An electron sits in quantum state of a harmonic-oscillator potential of angular frequency

PHYSICS · Multi-part · Fermi University Physics · Mechanics & Dynamical Systems · Oscillations and normal modes · Hamiltonian mechanics and canonical structure

Action and Symmetry, Ch Adiabatic invariants, canonical perturbation theory and integrable systems

Problem

An electron sits in quantum state n=3n = 3 of a harmonic-oscillator potential of angular frequency ω1=2.00×1013 rad s−1\omega_1 = 2.00\times10^{13}\,\mathrm{rad\,s^{-1}}, so its energy is E1=(n+12)ℏω1E_1 = (n+\tfrac12)\hbar\omega_1 with ℏ≡hP/2π=1.055×10−34 J s\hbar \equiv h_{\rm P}/2\pi = 1.055\times10^{-34}\,\mathrm{J\,s}. The confining potential is then slowly stiffened, over many oscillation periods, until the frequency reaches ω2=3.00×1013 rad s−1\omega_2 = 3.00\times10^{13}\,\mathrm{rad\,s^{-1}}. Someone argues: "Energy is the quantity that gets quantised in quantum mechanics, so it must be the energy EE itself, not the action, that is fixed under this slow change: EE stays at E1E_1 throughout." ((a)) Take, with Ehrenfest's guiding argument, that a slow change cannot make the quantum number jump, so nn stays at 3. Show that this is the same as holding the adiabatic invariant E/ωE/\omega fixed, and find the electron's energy E2E_2 once ω2\omega_2 is reached. ((b)) Following the stated (wrong) claim that EE itself stays fixed at E1E_1, compute the "quantum number" n′n' this would require once ω=ω2\omega = \omega_2, and use the result to say why the claim must be rejected.

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