Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 3 / 33B / b / Solution

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 ${\epsilon }^2$.