I was wondering whether anything is known on the following: Let $h_k (x)= (-1)^k e^{x^2/2} \frac {d^k}{dx^k} \, e^{-x^2/2}$, $k \geq 0$, be the classical Hermite polynomials ($h_0(x) = 1$, $h_1(x) = x$, $h_2(x) = x^2 -1$, $h_3(x) = x^3 - 3x$, ...). Is there anything known on the asymptotic distributional behavior (if any), as $k \to \infty$, of the sequence of random variables $X_k$, $k \geq 0$, such that the law of $X_k$ has a density proportional to $h_k^2 \, e^{-x^2/2}$, $x \in \mathbb{R}$, with respect to the Lebesgue measure?
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$\begingroup$ Thank you for the indications! Would this be similar for other orthogonal polynomials (Chebyshevn Jacobi, Laguerre...)? $\endgroup$– Sunia CortezCommented Feb 8, 2023 at 8:02
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$\begingroup$ the approach I sketched in the answer box generalises to orthogonal polynomials that are eigenfunctions of a second order differential equation; we can then apply the WKB method to find the asymptotics at large eigenvalue; if the differential equation is of the form $-\psi''(x)+V(x)\psi(x)=E\psi(x)$, then the asymptotic probability density is $\propto [E-V(x)]^{-1/2}$. (For the Hermite polynomials of your question one has $E=4k$ and $V(x)=x^2$.) $\endgroup$– Carlo BeenakkerCommented Feb 8, 2023 at 11:52
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$\begingroup$ Dear Professor Beenakker, thank you very much indeed for your detailed answer, and all the comments. I am not so familiar with physics arguments, so I need some time to fully understand the picture. I do not know how to technically "accept the answer"? (About generalizations to other orthogonal polynomials, I was also wondering similarly about discrete examples (Meixner, Krawtchouk...)) $\endgroup$– Sunia CortezCommented Feb 9, 2023 at 8:20
1 Answer
This is related to the probability density of the quantum harmonic oscillator. (The Hermite polynomials are the eigenstates of that problem.) For large $k$ the density $$P_k(x)=\frac{1}{\sqrt{2\pi}k!} [h_k(x)]^2 \, e^{-x^2/2},$$ normalized to unity, has the approximation $$p_k(x)=\frac{1}{\pi}(4k-x^2)^{-1/2},\;\;|x|<2\sqrt{k}.$$ The boundaries $\pm 2\sqrt{k}$ are the classical turning points of the harmonic oscillator. In that context $E=4k$ is the energy of the particle and $V(x)=x^2$ is the harmonic potential in which it moves. The large-$k$ approximation is known as the WKB approximation. More generally, for an arbitrary potential $V(x)$ the density varies as $[E-V(x)]^{-1/2}$.
Alternatively, one can use the asymptotic expansion of the Hermite polynomials, $$e^{-x^2/4}h_k(x) \propto \cos \left(x \sqrt{ k}- \pi k/2 \right)\left(4k-x^2\right)^{-1/4}.$$ Squaring and averaging over the rapidly oscillating $\cos^2$ gives $p_k(x)$.
The plot compares $P_k$ and $p_k$ for $k=100$. The function $p_k$ is a smoothed version of $P_k$, suitable for averaging quantities that vary little on the scale of $1/\sqrt{k}$. For example, one can check by explicit computation that the average of $x^n$, $n\in\mathbb{N}$, agrees to leading order in $k$, $$\lim_{k\rightarrow\infty} k^{-n}\int P_k(x)x^{2n}\,dx=\frac{2^{2n}\,\Gamma(n+1/2)}{\sqrt{\pi}\,\Gamma(n+1)}=k^{-n}\int p_k(x) x^{2n}\,dx.$$
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$\begingroup$ Do you have a proof of this approximation? $\endgroup$ Commented Feb 7, 2023 at 20:31
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$\begingroup$ added two ways to arrive at the result, from the WKB approximation of the harmonic oscillator, or from the asymptotic expansion of the Hermite polynomials. $\endgroup$ Commented Feb 7, 2023 at 20:56
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$\begingroup$ I do not see your approximation on the WKB approximation page. As for asymptotic expansions on the Hermite polynomials Wikipedia page, I see there clearly different, apparently mutually incompatible expansions, without a statement on the conditions for, or rigorous meaning of, those expansions. In your post too, it is unclear in what sense the approximation holds and what the sign $\propto$ means. $\endgroup$ Commented Feb 7, 2023 at 21:31
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$\begingroup$ the sense in which $p_k(x)$ approximates $P_k(x)$ for large $k$ is as an integral kernel, I have added an example that hopefully explains this. $\endgroup$ Commented Feb 7, 2023 at 22:25
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$\begingroup$ Can you detail the "explicit computation"? Also, a limit sign in the last display is apparently missing. $\endgroup$ Commented Feb 7, 2023 at 22:33