ and $|-⟩$. Its unread Z measurement gives $I/2$, having removed all information contained in its relative phase.

For a uniformly distributed pure qubit on the Bloch sphere, rotational symmetry gives $⟨r_z^2⟩=1/3$. ThusEquivalently, $p=|\alpha {|}^2$ is uniform on $[0,1]$ under this measure and ${\int }_0^1[p^2+(1-p{)}^2]dp=2/3$. The three positive coordinate-axis states mentioned in the source give $F_Z=1$, $F_X=F_Y=1/2$, whose average is the same because this fidelity depends only on the quadratic component $r_z^2$. This is a specified uniform-state average, not a statement about every possible prior ensemble.