= Solution
For $H=H_0+H_{\mathrm{int}}$, the <interaction picture> has $H_I(t)=e^{iH_0(t-t_0)}H_{\mathrm{int}}(t)e^{-iH_0(t-t_0)}$ and
$$
i\frac{d}{dt}|\psi(t)\rangle_I=H_I(t)|\psi(t)\rangle_I.
$$
The trial solution $|\psi(t)\rangle_I=U(t,t_0)|\psi(t_0)\rangle_I$ works for arbitrary initial states precisely when
$$
\boxed{i\partial_tU(t,t_0)=H_I(t)U(t,t_0),\qquad U(t_0,t_0)=I.}
$$
For $t\geq t_0$, the <Dyson series> writes
$$
U(t,t_0)=T\exp\!\left(-i\int_{t_0}^tH_I(s)ds\right)
=I+\sum_{n\geq1}(-i)^n\int_{t_0}^t ds_1\int_{t_0}^{s_1}ds_2\cdots\int_{t_0}^{s_{n-1}}ds_n\,H_I(s_1)\cdots H_I(s_n).
$$
It is also $\sum_{n\geq0}(-i)^n/n!\int_{[t_0,t]^n}d^ns\,T\{H_I(s_1)\cdots H_I(s_n)\}$: 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 $\partial_tU=-iH_I(t)U$ 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>, $\partial_t(U^\dagger U)=iU^\dagger H_IU-iU^\dagger H_IU=0$, 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.
Back to article page
