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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/ib/paper-4/6b/solution

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Past exam of the mathematics course of the University of Cambridge / 2009 / ib / Paper 4 / 6B / Solution by

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

Put

D = 1 + i ℏ t / m

and write the Gaussian wave packet as

ψ = π^{- 1/4} D^{- 1/2} e^{- x^{2} / (2 D)}

. Differentiation gives

\frac{ψ _{t}}{ψ} = \frac{i ℏ}{m} (- \frac{1}{2 D} + \frac{x ^{2}}{2 D ^{2}}), \frac{ψ _{xx}}{ψ} = \frac{x ^{2}}{D ^{2}} - \frac{1}{D} .

Hence

i ℏ ψ_{t} = - (ℏ^{2} /2 m) ψ_{xx}

, the free Time-dependent Schrödinger equation.

With

τ = ℏ t / m

, its probability density is

∣ ψ ∣^{2} = \frac{1}{π ( 1 + τ ^{2} )} exp (- \frac{x ^{2}}{1 + τ ^{2}}) .

The stated Gaussian integral shows its integral is one. Symmetry gives

⟨ x ⟩ = 0

; differentiating that Gaussian integral with respect to its coefficient gives

⟨ x^{2} ⟩ = (1 + τ^{2}) /2

. Thus the position standard deviation is

Δ x = \frac{1 + ( ℏ t / m ) ^{2}}{2} .

It is not a stationary state: its probability density changes with time as the packet spreads. A stationary state would have only an overall time-dependent phase and a fixed position density.

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