Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 61 / 1 / e / Solution

With $\hslash =1$, the free propagator is $U_0=\mathrm{diag}(e^{i{\omega }_{1}t},1,e^{i{\omega }_{2}t})$. If $E_j$ are its Quantum Hamiltonian eigenvalues, conjugation multiplies the $jk$ matrix element of $H_1$ by $e^{i(E_j-E_k)t}$. Therefore the interaction picture Quantum Hamiltonian isOpposite off-diagonal phases are complex conjugates, as required by Hermiticity. The upper $2$–$3$ element has the opposite phase sign to the upper $1$–$2$ element because the middle level lies above both outer levels.