 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ib/paper-4/15c/ii/solution

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

ξ_{j} = mω /ℏ x_{j}

. Separation into three one-dimensional oscillators gives

Ψ_{n_{1} n_{2} n_{3}} = h_{n_{1}} (ξ_{1}) h_{n_{2}} (ξ_{2}) h_{n_{3}} (ξ_{3}) e^{- mω r^{2} / (2ℏ)},

E_{N} = ℏ ω (N + \frac{3}{2}), N = n_{1} + n_{2} + n_{3} .

E_{0} = \frac{3}{2} ℏ ω, Ψ_{0} = e^{- mω r^{2} / (2ℏ)} .

The number of triples of nonnegative integers summing to

N

(2 N + 2),

which is the degeneracy of

E_{N}

Finally,

χ = (x_{1} + i x_{2})^{2} e^{- mω r^{2} / (2ℏ)}

is a pure level-

N = 2

oscillator state: the constant terms in the two degree-two Hermite contributions cancel. Part (i) gives

L_{3} χ = 2ℏ χ

, while

N = 2

gives

Hχ = (7/2) ℏ ω χ

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