 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ib/paper-3/16b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

- \frac{ℏ ^{2}}{2} ψ^{''} = E ψ, ψ (0) = ψ (π) = 0.

Its normalized eigenstates and energies are

ψ_{n} (x) = \frac{2}{π} sin (n x), E_{n} = \frac{ℏ ^{2} n ^{2}}{2}, n = 1, 2, \dots .

The general normalized time-dependent solution is

Ψ (x, t) = n = 1 \sum \infty c_{n} ψ_{n} (x) e^{- i E_{n} t /ℏ}, n = 1 \sum \infty ∣ c_{n} ∣^{2} = 1 .

Immediately after removing the barrier, the normalized wavefunction is

Ψ (x, 0) = {4/ π sin (2 x), 0, 0 \leq x \leq π /2, π /2 < x \leq π .

c_{n} = \frac{8}{π} \int_{0}^{π /2} sin (n x) sin (2 x) d x .

They are

c_{2} = \frac{1}{2}, c_{n} = \frac{4 2}{π} \frac{s i n ( nπ /2 )}{4 - n ^{2}} (n \neq = 2) .

The Born rule therefore gives

P (E_{n}) = ⎩ ⎨ ⎧ \frac{1}{2}, \frac{32}{π ^{2} ( n ^{2} - 4 ) ^{2}}, 0, n = 2, n odd, n even and n \neq = 2.

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