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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-58/3/a/solution

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Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 58 / 3 / a / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-14

For a pure qubit

a ∣0 ⟩ + b ∣1 ⟩

, the Bloch vector is

(2 Re (a^{*} b), 2 Im (a^{*} b), ∣ a ∣^{2} - ∣ b ∣^{2})

. Substitution gives

z = (sin θ cos φ, sin θ sin φ, cos θ), \overline{z} = (sin θ cos φ, sin θ sin φ, - cos θ) .

(1)

Both lie on the unit Bloch sphere, at the same azimuth and opposite heights. They are reflections across its equatorial plane, not antipodal vectors. Choose

θ = π /3

and

φ = π /4

, so their coordinates are

(6 /4, 6 /4, \pm 1/2)

. The following original sketch shows those vectors and their common meridian:

Figure 1.
Bloch vectors at theta pi/3 and azimuth pi/4, reflected across the equatorial plane
. **Details**.

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