 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-4/59/b/ii/solution

∣ ψ_{n} (x) ∣^{2}

is symmetric about

a /2

, while

x - a /2

is antisymmetric. Therefore

E_{n}^{(1)} = \int_{0}^{a} ∣ ψ_{n} (x) ∣^{2} (x - a /2) d x = 0.

The use of a midpoint-zero electric potential matters: adding a constant potential would add the corresponding uniform energy shift.

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