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  "body": "\n## Definition\n\nLet $q \\in \\C$ with $q^k \\ne 1$ for all integers $k \\ge 1$, and let $x = \\cos\\theta$.\nFor an integer $n \\ge 0$, the *continuous $q$-Hermite polynomial* is\n$$H_n(x \\mid q) = \\sum_{k=0}^{n} \\binom{n}{k}_q\\, e^{i(n - 2k)\\theta},$$\nwhere $\\binom{n}{k}_q$ is the [[Gaussian binomial coefficient|YS09Z1YNEK]].\n\n## Remarks\n\nThe terms for $k$ and $n - k$ have the same coefficient, so they pair to $2\\binom{n}{k}_q \\cos((n - 2k)\\theta)$, and $H_n$ is a polynomial of degree $n$ in $x$ with leading coefficient $2^n$.\nThey satisfy $H_0 = 1$, $H_1 = 2x$ and the recurrence $2x\\, H_n(x \\mid q) = H_{n+1}(x \\mid q) + (1 - q^n)\\, H_{n-1}(x \\mid q)$ for $n \\ge 1$.\nFor $\\abs{q} \u003c 1$ and $\\abs{t} \u003c 1$ the generating function is $\\sum_{n \\ge 0} \\frac{H_n(x \\mid q)}{(q; q)_n} t^n = \\frac{1}{(t e^{i\\theta}; q)_\\infty\\, (t e^{-i\\theta}; q)_\\infty}$, with the [[q-Pochhammer symbol|Y1X8ZZQ7VM]].\n\nExample: $n = 2$.\nWith $\\binom{2}{1}_q = 1 + q$, $H_2 = e^{2i\\theta} + (1 + q) + e^{-2i\\theta} = 2\\cos 2\\theta + 1 + q = 4x^2 - 1 + q$.\nThe recurrence gives the same: $2x \\cdot 2x - (1 - q) \\cdot 1 = 4x^2 - 1 + q$.\n\nNon-example: the ordinary Hermite polynomial $4x^2 - 2$ is not $H_2(x \\mid q)$ for any $q$: matching constants would need $q = -1$, which is excluded since $q^2 = 1$.\n\nFor $-1 \u003c q \u003c 1$ they are orthogonal: $\\frac{1}{2\\pi} \\int_0^{\\pi} H_m(\\cos\\theta \\mid q)\\, H_n(\\cos\\theta \\mid q)\\, (e^{2i\\theta}, e^{-2i\\theta}; q)_\\infty \\, d\\theta = \\delta_{mn} \\frac{(q; q)_n}{(q; q)_\\infty}$.\nAs $q \\to 1^-$, $\\left(\\frac{2}{1 - q}\\right)^{n/2} H_n\\left( x \\sqrt{\\tfrac{1 - q}{2}} \\,\\middle|\\, q \\right)$ tends to the [[Hermite polynomial|MGACY98E9H]] $H_n(x)$.\n",
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