 Solution

ID: past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2021/paper-1/a2/a/ii/solution

Ψ = \sum_{n} c_{n} ϕ_{n}

, with

\hat{H} ϕ_{n} = E_{n} ϕ_{n}

E = \frac{\sum _{n} ∣ c _{n} ∣ ^{2} E _{n}}{\sum _{n} ∣ c _{n} ∣ ^{2}}, E - E_{0} = \frac{\sum _{n} ∣ c _{n} ∣ ^{2} ( E _{n} - E _{0} )}{\sum _{n} ∣ c _{n} ∣ ^{2}} \geq 0.

Every summand in the numerator is nonnegative because

E_{n} \geq E_{0}

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