 Solution

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

\hat{H}^{(1)} = x^{4}

E_{0}^{(1)} = \frac{1}{π} \int_{- \infty}^{\infty} x^{4} e^{- x^{2}} d x = \frac{1}{π} \frac{3 π}{4} = \frac{3}{4} .

The distinction between the coefficient and the actual first-order shift matters: the energy shift is $3 λ /4$ , so

E_{0} = 1/2 + 3 λ /4 + O (λ^{2})

in the stated atomic units.

 New to topics? Read the docs here!