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Past exam of the chemistry A course of the University of Cambridge / 2024 / Paper 1 / A2 / d / Solution 2026-10-20

∣ Ψ_{0} ∣^{2} = α / π e^{- α x^{2}}

is even. Hence

⟨ x ⟩ = 0

; also

Ψ_{0}^{'} = - αx Ψ_{0}

gives

⟨ p_{x} ⟩ = i ℏ α ⟨ x ⟩ = 0

.

To evaluate the second position moment, differentiate the Gaussian integral with respect to its positive parameter:

G (α) \int_{- \infty}^{\infty} x^{2} e^{- α x^{2}} d x ⟨ x^{2} ⟩ = \int_{- \infty}^{\infty} e^{- α x^{2}} d x = π α^{- 1/2}, = - G^{'} (α) = \frac{π}{2 α ^{3/2}}, = \frac{α}{π} \frac{π}{2 α ^{3/2}} = \frac{1}{2 α} = \frac{ℏ}{2 mω} .

For the momentum second moment, apply the momentum operator twice:

⟨ p_{x}^{2} ⟩ = - ℏ^{2} \int Ψ_{0}^{*} Ψ_{0}^{''} d x = ℏ^{2} (α - α^{2} ⟨ x^{2} ⟩) = \frac{ℏ ^{2} α}{2} = \frac{m ℏ ω}{2} .

Both means vanish, so

Δ x = \frac{ℏ}{2 mω}, Δ p_{x} = \frac{m ℏ ω}{2}, Δ x Δ p_{x} = \frac{ℏ}{2} .

The ground state attains the bound proved above and is therefore a minimum uncertainty state. A wider Gaussian reduces momentum spread while increasing position spread: the product remains at this lower limit. Other states, including excited oscillator eigenstates, generally give a larger product.

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