previous commutators the Heisenberg equations of motion can be cast into the form $$\begin{aligned}
\label{EquacaoR}
\frac{d{\bf R}(\tau)}{d\tau}&=&\frac{\mbox{\mathversion{bold}${\Pi}$}(\tau)}{m}\ , \\
 \frac{d\mbox{\mathversion{bold}${\Pi}$}(\tau)}{d\tau} &=& 2\omega
\mathbb{C}  \mbox{\mathversion{bold}${\Pi}$}(\tau) \,,
 \label{EquacaoPi}\end{aligned}$$ where $2\omega={eB}/{m}$ is the cyclotron frequency and we defined the anti-diagonal matrix $$\mathbb{C}=\left(
\begin{array}{cc}
 0 & 1 \\
-1 & 0
\end{array}
\right)
\label{Eq21}\ .$$ Integrating equation (\[EquacaoPi\]) we find $$\begin{aligned}
\mbox{\mathversion{bold}${\Pi}$}(\tau) &=&
 e^{2\omega\mathbb{C}\tau}\mbox{\mathversion{bold}${\Pi}$}(0)
\label{Eq23}\ .\end{aligned}$$ Substituting this solution in equation (\[EquacaoR\]) and integrating once more, we get $$\begin{aligned}
\label{Eq22}
\textbf{R}(\tau)-\textbf{R}(0)&=&\frac{\sin{(\omega\tau)}}{m\omega}
 e^{\omega\mathbb{C}\tau}\mbox{\mathversion{bold}${\Pi}$}(0)
\  ,\end{aligned}$$ where we used the following properties of $\mathbb{C}$ matrix: $\mathbb{C}^{2}\!=\!-\uma$; $\mathbb{C}^{-1}\!=\!-\mathbb{C}=\mathbb{C}^{T}$, $e^{\alpha\mathbb{C}} =
 \cos{(\alpha)}\uma + \sin{(\alpha)}\mathbb{C}$ with $\mathbb{C}^{T}$ being the transpose of $\mathbb{C}$. Combining equations (\[Eq22\]) and (\[Eq23\]) we can write $\mbox{\mathversion{bold}${\Pi}$}(0)$ in terms of the operators ${\bf R}(\tau)$ and ${\bf R}(0)$ as $$\label{Eq24}
\mbox{\mathversion{bold}${\Pi}$}(0)=\frac{m\omega}{\sin{(\omega\tau)}}
 e^{-\omega\mathbb{C}\tau}\biggl({\bf R}(\tau)-{\bf R}(0)\biggr).$$ In order to express $\mathcal{H}_\bot = (\pi_1^2 + \pi_2^2)/2m$ in terms of $\textbf{R}(\tau)$ and $\textbf{R}(0)$, we use (\[Eq24\]). In matrix notation, we have $$\begin{aligned}
\mathcal{H}_{\bot}&=&\frac{1}{2m}\,\mbox{\mathversion{bold}${\Pi}$}^{T}(0)\mbox{\mathversion{bold}${\Pi}$}(0)\nonumber\\
&=&\frac{m\omega^2}{2\sin^2{(\omega\tau)}}\biggl({\bf R}^{T}(\tau){\bf R}(\tau)+{\bf R}^{T}(0){\bf R}(0)+\nonumber\\
&\ &\hspace{1.5cm}-{\bf R}^{T}(\tau){\bf R}(0)-{\bf R}^{T}(0){\bf R}(\tau)\biggr)\ .
\label{Hnaoordenado}\end{aligned}$$ Last term on the r.h.s. of (\[Hnaoordenado\]) is not ordered appropriately as required in the step ([**ii**]{}). The correct ordering may be obtained as follows: first, we write $$\textbf{R}(0)^{T}  \textbf{R}(\tau)
 = \textbf{R}(\tau)^{T}\textbf{R}(0) +
\sum_{i=1}^{2}[X_{i}(0) 