a discretization scheme. Various schemes can be used like lattice approximation, but the nature of star product due to its oscillator exponential, makes the lattice approach not suitable without adaptations. We will use another discretization scheme in which our fields are approximated by finite matrices and the star product becomes the standard matrix multiplication. Using the identity $D_\mu\phi =\phi\star\tilde{X}_{B\mu}-\tilde{X}_{A\mu}\star\phi$ we can recast the action in the following form: $$\begin{aligned}
&S(\phi,\tilde{X}_A,\tilde{X}_B)&= \frac{1}{(1+\Omega^2)^2}\int d^4z\Bigg\{\left(\frac{\left(1-\Omega^2\right)^2}{2}-\frac{\left(1-\Omega^2\right)^4}{6\left(1+\Omega^2\right)^2}\right)\Big(\left[\tilde{X}_{A\mu},\tilde{X}_{A\nu}\right]_\star\left[\tilde{X}_{A}^\mu,\tilde{X}_{A}^\nu\right]_\star  \nonumber \\
& &+\left[\tilde{X}_{B\mu},\tilde{X}_{B\nu}\right]_\star\left[\tilde{X}_{B}^\mu,\tilde{X}_{B}^\nu\right]_\star\Big)
+ \left(\phi\star\bar{\phi}+\frac{4\Omega^2}{1+\Omega^2}\tilde{X}_A^\mu\star\tilde{X}_{A\mu}-\frac{\chi_{-1}}{\chi_0}\right)^2 \nonumber \\
& & +\left(\bar{\phi}\star\phi +\frac{4\Omega^2}{1+\Omega^2}\tilde{X}_B^\mu\star\tilde{X}_{B\mu}-\frac{\chi_{-1}}{\chi_0} \right)^2  \nonumber \\
& &+2(1+\Omega^2)\left(\phi\star\tilde{X}_{B\mu}-\tilde{X}_{A\mu}\star\phi\right)
\left(\bar{\phi}\star\tilde{X}_{A}^\mu-\tilde{X}_{B}^\mu\star\bar{\phi}\right)\Bigg\}(z) \nonumber \\
& &+\mathcal{O}(\chi_1)\label{S0}\end{aligned}$$As a first approach to the numerical simulation and forced by limited computation resource, we will consider the Monte Carlo simulation of the previous action around its minimum and the simulation will take $\sqrt{\frac{\chi_{-1}}{\chi_0}} $ as a positive parameter. In this setting the behavior of the simulations will be identical for the negative case and avoiding $\frac{\chi_{-1}}{\chi_0}$ to be negative we have not any problems about the thermalization. In order to define the previous action around the minimum we translate the fields $\phi$, $\tilde{X}_{A\mu}$, $\tilde{X}_{B\mu}$ using the following translated fields: $$\begin{aligned}
\phi &=& \psi +\sqrt{\frac{\chi_{-1}}{\chi_0}}\cos\alpha\textbf{I} \ \\
\tilde{X}_{A\mu}&=& Y_{A\mu} +\frac{1}{2}\sqrt{\frac{\chi_{-1}}{\chi_0}}\sqrt{\frac{2\Omega^2}{(1+\Omega^2)}}\textbf{I}_\mu\sin\alpha  \\
\tilde{X}_{B\mu}&=& Y_{B\mu} +\frac{1}{2}\sqrt{\frac{\chi_{-1}}{\chi_0}}\sqrt{\frac{2\Omega^2}{(1+\Omega^2)}}\textbf{I}_\mu\sin\alpha  \end{aligned}$$ Substituting the 