 Solution

ID: past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2022/paper-1/a1/b/ii/solution

Insert the expansions into the energy eigenvalue equation

\hat{H} Ψ_{i} = E_{i} Ψ_{i}

. Through first order,

\hat{H} Ψ_{i} E_{i} Ψ_{i} = \hat{H}^{(0)} Ψ_{i}^{(0)} + λ (\hat{H}^{(0)} Ψ_{i}^{(1)} + \hat{H}^{(1)} Ψ_{i}^{(0)}) + O (λ^{2}), = E_{i}^{(0)} Ψ_{i}^{(0)} + λ (E_{i}^{(0)} Ψ_{i}^{(1)} + E_{i}^{(1)} Ψ_{i}^{(0)}) + O (λ^{2}) .

The zeroth-order coefficients reproduce part (i). Equating the coefficients of

λ

and moving them to one side gives

(\hat{H}^{(0)} - E_{i}^{(0)}) Ψ_{i}^{(1)} + (\hat{H}^{(1)} - E_{i}^{(1)}) Ψ_{i}^{(0)} = 0.

Terms involving

\hat{H}^{(2)}

first enter at order

λ^{2}

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