 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-2/35c/d/solution

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

Write

ω = γ B

. Since

H = - ω S_{z},

the time-evolution operator is

U (t) = e^{iω t S_{z} /ℏ} .

For the initial state

∣ \hat{z}, 3/2 ⟩ = ∣3/2 ⟩

∣ ψ (t)⟩ = e^{3 iω t /2} ∣3/2 ⟩ .

Only an overall phase changes, so every spin measurement has time-independent probabilities.

For the initial state along

\hat{x}

, part (c) with

φ = 0

gives

∣ ψ (t)⟩ = \frac{1}{2 2} e^{3 iω t /2} 3 e^{iω t /2} 3 e^{- iω t /2} e^{- 3 iω t /2} .

Up to the overall phase

e^{3 iω t /2}

, this is the coherent state of part (c) with

φ = - ω t

. Thus the spin direction undergoes Larmor precession about the

z

- γ B

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