 Solution

ID: past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2024/paper-1/a1/a/iii/solution

Within the well, the potential is zero and the Hamiltonian is purely kinetic:

\hat{H} = - \frac{ℏ ^{2}}{2 m} \frac{d ^{2}}{d x ^{2}} .

The infinite walls are encoded in the operator's boundary conditions

Ψ (0) = Ψ (a) = 0

, rather than differentiated as part of the interior expression. Since

Ψ_{n}^{''} (x) = - (\frac{nπ}{a})^{2} Ψ_{n} (x),

we obtain

\hat{H} Ψ_{n} = \frac{n ^{2} π ^{2} ℏ ^{2}}{2 m a ^{2}} Ψ_{n}, E_{n} = \frac{n ^{2} h ^{2}}{8 m a ^{2}} .

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