Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 3 / 33B (Principles of Quantum Mechanics)

For a normalized eigenstate $|n⟩$ with a nondegenerate energy eigenvalue, first-order nondegenerate perturbation theory and second-order nondegenerate perturbation theory give

Put $E_1=E_2=E$. Within the degenerate eigenspace spanned by $|1⟩,|2⟩$, the perturbation isDegenerate perturbation theory diagonalizes this matrix. Its normalized eigenstates $(|1⟩\pm |2⟩)/\sqrt{2}$ have first-order shifts $\pm \epsilon V_0$. The coupling to $|3⟩$ has order ${\epsilon }^2$, so it cannot affect an energy through order $\epsilon$. Therefore

Now put $E_2=E_3=E$ and $E_1\ne E$. All diagonal matrix elements of $V$ vanish. The leading coupling is $⟨1|V|2⟩=\epsilon V_0$, whereas $⟨1|V|3⟩={\epsilon }^2{V}_0$. Applying the second-order formula givesThe degeneracy causes no singular denominator through this order because $⟨2|V|3⟩=0$. Equivalently, the effective perturbation on the degenerate subspace begins asIts off-diagonal mixing is only order ${\epsilon }^3$, so degenerate perturbation theory is unnecessary for energies through order ${\epsilon }^2$.

The states $|2⟩$ and $|3⟩$ are degenerate and couple to $|1⟩$ with amplitudes $\epsilon V_0$ and ${\epsilon }^2{V}_0$. The bright and dark state reduction of a star-coupled Hamiltonian gives one exact dark eigenstate proportional to $-\epsilon |2⟩+|3⟩$, with energy $E$. The orthogonal bright state couples to $|1⟩$ with magnitudeDiagonalizing the remaining two-dimensional Hermitian matrix gives the exact spectrumWriting $\Delta =E_1-E\ne 0$ and expanding the branches according to the eigenvalue to which they converge,These are exactly the three perturbative energies from part b through order ${\epsilon }^2$.