 Solution

ID: past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2022/paper-1/a4/b/ii/solution

Use the four supplied opposite-site combinations in the order

(ϕ_{1}, ϕ_{2}, ϕ_{3}, ϕ_{4})

H_{E_{g}} = α β_{S} 0 - β_{L} β_{S} α β_{L} 0 0 β_{L} α β_{S} - β_{L} 0 β_{S} α .

The negative corner is essential. For example,

⟨ ϕ_{4} ∣ H ∣ ϕ_{1} ⟩

includes the long bonds

4

–

5

and

8

–

1

, each with opposite coefficient signs, giving

- β_{L}

Because the basis is orthonormal, the secular determinant is

α - E β_{S} 0 - β_{L} β_{S} α - E β_{L} 0 0 β_{L} α - E β_{S} - β_{L} 0 β_{S} α - E = 0.

It factors as

[(α - E)^{2} - β_{S}^{2} - β_{L}^{2}]^{2} = 0

, giving two doubly degenerate energies,

α \pm β_{S}^{2} + β_{L}^{2}

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