11.1: Exercises
( \newcommand{\kernel}{\mathrm{null}\,}\)
- Consider the two-state system investigated in Section 1.3. Show that the most general expressions for the perturbed energy eigenvalues and eigenstates are E′1=E1+e11+|e12|2E1−E2+O(ϵ3),E′2=E2+e22−|e12|2E1−E2+O(ϵ3), and ψ′1=ψ1+e∗12E1−E2ψ2+O(ϵ2),ψ′2=ψ2−e12E1−E2ψ1+O(ϵ2), respectively. Here, ϵ=|e12|/(E1−E2)≪1. You may assume that |e11|/(E1−E2), |e22|/(E1−E2)∼O(ϵ).
- Consider the two-state system investigated in Section 1.3. Show that if the unperturbed energy eigenstates are also eigenstates of the perturbing Hamiltonian then E′1=E1+e11,E′2=E2+e22, and ψ′1=ψ1ψ′2=ψ2 to all orders in the perturbation expansion.
- Consider the two-state system investigated in Section 1.3. Show that if the unperturbed energy eigenstates are degenerate, so that E1=E2=E12, then the most general expressions for the perturbed energy eigenvalues and eigenstates are E±=E12+e±, and ψ±=⟨1|ψ±⟩ψ1+⟨2|ψ±⟩ψ2, respectively, where e±=12(e11+e22)±12[(e11−e22)2+4|e12|2]1/2, and ⟨1|ψ±⟩⟨2|ψ±⟩=−(e12e11−e±)=−(e22−e±e∗12). Demonstrate that the ψ± are the simultaneous eigenstates of the unperturbed Hamiltonian, H0, and the perturbed Hamiltonian, H1, and that the e± are the corresponding eigenvalues of H1.
- Calculate the lowest-order energy-shift in the ground state of the one-dimensional harmonic oscillator when the perturbation V=λx4 is added to H=p2x2m+12mω2x2.
Contributors and Attributions
Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)