 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-2/35b/b/solution

 Top articles  Latest articles New article in topic

Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 2 / 35B / b / Solution by

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

Since

X = L (a + a^{†})

, the perturbation is

λ ℏ ω (a + a^{†})^{3} .

Every diagonal matrix element vanishes by parity, so all first-order shifts are zero. For the ground state,

(a + a^{†})^{3} ∣0 ⟩ = 3∣1 ⟩ + 6 ∣3 ⟩ .

The second-order shift is therefore

λ^{2} (ℏ ω)^{2} (\frac{9}{- ℏ ω} + \frac{6}{- 3ℏ ω}) = - 11 λ^{2} ℏ ω .

Thus

E_{0} = ℏ ω (\frac{1}{2} - 11 λ^{2}) + O (λ^{3}) .

For every nonzero real

λ

, the cubic potential is unbounded below on one side, so there are no genuine bound-state eigenvalues continuously connected to the harmonic levels. The formal perturbation series does not converge; it is at best asymptotic to resonance energies.

 Read the full article

 New to topics? Read the docs here!