Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 34C / c

Define $|\pm {⟩}_1=(|0{⟩}_1\pm |1{⟩}_1)/\sqrt{2}$. Since $B_1+B_1^{†}$ is the Pauli $X$ operator on the first oscillator, the exact eigenvectors arewith energies

Within each degenerate unperturbed eigenspace of fixed $n_2$, the perturbation has matrixIts eigenvectors are $|+{⟩}_1$ and $|-{⟩}_1$, with first-order shifts $+g$ and $-g$. Thus degenerate perturbation theory gives exactly the states and energies found in part (i); again, all higher corrections vanish because the perturbation commutes with the second-oscillator Hamiltonian and is exactly diagonalized within each degenerate block.