 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ib/paper-3/6a/iii/solution

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

Positivity of the squared norm of the supplied state gives, for every real

s

0 \leq ∥(p - i s x) ψ ∥^{2} = ⟨(p + i s x) (p - i s x)⟩ = ⟨ p^{2} ⟩ + s^{2} ⟨ x^{2} ⟩ + i s ⟨[x, p]⟩ = (Δ p)^{2} + s^{2} (Δ x)^{2} - s ℏ,

where the stated zero means were used in the last line. This quadratic in

s

is nonnegative for every real

s

, so its discriminant is nonpositive:

ℏ^{2} - 4 (Δ x)^{2} (Δ p)^{2} \leq 0.

Δ x Δ p \geq \frac{ℏ}{2} .

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