 Solution

ID: past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2025/paper-1/a2/c/solution

J = \iint \frac{∣ a ( 1 ) ∣ ^{2} ∣ b ( 2 ) ∣ ^{2}}{r _{12}} d^{3} r_{1} d^{3} r_{2}, K = \iint \frac{a ^{*} ( 1 ) b ( 1 ) b ^{*} ( 2 ) a ( 2 )}{r _{12}} d^{3} r_{1} d^{3} r_{2} .

Expanding the four products in the expectation of

r_{12}^{- 1}

yields

⟨ Φ_{\pm} ∣ r_{12}^{- 1} ∣ Φ_{\pm} ⟩ = \frac{1}{2} (J + J \pm K \pm K) = J \pm K .

The one-electron terms have the same value for these two states. Within this common-orbital approximation,

E_{S} = ϵ_{1 s} + ϵ_{2 s} + J + K, E_{T} = ϵ_{1 s} + ϵ_{2 s} + J - K,

E_{S} - E_{T} = 2 K

. The exchange integral is nonnegative: it is the positive Coulomb quadratic form of the transition density

a^{*} b

, or equivalently has a nonnegative Fourier representation. The triplet is lower. Actual orbital relaxation and correlation make the common-orbital

2 K

an approximate physical splitting.

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