Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 58 / 2 / c

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.