 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2a/section-b/24/b/ii/solution

Let

S_{12} = ⟨ ϕ_{1} ∣ ϕ_{2} ⟩

be real, with each atomic orbital normalized. Then

⟨ ψ ∣ ψ ⟩ = c_{1}^{2} + c_{2}^{2} + 2 c_{1} c_{2} S_{12} = 1.

Together with orthonormal spin functions, this also normalizes the Slater determinant: its norm squared is

⟨ ψ ∣ ψ ⟩^{2}

. Under the subsequent zero-overlap assumption,

c_{1} = 1 - c_{2}^{2}, - 1 \leq c_{2} \leq 1,

using the specified nonnegative branch for

c_{1}

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