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'Hermite polynomials'
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| Title of object: |
Hermite polynomials |
| Canonical Name: |
HermitePolynomials |
| Type: |
Definition |
| Created on: |
2005-05-17 09:11:41 |
| Modified on: |
2005-05-19 16:00:05 |
| Classification: |
msc:12D99, msc:26A09, msc:26C05, msc:33B99, msc:33E30 |
Preamble:
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\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
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%\usepackage{psfrag}
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%\usepackage{graphicx}
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\usepackage{amsthm}
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%\usepackage{xypic}
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\newcommand{\sijoitus}[2]%
{\operatornamewithlimits{\Big/}_{\!\!\!#1}^{\,#2}}
\theoremstyle{definition}
\newtheorem*{thmplain}{Theorem} |
Content:
The polynomial solutions of the Hermite differential equation, with $n$ a non-negative integer, are usually normed so that the highest \PMlinkname{degree}{PolynomialRing} \PMlinkescapetext{term} is $(2z)^n$ and called the {\em Hermite polynomials} $H_n(z)$. \,The Hermite polynomials may be defined explicitely 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 \PMlinkescapetext{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 \PMlinkescapetext{polynomial form} is
$H_n(z) \equiv (2z)^n-\frac{n(n-1)}{1!}(2z)^{n-2}
+\frac{n(n-1)(n-2)(n-3)}{2!}(2z)^{n-4}-+...$
Differentiating this termwise gives
$H'_n(z) = 2n[(2z)^{n-1}-\frac{(n-1)(n-2)}{1!}(2z)^{n-3}+
\frac{(n-1)(n-2)(n-3)(n-4)}{2!}(2z)^{n-5}-+...],$
i.e.
\begin{align}
H'_n(z) = 2nH_{n-1}(z).
\end{align}
We shall now show that the Hermite polynomials form an \PMlinkname{orthogonal set}{OrthogonalPolynomials} on the interval $(-\infty,\,\infty)$ with the \PMlinkname{weight factor}{OrthogonalPolynomials} $e^{-x^2}$. \,Let \,$m < n$; \,using (1) and \PMlinkname{integrating by parts}{IntegrationByParts} 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
$$\int_{-\infty}^\infty H_m(x)H_n(x)e^{-x^2}\,dx =
2^m(-1)^{m+n}m!\int_{-\infty}^\infty H_0(x)\frac{d^{n-m}e^{-x^2}}{dx^{n-m}}\,dx
=$$ $$= 2^m(-1)^{m+n}m!\sijoitus{-\infty}{\quad\infty}\frac{d^{n-m-1}e^{-x^2}}{dx^{n-m-1}} = 0,$$
but 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 \PMlinkescapetext{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
$$\xi \mapsto \Psi_n(\xi) = C_nH_n(\xi)e^{-\frac{\xi^2}{2}}.$$ |
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