 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/ib/paper-1/15b/a/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

ϕ_{n} (x) = \frac{2}{a} sin \frac{nπ x}{a}, E_{n} = \frac{n ^{2} π ^{2} ℏ ^{2}}{2 m a ^{2}}, 0 < x < a .

The sine identity expands the initial wavefunction as

ψ_{0} = \frac{3 ϕ _{1} - ϕ _{3}}{10} .

The normalization factor

4/ 5 a

9/10

for

E_{1}

and

1/10

for

E_{3} = 9 E_{1}

. Thus the energy expectation is

⟨ E ⟩ = \frac{9}{10} E_{1} + \frac{1}{10} E_{3} = \frac{9 π ^{2} ℏ ^{2}}{10 m a ^{2}} .

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