creation operators $a_\mu$, $a^\dagger_\nu$ satisfying for $\mu,\nu=1,\cdots,n$: $$\begin{aligned}
&[a_\mu,a_\nu]=[a^\dagger_\mu,a^\dagger_\nu]=0, & [a_\mu,a^\dagger_\nu]=\delta_{\mu\nu} \label{com1} \\ 
&\{b_\mu,b_\nu\}=\{b_\mu^\dagger ,b_\nu^\dagger\}=0, & \{b_\mu,b_\nu^\dagger\}=\delta_{\mu\nu} \label{anti1}\end{aligned}$$ Where $a_\mu =\frac{1}{\sqrt{2\omega}}  (\omega x_\mu + \partial_\mu ), \ a_\mu^\dagger = \frac{1}{\sqrt{2\omega}}(\omega x_\mu-\partial_\mu ) $. Using this operators is possible to construct a Dirac operator as: $$\mathcal{D}_n=-i\sqrt{2\omega}\delta^{\mu\nu}a_\mu^\dagger\otimes b_\nu + i\sqrt{2\omega}\delta^{\mu\nu}a_\mu\otimes b_\nu^\dagger=i\frac{d}{dx_\mu}\otimes(b_\mu+b_\mu^\dagger)+ i\omega x^\mu\otimes(b_\mu-b^\dagger_\mu) \label{dirac2}$$ summed over repeated index. We can define the fermionic part of the Hilbert space on which the Dirac operator acts starting from the vacuum state by subsequent applications of the fermionic creation operators $b_\nu^\dagger$ on the vacuum $b|0\rangle=0$, using the anti-commutation relations defining $\Lambda(\mathbb{C}^n)$. The complete Hilbert space is $\mathcal{H}_n=S(\mathbb{R}^n)\otimes\Lambda(\mathbb{C}^n)$. Beside, we can define a grading operator $\chi_n $ as: $$\begin{aligned}
\chi_n= \textbf{I}\otimes \prod^n_{\mu=1}(b_\mu b^\dagger_\mu-b_\mu b^\dagger_\mu )\end{aligned}$$ Using the relations - we can compute the square the Dirac operator as: $$\mathcal{D}_n^2=2\omega a^\dagger_\mu a^\mu\otimes \textbf{I} -2\omega \textbf{I} \otimes b^\dagger_\mu b^\mu =2\omega N_B\otimes \textbf{I} -2\omega \textbf{I}\otimes N_F$$ Where $N_F$ and $N_B$ are the number operators. In this form it easy to see that $\mathcal{D}^2_n $, being a “difference” between fermionic and bosonic number operator, has only one zero mode corresponding to the vacuum state. For practical reasons it is convenient write $\mathcal{D}^2_n $ as: $$\mathcal{D}_n^2=\omega \delta^{\mu\nu}(a^\dagger_\mu a_\nu + a_\nu a^\dagger_\mu)\otimes 