the kernel $$\label{H7}
K_s({\mathbf{q}}_1,\overline{{\mathbf{q}}_2})=\sum_{n=0}^{\infty}H_n^s({\mathbf{q}}_1)\overline{H_n^s({\mathbf{q}}_2)},$$ then from (\[H4\]) $K_s({\mathbf{q}}_1,{\mathbf{q}}_2)$ is a reproducing kernel and the corresponding reproducing kernel Hilbert space is $$\label{C1}
A_s=\overline{\text{span}\{H_n^s({\mathbf{q}}) \mid n\in\mathbb{N}\}},$$ the span and its closure being taken under left multiplication by quaternionic constants. Note that the above reproducing kernel is the quaternionic equivalent of the kernel (\[H4\]) and the Hilbert space $A_s$ the equivalent of $\mathfrak H_{hol}^s$, generated by that kernel.

\[L1\] $K_s({\mathbf{q}},\overline{{\mathbf{q}}})=K_s(z,\overline{z})\mathbb{I}_2.$

Since $$H_n({\mathbf{q}})=u_{{\mathbf{q}}}\left(\begin{array}{cc}H_n(z)&0\\0&H_n(\overline{z})\end{array}\right)
u_{{\mathbf{q}}}^{\dagger},$$ we have $$H_n({\mathbf{q}})\overline{H_n({\mathbf{q}})}=u_{{\mathbf{q}}}\left(\begin{array}{cc}H_n(z)\overline{H_n(z)}&0\\0&H_n(z)\overline{H_n(z)}\end{array}\right)u_{{\mathbf{q}}}^{\dagger}.$$ Since $b_n(s)$ are real numbers, $$\begin{aligned}
b_n(s)^{-1}H_n({\mathbf{q}})\overline{H_n({\mathbf{q}})}&=&u_{{\mathbf{q}}}\left(\begin{array}{cc}b_n(s)^{-1}H_n(z)\overline{H_n(z)}&0\\
0&b_n(s)^{-1}H_n(z)\overline{H_n(z)}\end{array}\right)u_{{\mathbf{q}}}^{\dagger}\\
&=&u_{{\mathbf{q}}}\left(\begin{array}{cc}h_{n,s}(z)\overline{h_{n,s}(z)}&0\\0&h_{n,s}(z)\overline{h_{n,s}(z)}\end{array}\right)u_{{\mathbf{q}}}^{\dagger}.\end{aligned}$$ Therefore $$\begin{aligned}
\sum_{n}b_n(s)^{-1}H_n({\mathbf{q}})\overline{H_n({\mathbf{q}})}&=&u_{{\mathbf{q}}}\left(\begin{array}{cc}\sum_nh_{n,s}(z)\overline{h_{n,s}(z)}&0\\0&\sum_nh_{n,s}(z)\overline{h_{n,s}(z)}\end{array}\right)u_{{\mathbf{q}}}^{\dagger}\\
&=&u({\mathbf{q}})\left(\begin{array}{cc}K_s(z,z)&0\\0&K_s(z,z)\end{array}\right)u_{{\mathbf{q}}}^{\dagger}\\
&=&K_s(z,\overline{z})\mathbb{I}_2.\end{aligned}$$ That is, $$\label{C2}
K_s({\mathbf{q}},\overline{{\mathbf{q}}})=K_s(z,\overline{z})\mathbb{I}_2.$$

The quaternionic Hermite polynomials $H_{n,m}({\mathbf{q}},\overline{{\mathbf{q}}})$
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The exponential series $$e^{{\mathbf{q}}}=\sum_{n=0}^{\infty}\frac{{\mathbf{q}}^n}{n!},\quad {\mathbf{q}}\in H,$$ converges absolutely, and uniformly on compact subsets with respect to the norm of $H$ [@Ebb] (p 204). Thereby $$e^{-|{\mathbf{q}}|^2}=e^{-{\mathbf{q}}\overline{{\mathbf{q}}}}=\sum_{n=0}^{\infty}(-1)^n\frac{({\mathbf{q}}\overline{{\mathbf{q}}})^n}{n!}$$ converges uniformly. Further ${\mathbf{q}}$ and $\overline{{\mathbf{q}}}$ commute and real numbers commute with quaternions. Therefore, as in the complex case, as an extension of complex hermite polynomials, using Definition \[D3\] we get $$\begin{aligned}
\label{E8}
h_{n,m}({\mathbf{q}},\overline{{\mathbf{q}}})&=&(-1)^{n+m}e^{|{\mathbf{q}}|^2}\frac{\partial^{n+m}}
{\partial{\mathbf{q}}^n\partial\overline{{\mathbf{q}}}^m}e^{-|{\mathbf{q}}|^2}=\sum_{i=0}^{n}\sum_{j=0}^{m}a_{ij}{\mathbf{q}}^i\overline{{\mathbf{q}}}^j\\
 &=& n!m!\sum_{j=0}^{\text{min}\{n,m\}}\frac{(\overline{{\mathbf{q}}})^{n-j}}{(n-j)!}\frac{{\mathbf{q}}^{m-j}}{(m-j)!},\end{aligned}$$ where the $a_{ij}$ are real coefficients and the derivatives should be understood in the Cullen sense. Further from (\[E7\]) we have $$a_{ij}{\mathbf{q}}^i\overline{{\mathbf{q}}}^j=u_{{\mathbf{q}}}\left(\begin{array}{cc}
a_{ij}z^i{\overline{z}}^j&0\\
0&a_{ij}{\overline{z}}^iz^j\end{array}\right)u_{{\mathbf{q}}}^{\dagger}.$$ Therefore $$\sum_{i=0}^{n}\sum_{j=0}^{m}a_{ij}{\mathbf{q}}^i\overline{{\mathbf{q}}}^j=u_{{\mathbf{q}}}\left(\begin{array}{cc}
\sum_{i=0}^{n}\sum_{j=0}^{m}a_{ij}z^i{\overline{z}}^j&0\\
0&\sum_{i=0}^{n}\sum_{j=0}^{m}a_{ij}{\overline{z}}^iz^j\end{array}\right)u_{{\mathbf{q}}}^{\dagger}.$$ Thereby from (\[E8\]) we get $$\label{E9}
h_{n,m}({\mathbf{q}},\overline{{\mathbf{q}}})=u_{{\mathbf{q}}}\left(\begin{array}{cc}
h_{n,m}(z,{\overline{z}})&0\\
0&h_{m,n}(z,{\overline{z}})\end{array}\right)u_{{\mathbf{q}}}^{\dagger}.$$ We can see from 