Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 1 / 34B (Principles of Quantum Mechanics)

In units with $\hslash =1$, the quantum harmonic oscillator Hamiltonian and number operator areThe nontrivial commutators among the creation and annihilation operators and $N$ areTheir reversed commutators have the opposite signs, while the commutator of any operator with itself is zero.

The operator $N$ is self-adjoint becauseIts eigenvalues are therefore real. Moreover, for every state $|\psi ⟩$,so every eigenvalue is nonnegative.

If $N|\nu ⟩=\nu |\nu ⟩$, the commutator with $A$ givesandRepeated application of $A$ lowers the eigenvalue by one. It must terminate before producing a negative eigenvalue. If $m$ is the last occupied rung, then $A|\nu -m⟩=0$, and the norm identity forces $\nu -m=0$. HenceThis is the integer spectrum of the number operator.

For occupation numbers $n,m\ge 0$, the Hamiltonian of the two commensurate quantum harmonic oscillators isThe five lowest distinct levels and their product-state bases areThe repeated levels exhibit quantum degeneracy.

The degenerate $E_0=9/2$ subspace has ordered basisUsing the ladder actions,Thus degenerate perturbation theory asks us to diagonalizeIts normalized eigenvectors and first-order energies areThere is no order-$\lambda$ admixture from other unperturbed levels.

In fact, $H^{\prime }$ preserves $n+2m$, so this two-dimensional eigenspace of $H_0$ is invariant. Directly,Thereforefor every real $\lambda$, with no omitted higher-order terms. This is exact diagonalization within a degenerate subspace, specifically the resonant two-to-one oscillator coupling.