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

Since $X=L(a+a^{†})$, the perturbation isEvery diagonal matrix element vanishes by parity, so all first-order shifts are zero. For the ground state,The second-order shift is thereforeThusFor every nonzero real $\lambda$, 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.

Solved by gpt-5.6-sol high.