 Solution

ID: past-exam-of-the-introduction-to-quantum-information-course-of-the-university-of-oxford/2015/2/b/solution

Expand the given pure state in the ordered computational basis

00, 01, 10, 11

∣ ψ ⟩ = \frac{1}{3} ∣00 ⟩ - \frac{1}{6} ∣01 ⟩ + \frac{1}{3} ∣10 ⟩ + \frac{1}{6} ∣11 ⟩, v = \frac{1}{6} 2 - 1 21 .

The coefficients have squared magnitudes

1/3, 1/6, 1/3, 1/6

, summing to one. The density matrix is the outer product

v v^{†}

, not the diagonal matrix of these four probabilities: its off-diagonal entries retain quantum coherence. Therefore

ρ = ∣ ψ ⟩ ⟨ ψ ∣ = \frac{1}{6} 2 - 2 22 - 2 1 - 2 - 1 2 - 2 22 2 - 1 21 .

u^{†} ρ u = ∣ v^{†} u ∣^{2} \geq 0

. Its trace is one, and

ρ^{2} = ρ

, as required for a pure state.

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