Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 34C (Principles of Quantum Mechanics)

Using $B^2=0$ and $B{B}^{†}=1-B^{†}B$,Thus $H$ is a projection and its eigenvalues are $0$ and $1$.

The state $B|1⟩$ has normand satisfies $HB|1⟩=B^{†}{B}^2|1⟩=0$. Hence it is the normalized ground state, denoted $|0⟩$. Alsoso

In the ordered basis $(|0⟩,|1⟩)$,

The product states $|n_1{n}_2⟩$, $n_a\in \{0,1\}$, are exact eigenvectors, withThus the energies attached to $|00⟩,|01⟩,|10⟩,|11⟩$ are respectively

In the ordered basis $(|00⟩,|01⟩,|10⟩,|11⟩)$,$B_1^{†}$ and $B_2^{†}$ are their transposes, and

For $H_0=E_2{B}_2^{†}{B}_2$, the eigenspaces of energies $0$ and $E_2$ are each two-dimensional because $n_1$ is free. Degenerate perturbation theory requires diagonalizing $E_1{B}_1^{†}{B}_1$ within each space. It is already diagonal in the product basis, with eigenvalues $0,E_1$. Therefore the first-order eigenvectors remainand their energies are $0,E_1,E_2,E_2+E_1$, respectively. These are the exact answers, so all higher corrections vanish.

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.