11.4: Non-Degenerate Perturbation Theory
( \newcommand{\kernel}{\mathrm{null}\,}\)
Let us now generalize our perturbation analysis to deal with systems possessing more than two energy eigenstates. Consider a system in which the energy eigenstates of the unperturbed Hamiltonian, H0, are denoted H0ψn=Enψn, where n runs from 1 to N. The eigenstates are assumed to be orthonormal, so that ⟨m|n⟩=δnm, and to form a complete set. Let us now try to solve the energy eigenvalue problem for the perturbed Hamiltonian: (H0+H1)ψE=EψE. It follows that ⟨m|H0+H1|E⟩=E⟨m|E⟩, where m can take any value from 1 to N. Now, we can express ψE as a linear superposition of the unperturbed energy eigenstates: ψE=∑k⟨k|E⟩ψk, where k runs from 1 to N. We can combine the previous equations to give
(Em−E+emm)⟨m|E⟩+∑k≠memk⟨k|E⟩=0, where emk=⟨m|H1|k⟩.
Let us now develop our perturbation expansion. We assume that emkEm−Ek∼O(ϵ) for all m≠k, where ϵ≪1 is our expansion parameter. We also assume that emmEm∼O(ϵ) for all m. Let us search for a modified version of the nth unperturbed energy eigenstate for which E=En+O(ϵ), and ⟨n|E⟩=1,⟨m|E⟩=O(ϵ) for m≠n. Suppose that we write out Equation ([e12.45]) for m≠n, neglecting terms that are O(ϵ2) according to our expansion scheme. We find that (Em−En)⟨m|E⟩+emn≃0, giving ⟨m|E⟩≃−emnEm−En. Substituting the previous expression into Equation ([e12.45]), evaluated for m=n, and neglecting O(ϵ3) terms, we obtain (En−E+enn)−∑k≠n|enk|2Ek−En≃0. Thus, the modified nth energy eigenstate possesses an eigenvalue
E′n=En+enn+∑k≠n|enk|2En−Ek+O(ϵ3) and a wavefunction
ψ′n=ψn+∑k≠neknEn−Ekψk+O(ϵ2). Incidentally, it is easily demonstrated that the modified eigenstates remain orthonormal to O(ϵ2).
Contributors and Attributions
Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)