Talk:PlanetPhysics/Hermite Polynomials

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The polynomial solutions of the Hermite \htmladdnormallink{differential equation}{http://planetphysics.us/encyclopedia/DifferentialEquations.html}, with $n$ a non-negative integer, are usually normed so that the highest degree term is $(2z)^n$ and called the {\em Hermite polynomials} $H_n(z)$.\, The Hermite polynomials may be defined explicitly by

\begin{align} H_n(z) := (-1)^ne^{z^2}\frac{d^n}{dz^n}e^{-z^2}, \end{align} since this is a polynomial having the highest degree term $(2z)^n$ and satisfying the \htmladdnormallink{Hermite equation}{http://planetphysics.us/encyclopedia/HermiteEquation.html}.\, The first six Hermite polynomials are

$H_0(z) \equiv 1,$\\ $H_1(z) \equiv 2z,$\\ $H_2(z) \equiv 4z^2-2,$\\ $H_3(z) \equiv 8z^3-12z,$\\ $H_4(z) \equiv 16z^4-48z^2+12,$\\ $H_5(z) \equiv 32z^5-160z^3+120z,$

and the general polynomial form is

$H_n(z) \equiv (2z)^n-\frac{n(n-1)}{1!}(2z)^{n-2} +\frac{n(n-1)(n-2)(n-3)}{2!}(2z)^{n-4}-+\cdots.$\\

Differentiating this termwise gives $H'_n(z) = 2n\left[(2z)^{n-1}-\frac{(n-1)(n-2)}{1!}(2z)^{n-3}+ \frac{(n-1)(n-2)(n-3)(n-4)}{2!}(2z)^{n-5}-+\cdots\right],$ i.e. \begin{align} H'_n(z) = 2nH_{n-1}(z). \end{align}

We shall now show that the Hermite polynomials form an orthogonal set on the interval\, $(-\infty,\,\infty)$\, with the weight factor $e^{-x^2}$.\, Let\, $m < n$;\, using (1) and integrating by parts we get $$(-1)^n\int_{-\infty}^\infty H_m(x)H_n(x)e^{-x^2}\,dx = \int_{-\infty}^\infty H_m(x)\frac{d^ne^{-x^2}}{dx^n}\,dx =$$ $$= \sijoitus{-\infty}{\quad\infty}H_m(x)\frac{d^{n-1}e^{-x^2}}{dx^{n-1}} -\int_{-\infty}^\infty H'_m(x)\frac{d^{n-1}e^{-x^2}}{dx^{n-1}}\, dx.$$ The substitution portion here equals to zero because $e^{-x^2}$ and its derivatives vanish at $\pm\infty$.\, Using then (2) we obtain $$\int_{-\infty}^\infty H_m(x)H_n(x)e^{-x^2}\,dx = 2(-1)^{1+n}m\int_{-\infty}^\infty H_{m-1}(x)\frac{d^{n-1}e^{-x^2}}{dx^{n-1}}\,dx.$$ Repeating the integration by parts gives the result $$\int_{-\infty}^\infty H_m(x)H_n(x)e^{-x^2}\,dx = 2^m(-1)^{m+n}m!\int_{-\infty}^\infty H_0(x)\frac{d^{n-m}e^{-x^2}}{dx^{n-m}}\,dx =$$ $$= 2^m(-1)^{m+n}m!\sijoitus{-\infty}{\quad\infty}\frac{d^{n-m-1}e^{-x^2}}{dx^{n-m-1}} = 0,$$ whereas in the case\, $m = n$\, the result $$\int_{-\infty}^\infty (H_n(x))^2e^{-x^2}\,dx = 2^n(-1)^{2n}n!\int_{-\infty}^\infty e^{-x^2}\,dx = 2^nn!\sqrt{\pi}$$ (see the area under Gaussian curve). The results mean that the \htmladdnormallink{functions\,}{http://planetphysics.us/encyclopedia/Bijective.html} $x \mapsto\frac{H_n(x)}{\sqrt{2^nn!\sqrt{\pi}}}e^{-\frac{x^2}{2}}$\, form an orthonormal set on\, $(-\infty,\,\infty)$.\\

The Hermite polynomials are used in the quantum mechanical treatment of a harmonic oscillator, the \htmladdnormallink{wave}{http://planetphysics.us/encyclopedia/CosmologicalConstant2.html} functions of which have the form $$\xi \mapsto \Psi_n(\xi) = C_nH_n(\xi)e^{-\frac{\xi^2}{2}}.$$

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