Questions: The Schrödinger, Heisenberg, and Interaction Pictures
4 questions to test your understanding
Score: 0 / 4
Question 1 True / False
The Schrödinger and Heisenberg pictures predict the same expectation value for every observable at every time.
TTrue
FFalse
Answer: True
⟨ψ(t)|A|ψ(t)⟩ = ⟨ψ(0)|U†AU|ψ(0)⟩ = ⟨ψ_H|A_H(t)|ψ_H⟩. Moving U from the state to the operator does not change the number.
Question 2 Multiple Choice
An operator A has no explicit time dependence and commutes with the Hamiltonian. What happens to A_H(t) in the Heisenberg picture?
AIt grows linearly in time
BIt oscillates at frequency H/ħ
CIt stays constant, so its expectation value is conserved
DIt becomes zero
dA_H/dt = (i/ħ)[H, A_H] = 0. The operator does not change, so A is a constant of the motion.
Question 3 Multiple Choice
For a harmonic oscillator, which is the Heisenberg-picture position operator?
Ax_H(t) = x(0) + (p(0)/m)t
Bx_H(t) = x(0) cos ωt + (p(0)/mω) sin ωt
Cx_H(t) = x(0)e^(−iωt)
Dx_H(t) = x(0)
The Heisenberg equations give dx_H/dt = p_H/m and dp_H/dt = −mω²x_H. Their solution has the same form as the classical solution, with operators in place of numbers.
Question 4 Short Answer
In the interaction picture with H = H₀ + V, what governs the time evolution of the state vector?
Think about your answer, then reveal below.
Model answer: The interaction alone, in the form V_I(t) = e^(iH₀t/ħ) V e^(−iH₀t/ħ), through iħ d|ψ_I⟩/dt = V_I(t)|ψ_I⟩.
The free evolution under H₀ is moved onto the operators. If V = 0, the interaction-picture state does not change at all, which makes the effect of V easy to isolate.