In the article "Two integral operators in Clifford analysis" , https://www.sciencedirect.com/science/article/pii/S0022247X08012262, it said that
$\langle V_{\alpha},V_{\alpha'}\rangle = \int_{\mathbb{R}^{n+1}} (\bar{V_{\alpha}}V_{\alpha'})_0e^{-n|x|^2} \,dx = 0$ for $\alpha\neq\alpha' $
where $V_{\alpha}$ is CK-extension of $\frac{1}{\alpha!}x^{\alpha}$ in multi-index notation.
However, when I tried for $n = 2 , \alpha=(2,0), \alpha' =(0,2)$, I get a counterexample. What I did is as follow:
1.$ 2 V_{2e^i} = z_i^2 = (x_i -x_0e_i)^2 = x_i^2 - x_0^2 -2x_ix_0e_i$
2.Let $x =x_1 ,y=x_2$ and $z= x_0$.
3.\begin{align*} \langle V_{\alpha},V_{\alpha'}\rangle &= \int_{\mathbb{R}^{3}} (\bar{V_{\alpha}}V_{\alpha'})_0e^{-2|x|^2} \,dx = \frac 1 4\int_{\mathbb{R}^3} (\bar{z_1^2}z_2^2)_0e^{-2(x^2 +y^2 +z^2)} \,dxdydz\\ &=\frac 1 4\int_{\mathbb{R}^3}(x^2-z^2)(y^2-z^2) e^{-2(z^2 +x^2+y^2)} \,dxdydz \\ &=\frac 1 42^{\frac 1 2}\int_{\mathbb{R}^3}(x^2-z^2)(y^2-z^2)e^{-(z^2 +x^2+y^2)} \,dxdydz\\ &= \frac 1 42^{\frac 1 2}\left(\Gamma(3/2)^2\sqrt{\pi} - \Gamma(3/2)^2\sqrt{\pi} -\Gamma(3/2)^2\sqrt{\pi} + \Gamma(5/2)^2\pi\right)\\ &= \frac 1 42^{\frac 1 2}(\frac 1 2 \pi\sqrt{\pi}) \neq 0 \end{align*}
Now, I wonder what is the mistake here. Please tell me anything.
Thanks in advance.