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It is well known that the q-Hermite polynomials defined by $$H_n(\theta; q)= \sum\limits_{k=0}^n \frac{(q;q)_n}{(q;q)_k(q;q)_{n-k}}e^{i(n-2k)\theta}$$ are orthogonal in $\theta \in [0, \pi]$ with respect to the measure $(e^{i2\theta},e^{-2i\theta}; q)_{\infty}d\theta=d\mu$.

Let me define now the modified q-Hermite polynomials $$H_n^{\lambda}(\theta; q)= \sum\limits_{k=0}^n \frac{(q;q)_n}{(q;q)_k(q;q)_{n-k}}\lambda^ke^{i(n-2k)\theta} $$. Is it still true that the polynomials $\{H_n^{\lambda}(\theta; q)\}_{n \in \mathbb{N}}$ are orthogonal with respect to some other measure $d\mu_{\lambda}$ in $[0, \pi]$? If Yes, which is their norm?

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  • $\begingroup$ I think you can incorporate $\ln\lambda$ into the argument $\theta$, just a linear shift. - Well there is a factor $i$ that may mess it all up. $\endgroup$
    – Wolfgang
    Commented Mar 20, 2015 at 16:23
  • $\begingroup$ Can you explain me a little better your observation and how can it be useful in terms of guessing the measure that make the sequence orthogonal? $\endgroup$
    – Matteo
    Commented Mar 21, 2015 at 16:34
  • $\begingroup$ I meant you can write $H_n^{\lambda}(\theta; q)=\lambda^{n/2 }H_n (\theta+\frac i2 \log\lambda; q)$ but not sure if that helps, as the argument is no more real. $\endgroup$
    – Wolfgang
    Commented Mar 21, 2015 at 20:43

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