Prove time-dependent hamiltonian is hermitian from unitarity of time-evolution operator

Question

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?

Answer 1

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.

Answer 2

Hermiticity of $H$ for all three cases combined together can be shown directly from the Schrödinger equation. To do this, first take the derivative of the unitarity relation $U(t,t_{0}) U^{\dagger}(t,t_{0}) = I$ with respect to $t$, which gives \begin{equation} U(t,t_{0}) \left[\frac{\partial}{\partial t}U^{\dagger}(t,t_{0}) \right] = -\left[\frac{\partial}{\partial t}U(t,t_{0}) \right]U^{\dagger}(t,t_{0}). \end{equation} Then, consider the Hermitian conjugate of \begin{equation} H = i\hbar \left[\frac{\partial}{\partial t}U(t,t_{0}) \right] U^{\dagger}(t,t_{0}), \end{equation} which reads \begin{equation} \begin{split} H^{\dagger} &= -i\hbar\, U(t,t_{0})\left[\frac{\partial}{\partial t}U^{\dagger}(t,t_{0}) \right] \\ &= i\hbar\, \left[\frac{\partial}{\partial t}U(t,t_{0}) \right]U^{\dagger}(t,t_{0}). \end{split} \end{equation} Therefore, $H = H^\dagger$, and hence $H$ is Hermitian.

Answer 3

I think that the unitarity of all the $\bigl(U(t,t_0)\bigr)_{(t,t_0)\in\mathbb{R}^2}$ does not give you so much. The unitarity is actually encoded in the (two-parameter) group property i.e. $U(t,t_0)U(t_0,t_1)=U(t,t_1)$ for any $t,t_0,t_1\in\mathbb{R}$ (with $U(t,t)=1$ for any $t$). Let's suppose that $U(t,t_0)$ is differentiable on $t$ and $t_0$ on suitable domains for any $t,t_0$ (the domains may depend on time); and that differentiating we get the equations \begin{align*} i\partial_t U(t,s)&=H(t)U(t,s)\\ i\partial_{s} U(t,s)&=-U(t,s)H'(s)\; ; \end{align*} for some families $\bigl(H(t)\bigr)_{t\in\mathbb{R}}$ and $\bigl(H'(s)\bigr)_{s\in\mathbb{R}}$ of operators. Now unitarity implies that $U(t,s)^* = U(s,t)$, therefore taking the adjoint of the first equation we get that $$-i\partial_t U(s,t)=U(s,t)H(t)^*\; ,$$ and therefore it follows that for any $t\in \mathbb{R}$, the "right generators" $H'(t)$ are related to the "left generators" by $H'(t)=H(t)^*$. In other words, the two derivatives above now read \begin{align*}\tag{$*$} i\partial_t U(t,s)&=H(t)U(t,s)\\ i\partial_{s} U(t,s)&=-U(t,s)H(s)^*\; . \end{align*}

If we now take the double adjoint, it also follows that each operator $H(t)$ must be closed (because we get $H(t)^{**}=H(t)$, and that each adjoint is densely defined).

In conclusion, we see that to have a densely differentiable unitary two-paprameter group of evolution $\bigl(U(t,t_0)\bigr)_{(t,t_0)\in\mathbb{R}^2}$ it is necessary that there is a family of closed (densely defined) operators $\bigl(H(t)\bigr)_{t\in\mathbb{R}}$ that "generate" the group on the left and on the right via the equations ($*$). This however, do not yield the hermiticity of the generators.