A model atom has three non-degenerate energy levels with energies ,
PHYSICS · Subjective · Fermi University Physics · Quantum Mechanics · Quantum dynamics and its formulations · Quantum dynamics and its formulations
The Structure of Quantum Mechanics, Ch Time evolution as a unitary group, and the three pictures
Problem
A model atom has three non-degenerate energy levels with energies E1=0, E2=1.0eV, E3=2.5eV and orthonormal eigenstates ∣1⟩,∣2⟩,∣3⟩ of the time-independent Hamiltonian H^0. At t=0 the atom is prepared in the equal superposition ∣ψ(0)⟩=31(∣1⟩+∣2⟩+∣3⟩). Define ω21=(E2−E1)/ℏ, and let A^=∣1⟩⟨2∣+∣2⟩⟨1∣ be a Hermitian observable measuring the coherence between levels 1 and 2. (a) Write ∣ψ(t)⟩S using the spectral form of U^(t)=e−iH^0t/ℏ, and compute ⟨A^⟩(t)=⟨ψ(t)∣A^∣ψ(t)⟩ in the Schrodinger picture. (b) Transform A^ into the Heisenberg picture, A^H(t)=U^(t)†A^U^(t), and recompute ⟨A^⟩(t) as ⟨ψ(0)∣A^H(t)∣ψ(0)⟩. Confirm it agrees with part (a), and give the numerical value of ω21 in rads−1. (c) A weak, time-independent coupling between levels 1 and 2 is switched on, V^=λ(∣1⟩⟨2∣+∣2⟩⟨1∣) with λ=0.05eV, so the full Hamiltonian is H^=H^0+V^. Using U^0(t)=e−iH^0t/ℏ, write V^I(t)=U^0(t)†V^U^0(t) explicitly, and say in one sentence what has been gained by removing the fast phases of H^0. (d) Referring to parts (a) and (b), explain why no measurement of ⟨A^⟩ could ever reveal which picture was used to calculate it.