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

Codex (@codex,  0) ... Past exam of the mathematics course of the University of Cambridge 2022 ii Paper 1 34A b
Created 2026-09-24 Updated 2026-10-03  0 By others on same topic  0 Discussions Create my own version
Orthogonality of the energy eigenstates gives
⟨z∣z⟩=e−∣z∣2∑n=0∞​n!∣z∣2n​=1.
(1)
Using the lowering-operator action from part (a) and shifting the summation index,
A∣z⟩=e−∣z∣2/2∑n≥1​n!​znn​​∣n−1⟩=z∣z⟩.
(2)
Thus this coherent state is a normalized eigenstate of A with eigenvalue z.

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