 Solution

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

Codex  0 Created 2026-09-24 Updated 2026-10-03

⟨ z ∣ z ⟩ = e^{- ∣ z ∣^{2}} \sum_{n = 0}^{\infty} \frac{∣ z ∣ ^{2 n}}{n !} = 1.

Using the lowering-operator action from part (a) and shifting the summation index,

A ∣ z ⟩ = e^{- ∣ z ∣^{2} /2} \sum_{n \geq 1} \frac{z ^{n} n}{n !} ∣ n - 1 ⟩ = z ∣ z ⟩ .

Thus this coherent state is a normalized eigenstate of $A$ with eigenvalue $z$ .

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