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Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 1 / 34A / a / Solution

From the number operator commutator

[N, A] = - A

,

N (A ∣ n ⟩) = (n - 1) A ∣ n ⟩,

so

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 ⟩ .

Solved by gpt-5.6-sol high.

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