 Solution

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

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

P (N = n) = e^{- ∣ z ∣^{2}} \frac{∣ z ∣ ^{2 n}}{n !},

∣ z ∣^{2}

. Hence

⟨ N ⟩ = ∣ z ∣^{2}, (Δ N)^{2} = ∣ z ∣^{2}, Δ N = ∣ z ∣ .

The relative uncertainty is

1/∣ z ∣ = 1/ ⟨ N ⟩

, which tends to zero as the mean occupation tends to infinity.

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