 Solution

ID: past-exam-of-the-computer-science-course-of-the-university-of-oxford/2021/quantum-information/3/b/ii/solution

∣ t_{j} ⟩ = \frac{∣0 ⟩ + ω ^{j} ∣1 ⟩}{2}, p_{j} = \frac{1}{3}, j = 0, 1, 2, ω = e^{2 πi /3} .

All three states are normalized. Since

1 + ω + ω^{2} = 0

\frac{1}{3} \sum_{j = 0}^{2} ∣ t_{j} ⟩ ⟨ t_{j} ∣ = \frac{1}{6} \sum_{j = 0}^{2} (1 ω^{j} ω^{- j} 1) = \frac{I}{2} .

Geometrically, their unit Bloch vectors form an equilateral triangle on the equator and sum to zero. Orthogonality is unnecessary for an ensemble decomposition of a mixed state: three distinct, nonorthogonal pure states can give the same maximally mixed average as two orthogonal states.

 New to topics? Read the docs here!