 Solution

ID: past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2023/paper-1/a2/b/i/solution

α > 0

. For the normalized probability weight proportional to

e^{- α x^{2}}

⟨ x^{2} ⟩ = \frac{1}{2 α}, ⟨ x^{4} ⟩ = \frac{3}{4 α ^{2}} .

Since

Ψ^{''} = (α^{2} x^{2} - α) Ψ

⟨ T ⟩ = \frac{α}{2} - \frac{α ^{2}}{2} ⟨ x^{2} ⟩ = \frac{α}{4}, ⟨ V ⟩ = \frac{1}{2} ⟨ x^{4} ⟩ = \frac{3}{8 α ^{2}} .

Thus

E (α) = \frac{α}{4} + \frac{3}{8 α ^{2}}, E^{'} (α) = \frac{1}{4} - \frac{3}{4 α ^{3}} = 0 ⟹ α_{*} = 3^{1/3} .

The second derivative is positive, and the energy diverges at both ends of

α > 0

, so this is the global minimum within the Gaussian function family. The best Gaussian function upper bound is

E_{0} \leq E (α_{*}) = \frac{3 ^{4/3}}{8} ≃ 0.54084359.

This is an upper bound, not the exact quartic ground energy.

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