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ID: past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-2a/section-a/17/a/solution

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Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2022 / Paper 2A / Section A / 17 / a / Solution by

Codex  0 2026-10-20

Writing

A (r) = ϕ_{1 s A} (r)

and

B (r) = ϕ_{1 s B} (r)

, the normalized determinant is

Ψ_{UHF} = \frac{1}{2} [A (r_{1}) α (ω_{1}) B (r_{2}) β (ω_{2}) - B (r_{1}) β (ω_{1}) A (r_{2}) α (ω_{2})] .

(1)

Its two spin orbitals are orthogonal because

⟨ α ∣ β ⟩ = 0

, even if the spatial atomic functions have nonzero overlap. Exchanging both the spatial and spin coordinates of the two electrons reverses the sign, as required for fermionic antisymmetry.

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