 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-3/6c/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

t

the state is

χ (t) = \frac{1}{2} e^{- i E_{1} t /ℏ} χ_{1} + \frac{3}{2} e^{- i E_{2} t /ℏ} χ_{2} .

⟨ χ_{1} ∣ r ∣ χ_{1} ⟩ = \frac{3 a}{2}, ⟨ χ_{2} ∣ r ∣ χ_{2} ⟩ = 6 a,

and

⟨ χ_{1} ∣ r ∣ χ_{2} ⟩ = 4 π \int_{0}^{\infty} r^{3} χ_{1} (r) χ_{2} (r) d r = - \frac{64 a}{81 2} .

R (t) = \frac{39 a}{8} - \frac{32 a}{81} \frac{3}{2} cos (\frac{E _{2} - E _{1}}{ℏ} t) .

Thus

R (t)

oscillates sinusoidally about

39 a /8

ω = \frac{E _{2} - E _{1}}{ℏ} = \frac{3 m K ^{2}}{8 ℏ ^{3}} = \frac{3 K}{8 a ℏ} .

 New to topics? Read the docs here!