 Solution

For $H=H_0+H_{\mathrm{int}}$, the interaction picture has $H_I(t)=e^{i{H}_0(t-t_0)}{H}_{\mathrm{int}}(t)e^{-i{H}_0(t-t_0)}$ andThe trial solution $|\psi (t){⟩}_I=U(t,t_0)|\psi (t_0){⟩}_I$ works for arbitrary initial states precisely whenFor $t\ge t_0$, the Dyson series writesIt is also ${\sum }_{n\ge 0}(-i{)}^n/n!{\int }_{[t_0,t{]}^n}{d}^{n}sT\{H_I(s_1)\cdots H_I(s_n)\}$: the cube consists of $n!$ ordered simplexes with the same time-ordered integrand. Differentiating the nested integrals makes the latest operator $H_I(t)$ the leftmost factor, leaving the $(n-1)$-fold integral. Thus ${\partial }_{t}U=-i{H}_I(t)U$ and at $t=t_0$ every nonconstant term vanishes. This proves the differentiation of the Dyson time-ordered exponential identity without incorrectly treating the generally noncommuting Hamiltonians as scalars.

For a Hermitian Hamiltonian operator, ${\partial }_t(U^{†}U)=i{U}^{†}{H}_{I}U-i{U}^{†}{H}_{I}U=0$, so the initial condition makes $U$ a unitary operator. Evolution backwards in time is its inverse, with anti-time ordering. For bounded norm-continuous $H_I$ the series converges in operator norm and the differentiation is justified by the factorial estimates. In quantum field theory, unbounded operators and products of fields require domain or regulator assumptions; this computation is the formal perturbative identity unless those analytic assumptions are supplied.