Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 2 / 32D (Principles of Quantum Mechanics)

Let $H|n⟩=E_n|n⟩$, with nondegenerate eigenvalues, and expand the perturbed state as $|n⟩+\lambda |n^{(1)}⟩+\cdots$, choosing $⟨n|n^{(1)}⟩=0$. The first-order eigenvalue equation gives $E_n^{(1)}=V_{nn}$ and, after projection onto $m\ne n$, $⟨m|n^{(1)}⟩=V_{mn}/(E_n-E_m)$. Projecting the second-order equation onto $n$ then gives the second-order nondegenerate perturbation theory resultFor angular momentum, $j=0,1/2,1,\dots$, $m=-j,-j+1,\dots ,j$, and $J^2|jm⟩=j(j+1)|jm⟩$, $J_3|jm⟩=m|jm⟩$ in units $\hslash =1$.

Take $H_0=-\gamma B_3{J}_3$ with $\gamma B_3\ne 0$ and perturbation $-\gamma B_1{J}_1$. The diagonal matrix element of $J_1=(J_{+}+J_{-})/2$ vanishes. Only $m\pm 1$ contribute at second order. Their squared matrix elements are $[j(j+1)-m(m\pm 1)]/4$, and their unperturbed energy denominators are respectively $\gamma B_3$ and $-\gamma B_3$. ThusHence $\boxed{E_m=-\gamma B_{3}m-\gamma m{B}_1^2/(2B_3)+O(B_1^4)}$. Rotating the quantization axis to the field gives exact energies $-\gamma m_B\sqrt{B_3^2+B_1^2}$. Labeling each branch continuously from its original $m$ gives $E_m=-\gamma m{B}_3\sqrt{1+B_1^2/B_3^2}$, whose expansion agrees, including negative $B_3$. If $B_3=0$, the nondegenerate starting point fails and this expansion cannot be used.