 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/iii/paper-48/5/d/solution

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

The two displayed amplitudes have modulus

1/ 2

. Absorb their phases into Bob's basis rather than use complex Schmidt coefficients. Take

∣ u_{0} ⟩ = ∣0 ⟩

∣ u_{1} ⟩ = ∣1 ⟩

, and

∣ v_{0} ⟩ = e^{- iπ /4} ∣1 ⟩, ∣ v_{1} ⟩ = e^{iπ /4} ∣0 ⟩ .

Both local bases are orthonormal. A Schmidt decomposition is therefore

∣ ψ ⟩_{A B} = \frac{1}{2} ∣ u_{0} ⟩ ∣ v_{0} ⟩ + \frac{1}{2} ∣ u_{1} ⟩ ∣ v_{1} ⟩ .

1/ 2

, the Schmidt number is two, and both reduced density operators are

I /2

. Their equal eigenvalues make this decomposition nonunique.

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