and by the property: $$\label{compositionrule}
 h_1h_2 \triangleright f(x) =h_1\triangleright (h_2\triangleright f(x)),$$ where $h_1,\,h_2 \in H$. Hence, $H \triangleright {\mathcal{A}} = {\mathcal{A}} $ is a $H$-module. The Leibniz rule $h \triangleright (f(x)g(x)),$ for any $h \in H,$ is obtained by using Eq.(\[compositionrule\]) and from $x_{\mu } \triangleright (f(x)g(x)) =
\alpha (x_{\mu } f(x))g(x) + (1-\alpha)f(x) (x_{\mu } g(x)) $ and $p_{\mu } \triangleright (f(x)g(x)) =
 (p_{\mu } \triangleright f(x))g(x) + f(x) (p_{\mu } \triangleright g(x)), $ where $\alpha$ is an arbitrary real number.

The guiding principle in this paper is the idea that the Heisenberg algebra $H$ can be endowed with a coalgebra structure. Namely, the Leibniz rule and the coproduct $\Delta_0 $ are related by: $$\label{compositionrule1}
 h \triangleright (f(x)g(x)) = m_0 (\Delta_0 h) \triangleright (f(x)
 \otimes g(x) ),$$ for every $h \in H$, and $m_0$ is the multiplication map. From the Leibniz rule for $x_{\mu }$ it follows: $$\begin{aligned}
\label{copx}
 \Delta_0 x_{\mu} = \alpha x_{\mu} \otimes 1 + (1-\alpha ) 1 \otimes  x_{\mu}
    = \left[ x_{\mu } \otimes 1 \right] = \left[1 \otimes x_{\mu }
    \right] ~ \in {\mathcal{A}} \otimes {\mathcal{A}} / {\mathcal{R}}_0,\end{aligned}$$ where $\left[ x_{\mu } \otimes 1 \right] = \left[1 \otimes x_{\mu } \right]$ is the equivalence class 