Consider the set of Hermite functions $\{\phi_{n}(x,\varepsilon_{1})\}_{n}:= A$ defined below. \begin{equation} \label{eqn:funcs} \phi_{n}(x,\varepsilon_{1}) = \frac{\sqrt[8]{1+\big(\frac{2\varepsilon_{1}}{\gamma}\big)^{2}}}{\sqrt{2^{n}n!}}e^{-\Big(\sqrt{1+\big(\frac{2\varepsilon_{1}}{\gamma}\big)^{2}}-1 \Big)\frac{\gamma^{2}x^{2}}{2}}H_{n}\Bigg( \sqrt[4]{1+\Big(\frac{2\varepsilon_{1}}{\gamma}\Big)^{2}}\gamma x\Bigg). \end{equation}
These functions form an orthonormal basis for $(L^{2}(\mathbb{R},d\mu)$ where the measure is $d\mu = \frac{\gamma}{\sqrt{\pi}}e^{-\gamma^{2}x^{2}}dx$
Now, imagine that we perturb the $\varepsilon_{1}$ dependent function$\sqrt{1+\big(\frac{2\varepsilon_{1}}{\gamma}\big)^{2}}$ slightly in the following way, keeping in mind that this term show up in all of the components of the definition for $\phi_{n}(x,\varepsilon_{1})$, $\phi_{n}(x,\varepsilon_{1})\rightarrow \phi_{n}(x, \varepsilon_{2})$ where $\varepsilon_{2} := \sqrt{(\frac{2\varepsilon_{1}}{\gamma})^{2}+\alpha}$.
Assuming we do such a transformation the new set $\{\phi_{n}(x,\varepsilon_{2})\}_{n}:=B$ is still an orthonormal basis for $(L^{2}(\mathbb{R}, d\mu).$
I would like to know if there are any exact results for the inner product of an element of A and an element of B. i.e. $$\int_{\mathbb{R}}\phi_{n}(x,\varepsilon_{1})\phi_{m}(x,\varepsilon_{2})\frac{\gamma}{\sqrt{\pi}}e^{-\gamma^{2}x^{2}}dx.$$
In the limit where $\alpha$ is neglibible $\varepsilon_{2}\approx \varepsilon_{1}$ and we should have approximate orthogonality between a fixed element $\phi_{n}(x,\varepsilon_{2})$ of B and any $\phi_{m}(x,\varepsilon_{1})$ of set A where $m\neq n$.
Thank you in advance.