Let be $\lambda>0$. Put
$$ L_{\lambda}=\Big[-\frac{\partial^{2}}{\partial z \partial \overline{z}}+\lambda^{2}|z|^{2} +\lambda\Big(\overline{z}\frac{\partial}{ \partial \overline{z}}-z\frac{\partial}{ \partial z}\Big)\Big] \quad\hbox{with}\quad z\in\mathbb{C}$$
and
$$ \widetilde{\Delta}=-\frac{\partial^{2}}{\partial z \partial \overline{z}}+\overline{z}\frac{\partial}{ \partial \overline{z}}.$$
Consider the operator $T_{\lambda}:L^{2}(\mathbb{C}, dz)\rightarrow L^{2}(\mathbb{C}, e^{-|z|^{2}}dz)$ defined by: $$ T_{\lambda}f(z)=\frac{1}{2\lambda}e^{\frac{|z|^{2}}{2}}f\Big( \frac{z}{\sqrt{2\lambda}} \Big) $$
We have by a simple calculation
$
T_{\lambda}o(L_{\lambda} -\lambda)oT^{-1}_{\lambda}=2\lambda\widetilde{\Delta}
$.
I want to extend it for $\lambda\in \mathbb{C}$, for this, put $\lambda=|\lambda|e^{i\theta}$, we obtain
$$
T_{\lambda}o(L_{\lambda} -|\lambda|)oT^{-1}_{\lambda}=(1+e^{i\theta})|\lambda|\overline{z}\frac{\partial f}{ \partial \overline{z}}+(1-e^{i\theta})|\lambda|z\frac{\partial f}{ \partial z}-2|\lambda|\frac{\partial^{2} f}{\partial z \partial \overline{z}}+\frac{|\lambda|}{2}(1-e^{2i\theta})|z|^{2}f(z)
$$
So, for $\theta=0$ we get $T_{\lambda}o(L_{\lambda} -\lambda)oT^{-1}_{\lambda}=2\lambda\widetilde{\Delta}$.
How can I extend it for $\lambda=a+ib$? Thank you in advance.
