system) endowed with a nondegenerate Hermitian scalar product $
\langle\cdot | \cdot \rangle\colon  {\mathcal{F}}\times{\mathcal{F}}\to \mathbb{C}.
$[^5] For any two instants of time $t_2$ and $t_1$ the corresponding state vectors are connected by the equality $$\label{2.1}
\psi(t_2) = {\mathcal{U}}(t_2,t_1)\psi(t_1)$$ where ${\mathcal{U}}$ is the *evolution operator* of the system [@Prugovecki-QMinHS chapter IV, Sect. 3.2]. It is supposed to be linear and unitary, i.e. $$\begin{aligned}
   \label{2.20}
{\mathcal{U}}(t_2,t_1)( \lambda\psi(t_1) + \mu\xi(t_1) )  &=
\lambda {\mathcal{U}}(t_2,t_1)(\psi(t_1)) + \mu {\mathcal{U}}(t_2,t_1)( \xi(t_1) ),
\ \ \ \ \ \ \ \         \\  \label{2.2}
{\mathcal{U}}^\dag(t_1,t_2) &= {\mathcal{U}}^{-1}(t_2,t_1),
    \end{aligned}$$ for any $\lambda,\mu\in\mathbb{C}$ and state vectors $\psi(t),\xi(t)\in{\mathcal{F}}$, and such that for any $t$ $$\label{2.3}
{\mathcal{U}}(t,t) = {\mspace{2mu}\mathsf{i}\mspace{-1mu}\mathsf{d}}_{\mathcal{F}}.$$ Here ${\mspace{2mu}\mathsf{i}\mspace{-1mu}\mathsf{d}}_X$ means the identity map of a set $X$ and the dagger ($\dag$) denotes Hermitian conjugation, i.e.  if $\varphi,\psi\in{\mathcal{F}}$ and ${\mathcal{A}}\colon {\mathcal{F}}\to{\mathcal{F}}$, then ${\mathcal{A}}^\dag$ is defined by $$\label{2.4}
\langle {\mathcal{A}}^\dag\varphi | \psi \rangle =
            \langle \varphi | {\mathcal{A}}\psi \rangle.$$ In particular ${\mathcal{U}}^\dag$ is defined by $
\langle {\mathcal{U}}^\dag(t_1,t_2)\varphi(t_2) | \psi(t_1) \rangle =
\langle \varphi(t_2) | {\mathcal{U}}(t_2,t_1)\psi(t_1) \rangle.
$ So, knowing $\psi(t_0)=\psi_0$ for some moment $t_0$, one knows the state vector for any moment $t$ as $
\psi(t)={\mathcal{U}}(t,t_0)\psi(t_0)={\mathcal{U}}(t,t_0)\psi_0.
$

Let ${\mathcal{H}}(t)$ be the Hamiltonian (function) of the system, i.e.  its total energy operator. 