 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/ib/paper-3/7b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

E = \frac{( Δ p ) ^{2}}{2 m} + \frac{m ω ^{2} ( Δ x ) ^{2}}{2} + \frac{⟨ p ⟩ ^{2}}{2 m} + \frac{m ω ^{2} ⟨ x ⟩ ^{2}}{2} .

The last two terms are nonnegative, so

E \geq \frac{( Δ p ) ^{2}}{2 m} + \frac{m ω ^{2} ( Δ x ) ^{2}}{2} .

E \geq 2 \frac{( Δ p ) ^{2}}{2 m} \frac{m ω ^{2} ( Δ x ) ^{2}}{2} = ω Δ p Δ x \geq \frac{ℏ ω}{2} .

The argument applies to any normalized state with finite oscillator energy; the stated stationary state is a particular case. Equality requires zero position and momentum means, minimum quantum uncertainty, and balanced kinetic and potential contributions, as in the oscillator ground state.

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