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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-58/1/a/solution

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Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 58 / 1 / a / Solution by

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

Let

{∣ u_{j} ⟩}_{j = 1}^{N}

be another complete orthonormal basis, and write

∣ u_{j} ⟩ = \sum_{n} U_{nj} ∣ n ⟩

. Orthonormality makes

U

a unitary matrix, so

U U^{†} = I

. Therefore

j \sum ∣ u_{j} ⟩ ⟨ u_{j} ∣ = n, k \sum (j \sum U_{nj} U_{kj}^{*}) ∣ n ⟩ ⟨ k ∣ = n \sum ∣ n ⟩ ⟨ n ∣ = I_{N} .

(1)

Thus the resolution of the identity is independent of the orthonormal basis. Equivalently, expanding any vector in the new complete basis shows that the sum of its orthogonal projections returns that vector.

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