bosons this is not the case. One can prove using Eq.  that for bosons, $$\frac{{\mathrm{d}}}{{\mathrm{d}}t}\bigl[{\hat{U}}^\ddag(t,t_0){\hat{U}}(t,t_0)\bigr]=0,
\label{res3}$$ where the pseudo-Hermitian conjugate of ${\hat{U}}(t,t_0)$ has been introduced, $${\hat{U}}^\ddag(t,t_0)={\hat{\sigma}}_3{\hat{U}}^\dagger(t,t_0){\hat{\sigma}}_3,
\label{re333}$$ with ${\hat{\sigma}}_3=\begin{pmatrix} 1 & 0 \cr 0 & -1 \end{pmatrix}$. Hence, accounting for the initial condition ${\hat{U}}(t_0,t_0)={\hat{I}}$, we obtain $${\hat{U}}^\ddag(t,t_0){\hat{U}}(t,t_0)={\hat{I}}.
\label{res4}$$ This means that ${\hat{U}}$ is the element of the $SU(1,1)$ group, discussed in Appendix \[SU(1,1)\]. Keeping this in mind, we shall derive now physical consequences of unitary vs. pseudo-unitary time evolution of particles created from the vacuum by a sequence of electric field pulses.

### Monodromy matrix {#mono}

For a train of $N_{\rm rep}$ identical electric field pulses, each of time duration $T$, both functions $\Omega_{\bm p}(t)$ and $\omega_{\bm p}(t)$ in  are periodic within the time interval $N_{\rm rep}T$, with a period $T$. Hence, the same applies to ${\hat{U}}(t,t_0)$. This property combined with the composition condition for the time-evolution operators results in $${\hat{U}}(N_{\rm rep}T+t_0,t_0)=\prod_{j=0}^{N_{\rm rep}-1}{\hat{U}}\bigl((j+1)T+t_0,jT+t_0\bigr)=\bigl[{\hat{U}}(T+t_0,t_0)\bigr]^{N_{\rm rep}},
\label{composition}$$ where we shall refer to ${\hat{U}}(T+t_0,t_0)$ as the [*monodromy matrix*]{} [@Yakubovich]. This matrix is evaluated at the period of the interaction with the external electric field. As a consequence of , it determines the system evolution under the influence of a finite sequence of well-separated 