Qubit fidelity along a Bloch meridian ($F=\cos (\theta /2),F_{\mathrm{sq}}={\cos }^2(\theta /2),D=\sin (\theta /2)$)

Compare the qubit pure statewith $|0⟩$, using the computational basis. Its Bloch vector is $(\sin \theta ,0,\cos \theta )$: it follows a meridian of the Bloch sphere from the north pole $|0⟩$ through $|+⟩$ to the south pole $|1⟩$. The inner product with $|0⟩$ is $\cos (\theta /2)$, nonnegative on this interval. Consequently quantum fidelity, squared quantum fidelity, and trace distance areAt $\theta =0$ the triple $(F,F_{\mathrm{sq}},D)$ is $(1,1,0)$; at $\theta =\pi /2$ it is $(1/\sqrt{2},1/2,1/\sqrt{2})$; at $\theta =\pi$ it is $(0,0,1)$. The angle between the Bloch vectors is $\theta$, while the quantum fidelity uses its half-angle. The pure-state trace distance and fidelity identity becomes the familiar relation ${\cos }^2(\theta /2)+{\sin }^2(\theta /2)=1$.

For equal prior probabilities, the Holevo–Helstrom theorem gives optimal quantum state discrimination success $P_{\mathrm{succ}}=(1+D)/2$. Along this meridian it rises from $1/2$ to one, with $(1+1/\sqrt{2})/2\approx 0.8536$ at $|+⟩$. This connects the geometric trace distance to the best possible decision from a single preparation; it is a different measurement task from the target-acceptance probability defining squared quantum fidelity.

// Target: quantum-information-theory.bigb