 Solution

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

Since

\int_{0}^{a} x^{2} (a - x)^{2} d x = \frac{a ^{5}}{30},

an explicitly normalized trial is

ψ (x) = \frac{30}{a ^{5}} x (a - x), 0 < x < a .

It is zero outside the box. The original expression

x (x - a)

differs by a minus sign, which changes only the global phase and leaves probabilities and expectation values unchanged.

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