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

Orthogonality of the energy eigenstates gives

⟨ 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

.

Solved by gpt-5.6-sol high.

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