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}
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}
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}
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?
Best Answer
You can prove this without looking at any of the specific cases by doing a first-order perturbation of the differential equation that defines the time-evolution operator.
We start with
$$ i \hbar \frac{\partial}{\partial t} U(t, t_0) = H U(t, t_0). $$
Or, rearranging some terms,
$$ \frac{\partial}{\partial t} U(t, t_0) = - \frac{i}{\hbar} H U(t, t_0). $$
At a time $t + \delta t$, the above equation tells us that to first order in $\delta t$ we have
$$ U(t + \delta t, t_0) = U(t, t_0) - \frac{i}{\hbar} H U(t, t_0) \delta t. $$
If $U$ is unitary, then we have
$$ I = U^{\dagger} U (t + \delta t, t_0) = \left[ U^{\dagger}(t, t_0) + \frac{i}{\hbar} U^{\dagger}(t, t_0) H^{\dagger} \delta t\right] \left[ U(t, t_0) - \frac{i}{\hbar} H U(t, t_0) \delta t \right]. $$
Expanding this and keeping only terms to first order in $\delta t$ yields
$$ I = U^{\dagger} U (t, t_0) - \frac{i}{\hbar} U^{\dagger} (t, t_0) H U (t, t_0) \delta t + \frac{i}{\hbar} U^{\dagger}(t, t_0) H^{\dagger} U (t, t_0) \delta t. $$
Since we demanded that $U$ is unitary for all time, we should also have $U^{\dagger} U (t, t_0) = I.$ Substituting this and canceling $I$ from both sides of the equation yields
$$ 0 = \frac{i}{\hbar} U^{\dagger}(t, t_0) \left(H^{\dagger} - H \right) U(t, t_0) \delta t. $$
We must therefore have $H^{\dagger} = H$. So we have shown that if $U$ is unitary, then $H$ must be Hermitian.
For the other direction, we return to our first order expansion:
$$ U^{\dagger} U (t + \delta t, t_0) = U^{\dagger} U (t, t_0) - \frac{i}{\hbar} U^{\dagger} (t, t_0) H U (t, t_0) \delta t + \frac{i}{\hbar} U^{\dagger}(t, t_0) H^{\dagger} U (t, t_0) \delta t. $$
If $H$ is Hermitian, then
$$ U^{\dagger} U (t + \delta t, t_0) = U^{\dagger} U(t, t_0). $$
So $U^{\dagger} U(t, t_0)$ is constant for all $t$. Since $U(t_0, t_0)$ is the identity, we must have $U^{\dagger} U(t, t_0) = I$ for all $t$. We have thus proven that if $H$ is Hermitian, then $U$ must be unitary.