It is well known that any homogeneous polynomial $f \in \mathbb R[x_1, \ldots, x_n]$ can be uniquely split as $f = f_0 + x^2 f_1$, where $x^2 \equiv (x_1)^2 + \ldots + (x_n)^2$ and $f_0$ harmonic: $\Delta f_0 \equiv \left( \frac{\partial^2}{\partial x_1{}^2} + \ldots + \frac{\partial^2}{\partial x_n{}^2} \right) f_0 = 0$.
Roughly speaking, whether the same is true for ''square roots'' of $\Delta$ and $x^2$? I mean the following.
Consider the space $V$ of irreducible representation of the complex Clifford algebra $e_i e_j + e_j e_i = 2 \delta_{ij}$, $i,j = 1, \ldots, n$. And consider the space $V'$ of polynomials in $x_1, \ldots, x_n$ with coefficients being vectors from the representation of the Clifford algebra. So we have a module of linear combinations of $v (x_1)^{k_1} \ldots (x_n)^{k_n}$ with $v \in V$ for the associative algebra with generators $e_1, \ldots, e_n, x_1, \ldots, x_n, \frac{\partial}{\partial x_1}, \ldots, \frac{\partial}{\partial x_n}$.
Consider the Dirac operator $D = \sum_{k=1}^n e_k \frac{\partial}{\partial x_k}$, which is a ''square root'' of $\Delta$: $D^2 = \Delta$. And consirer a conjugate (see below) operator $D^* = \sum_{k=1}^n e_k x_k$, which is a ''square root'' of $x^2$: $(D^*)^2 = x^2$.
Let us call $v \in V'$ homogeneous is $N v \equiv \sum_{k=1}^n x_k \frac{\partial}{\partial x_k} v = m v$ for some $m \in \mathbb N$.
Is it true that any homogeneous $v \in V'$ can be uniquely split as $v = v_0 + D^* v_1$, with $D v_0 = 0$? The uniqueness part can be easily proved by induction. But I'm stuck with the existence of such decomposition.
The classical proof of decomposition of a homogeneous polynomial to harmonic and proportional to $x^2$ relies to the existence of a natural inner product on homogeneous polynomials. Namely, $\langle f(x), g(x) \rangle = f(\partial) g(x)$, where $f(\partial)$ is the differential operator obtained from $f(x)$ by substitution $x_k \mapsto \frac{\partial}{\partial x_k}$. Then the conjugate to $x_k$ is $(x_k)^*= \frac{\partial}{\partial x_k}$. And $\Delta^* = x^2$.
We can introduce an inner product on $V'$ in a similar way. On homogeneous elements $v f(x)$ and $w g(x)$ ($v, w \in V'$) we define $\langle v f(x), w g(x) \rangle' = \langle v, w \rangle'' \langle f(x), g(x) \rangle$, where $\langle v, w \rangle''$ is the inner product on $V$ such that the multiplication by Clifford algenra generators $e_k$ is self conjugate: $\langle e_k v, w \rangle = \langle v, e_k w \rangle$. We see that then $D^*$ becomes conjugate to $D$. The problem is that such inner products are not positive/negative definite: there exists isotropic vectors. So I can't follow the same steps as for $V$.