Solve Equation 11.17 for the case of a time-independent perturbation, assuming that ca (0) = 1 and c_b(0) = 0. Check that |c_a(t)|² +  |c_b(t)|² = 1. Comment: Ostensibly, this system oscillates between “pure Ψ_a ” and “some Ψ_b.” Doesn’t this contradict my general assertion that no transitions occur for time-independent perturbations? No, but the reason is rather subtle: In this case Ψ_a and Ψ_b are not, and never were, eigenstates of the Hamiltonian—a measurement of the energy never yields E_a or Eb. In time-dependent perturbation theory we typically contemplate turning on the perturbation for a while, and then turning it off again, in order to examine the system. At the beginning, and at the end, Ψ_a and Ψ_b are eigenstates of the exact Hamiltonian, and only in this context does it make sense to say that the system underwent a transition from one to the other. For the present problem, then, assume that the perturbation was turned on at time t = 0, and off again at time T—this doesn’t affect the calculations, but it allows for a more sensible interpretation of the result.

Transcribed Image Text:

Ca Hab h == Habe-icol Cb, i =-=-Hex Cas H'ba eiwot (11.17)