 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-1/34a/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

[N, A] = - A

N (A ∣ n ⟩) = (n - 1) A ∣ n ⟩,

A ∣ n ⟩

is proportional to

∣ n - 1 ⟩

. Its squared norm is

∥ A ∣ n ⟩ ∥^{2} = ⟨ n ∣ A^{†} A ∣ n ⟩ = n .

With the conventional phase choice,

A ∣ n ⟩ = n ∣ n - 1 ⟩ .

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