When we solve the Schrodinger equation for the time-evolution operator:
\begin{equation} i\hbar\frac{\partial}{\partial t}U(t,t_{0})=HU(t,t_{0}), \end{equation}
We have three cases to be treated separately:
Case 1. The Hamiltonian operator $H$ is independent of time:
\begin{equation} U(t,t_{0})=exp\left[\frac{-iH(t-t_{0})}{\hbar}\right]; \end{equation}\begin{equation} U(t,t_{0})=\exp\left[\frac{-iH(t-t_{0})}{\hbar}\right]; \end{equation}
Case 2. The Hamiltonian operator $H$ is time-dependent but $H's$ at different times commute:
\begin{equation} U(t,t_{0})=exp\left[-\frac{i}{\hbar}\int_{t_{0}}^{t}dt^{'}H\left(t^{'}\right)\right]; \end{equation}\begin{equation} U(t,t_{0})=\exp\left[-\frac{i}{\hbar}\int_{t_{0}}^{t}dt^{'}H\left(t^{'}\right)\right]; \end{equation}
Case 3. The Hamiltonian operator $H$ is time-dependent and $H's$ at different times do not commute:
\begin{eqnarray} U(t,t_{0}) & = & 1+\overset{\infty}{\underset{n=1}{\sum}}\left[\left(\frac{-i}{\hbar}\right)^{n}\int_{t_{0}}^{t}dt_{1}\int_{t_{0}}^{t_{1}}dt_{2}...\int_{t_{0}}^{t_{n-1}}dt_{n}H(t_{1})H(t_{2})...H(t_{n})\right]\\ & = & \mathcal{T}\left\{ exp\left[-i\int_{t_{0}}^{t}dt^{'}H(t^{'})\right]\right\} \end{eqnarray}\begin{eqnarray} U(t,t_{0}) & = & 1+\overset{\infty}{\underset{n=1}{\sum}}\left[\left(\frac{-i}{\hbar}\right)^{n}\int_{t_{0}}^{t}dt_{1}\int_{t_{0}}^{t_{1}}dt_{2}...\int_{t_{0}}^{t_{n-1}}dt_{n}H(t_{1})H(t_{2})...H(t_{n})\right]\\ & = & \mathcal{T}\left\{ \exp\left[-i\int_{t_{0}}^{t}dt^{'}H(t^{'})\right]\right\} \end{eqnarray}
If we consider the case 1, the following statement is easy to prove:
The Hamiltonian operator $H$ is hermitian if and only if the time-evolution operator $U$ is unitary.
But how to prove this statement for time-dependent Hamiltonian cases?
