(\Delta_0 h_0) = \Delta_0 (h h_0) $ for all $h, h_0 \in H $.

$\kappa$-deformation of Heisenberg algebra and coalgebra by twist
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We shall be interested in a deformation of the Heisenberg algebra $H$. We consider the family of twists given by $$\mathcal{F} = \exp\left(i(\lambda x_k p_k\otimes A-(1-\lambda)A
    \otimes x_k p_k)\right),
  \label{twist}$$ where $A = - a \cdot p,$ $\lambda $ is a real parameter and $a$ is a deformation fourvector, whose only non-zero component is the time component $a_0.$ The deformation parameter $a_0$ can be used to model deformation at the quantum gravity scale, in which case it corresponds to the inverse of the Planck mass, $a_0 \sim \frac{1}{\kappa}. $ The given family of twists represents a subfamily of Abelian twists which satisfy all required properties, including a cocycle and counit condition [@gghmm08]. Now the deformed coalgebra structure can be obtained by using this twist operator $\mathcal{F}$ to get $$\Delta h = \mathcal{F}\Delta_0 h \mathcal{F}^{-1}, 
  \label{defcops}$$ for every $h$ in $H$. By the same twist operation applied to relations ${\mathcal{R}}_0,$ namely $ \mathcal{F} {\mathcal{R}}_0 \mathcal{F}^{-1} \equiv {\mathcal{R}}, $ they transform into $$\begin{aligned}
\label{idxi}
 x_i \otimes 1 & = & Z^{\lambda - 1} \otimes x_i Z^{-\lambda},\\
  x_0 \otimes 1 