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[parent] Hermite polynomials (Definition)

The polynomial solutions of the Hermite 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

$\displaystyle H_n(z) := (-1)^ne^{z^2}\frac{d^n}{dz^n}e^{-z^2},$ (1)
since this is a polynomial having the highest degree term $(2z)^n$ and satisfying the 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 $H'_n(z) = 2n\left[(2z)^{n-1}-\frac{(n-1)(n-2)}{1!}(2z)^{n-3}+ \frac{(n-1)(n-2)(n-3)(n-4)}{2!}(2z)^{n-5}-+\cdots\right],$ i.e.

$\displaystyle H'_n(z) = 2nH_{n-1}(z).$ (2)

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

$\displaystyle (-1)^n\int_{-\infty}^\infty H_m(x)H_n(x)e^{-x^2}\,dx = \int_{-\infty}^\infty H_m(x)\frac{d^ne^{-x^2}}{dx^n}\,dx =$

$\displaystyle = \operatornamewithlimits{\Big/}_{\!\!\!-\infty}^{\,\quad\infty}H... ...{dx^{n-1}} -\int_{-\infty}^\infty H'_m(x)\frac{d^{n-1}e^{-x^2}}{dx^{n-1}}\, dx.$
The substitution portion here equals to zero because $e^{-x^2}$ and its derivatives vanish at $\pm\infty$.  Using then (2) we obtain

$\displaystyle \int_{-\infty}^\infty H_m(x)H_n(x)e^{-x^2}\,dx = 2(-1)^{1+n}m\int_{-\infty}^\infty H_{m-1}(x)\frac{d^{n-1}e^{-x^2}}{dx^{n-1}}\,dx.$
Repeating the integration by parts gives the result

$\displaystyle \int_{-\infty}^\infty H_m(x)H_n(x)e^{-x^2}\,dx = 2^m(-1)^{m+n}m!\int_{-\infty}^\infty H_0(x)\frac{d^{n-m}e^{-x^2}}{dx^{n-m}}\,dx =$

$\displaystyle = 2^m(-1)^{m+n}m!\operatornamewithlimits{\Big/}_{\!\!\!-\infty}^{\,\quad\infty}\frac{d^{n-m-1}e^{-x^2}}{dx^{n-m-1}} = 0,$
whereas in the case  $m = n$  the result

$\displaystyle \int_{-\infty}^\infty (H_n(x))^2e^{-x^2}\,dx = 2^n(-1)^{2n}n!\int_{-\infty}^\infty e^{-x^2}\,dx = 2^nn!\sqrt{\pi}$
(see the area under Gaussian curve). The results mean that the functions  $x \mapsto\frac{H_n(x)}{\sqrt{2^nn!\sqrt{\pi}}}e^{-\frac{x^2}{2}}$  form an orthonormal set on  $(-\infty,\,\infty)$.

The Hermite polynomials are used in the quantum mechanical treatment of a harmonic oscillator, the wave functions of which have the form

$\displaystyle \xi \mapsto \Psi_n(\xi) = C_nH_n(\xi)e^{-\frac{\xi^2}{2}}.$



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Cross-references: wave, functions, Hermite equation, differential equation
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This is version 1 of Hermite polynomials, born on 2009-04-19.
Object id is 684, canonical name is HermitePolynomials.
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Classification:
Physics Classification: 02.30.Hq (Ordinary differential equations)
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