Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 58 / 4 / iii / Solution

The Pauli X gate interchanges the two basis quantum states. Because the bit-flip outcome is a classical unobserved error rather than a coherent superposition, the quantum bit-flip channel givesFor $|\psi ⟩=\alpha |0⟩+\beta |1⟩$, its explicit density matrix isFor a pure reference quantum state, the square of the paper's quantum fidelity is $F^2=⟨\psi |{\rho }_2|\psi ⟩$. ThereforeWith $|\psi ⟩=\cos (\theta /2)|0⟩+e^{i\varphi }\sin (\theta /2)|1⟩$, the expectation is $⟨X⟩=\sin \theta \cos \varphi$, the $x$ component of the Bloch vector. Hence“All possible quantum states” is interpreted as the rotationally invariant uniform pure-qubit average, with measure $d\mu =\sin \theta d\theta d\varphi /(4\pi )$. Thusso the average pure-qubit fidelity of a bit-flip channel isEquivalently, rotational symmetry gives $⟨r_x^2⟩=⟨r_y^2⟩=⟨r_z^2⟩=1/3$, since a pure-state Bloch vector has unit length. The $X$ eigenstates are unchanged as density matrices and have fidelity one, while a computational basis quantum state has $F^2=1-p$; these provide checks on the angular expression. A uniform distribution in $\theta$ rather than in solid angle would produce the wrong ensemble average. The requested quantity is the mean squared fidelity, not the square of the mean unsquared fidelity.