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For pure input $|\psi ⟩$, squared quantum fidelity after the bit-flip channel is $(1-p)+p|⟨\psi |X|\psi ⟩{|}^2$. The last factor is the square of the input's Bloch $x$ component. Rotational symmetry of the uniform pure-qubit average gives $⟨r_x^2⟩=1/3$, because $r_x^2+r_y^2+r_z^2=1$. Hence the average squared fidelity is $1-2p/3$. Averaging an unsquared fidelity or over a different ensemble gives a different quantity.

// Target: quantum-mechanics.bigb

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.

Use the rotation-invariant distribution over pure initial states, namely the uniform pure-qubit average with area element $\sin \theta d\theta d\phi /(4\pi )$. Reflection symmetry gives $\mathbb{E}[xz]=0$, and rotational symmetry together with $x^2+y^2+z^2=1$ gives $\mathbb{E}[y^2]=1/3$. Averaging the squared quantum fidelity from part (b) therefore yieldsIf the single-number packet $k=1$ is included, its corrected average isUniform polar angle without the sine weight would not be the Haar-uniform average over all pure qubit states.