PlanetPhysics/Hermite Polynomials

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

$\begin{array}{c}{H}_n(z):=(-1{)}^n{e}^{z^2}\frac{d^n}{d{z}^n}{e}^{-z^2},\end{array}$

since this is a polynomial having the highest degree term $(2z{)}^n$ and satisfying the [[../HermiteEquation/|Hermite equation]].\, 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 $H{\text{'}}_n(z)=2n[(2z{)}^{n-1}-\frac{(n-1)(n-2)}{1!}(2z{)}^{n-3}+\frac{(n-1)(n-2)(n-3)(n-4)}{2!}(2z{)}^{n-5}-+\cdots ],$ i.e.

$\begin{array}{c}H{\text{'}}_n(z)=2n{H}_{n-1}(z).\end{array}$

We shall now show that the Hermite polynomials form an orthogonal set on the interval\, $(-\mathrm{\infty },\mathrm{\infty })$\, with the weight factor $e^{-x^2}$.\, Let\, $m<n$;\, using (1) and integrating by parts we get $(-1{)}^n\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{H}_m(x)H_n(x)e^{-x^2}dx=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{H}_m(x)\frac{d^n{e}^{-x^2}}{d{x}^n}dx=$ Failed to parse (unknown function "\sijoitus"): {\displaystyle = \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 $\pm \mathrm{\infty }$.\, Using then (2) we obtain $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{H}_m(x)H_n(x)e^{-x^2}dx=2(-1{)}^{1+n}m\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{H}_{m-1}(x)\frac{d^{n-1}{e}^{-x^2}}{d{x}^{n-1}}dx.$ Repeating the integration by parts gives the result $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{H}_m(x)H_n(x)e^{-x^2}dx=2^m(-1{)}^{m+n}m!\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{H}_0(x)\frac{d^{n-m}{e}^{-x^2}}{d{x}^{n-m}}dx=$ Failed to parse (unknown function "\sijoitus"): {\displaystyle = 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 $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}(H_n(x){)}^2{e}^{-x^2}dx=2^n(-1{)}^{2n}n!\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-x^2}dx=2^{n}n!\sqrt{\pi }$ (see the area under Gaussian curve). The results mean that the [[../Bijective/|functions\,]] $x\mapsto \frac{H_n(x)}{\sqrt{2^{n}n!\sqrt{\pi }}}{e}^{-\frac{x^2}{2}}$\, form an orthonormal set on\, $(-\mathrm{\infty },\mathrm{\infty })$.\\

The Hermite polynomials are used in the quantum mechanical treatment of a harmonic oscillator, the [[../CosmologicalConstant2/|wave]] functions of which have the form $\xi \mapsto {\mathrm{\Psi }}_n(\xi )=C_n{H}_n(\xi )e^{-\frac{{\xi }^2}{2}}.$