# Hermite polynomials

The polynomial  solutions of the Hermite differential equation  , with $n$ a non-negative integer, are usually normed so that the highest degree (http://planetmath.org/PolynomialRing) is $(2z)^{n}$ and called the $H_{n}(z)$.  The Hermite polynomials may be defined explicitly by

 $\displaystyle H_{n}(z)\;:=\;(-1)^{n}e^{z^{2}}\frac{d^{n}}{dz^{n}}e^{-z^{2}},$ (1)

since this is a polynomial having the highest $(2z)^{n}$ and satisfying the Hermite equation.  The equation (1) is the Rodrigues’s formula for Hermite polynomials.  Using the Faà di Bruno’s formula, one gets from (1) also

 $H_{n}(x)\;=\;(-1)^{n}\!\sum_{m_{1}+2m_{2}=n}\frac{n!}{m_{1}!m_{2}!}(-1)^{m_{1}% +m_{2}}(2x)^{m_{1}}.$

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 is

 $H_{n}(z)\;\equiv\;(2z)^{n}-\frac{n(n\!-\!1)}{1!}(2z)^{n-2}+\frac{n(n\!-\!1)(n% \!-\!2)(n\!-\!3)}{2!}(2z)^{n-4}-+\ldots$

Differentiating this termwise gives

 $H^{\prime}_{n}(z)\;=\;2n\!\left[(2z)^{n-1}-\frac{(n\!-\!1)(n\!-\!2)}{1!}(2z)^{% n-3}+\frac{(n\!-\!1)(n\!-\!2)(n\!-\!3)(n\!-\!4)}{2!}(2z)^{n-5}-+\ldots\right]\!,$

i.e.

 $\displaystyle H^{\prime}_{n}(z)\;=\;2nH_{n-1}(z).$ (2)

The Hermite polynomials are sometimes scaled to such ones $\mathrm{He_{n}}$ which obey the differentiation rule

 $\displaystyle\mathrm{He}^{\prime}_{n}(z)\;=\;n\mathrm{He}_{n-1}(z).$ (3)

Such Hermite polynomials form an Appell sequence.

We shall now show that the Hermite polynomials form an orthogonal set (http://planetmath.org/OrthogonalPolynomials) on the interval$(-\infty,\,\infty)$  with the weight factor (http://planetmath.org/OrthogonalPolynomials) $e^{-x^{2}}$.  Let  $m;  using (1) and integrating by parts (http://planetmath.org/IntegrationByParts) we get

 $\displaystyle(-1)^{n}\!\int_{-\infty}^{\infty}H_{m}(x)H_{n}(x)e^{-x^{2}}\,dx$ $\displaystyle\;=\;\int_{-\infty}^{\infty}H_{m}(x)\frac{d^{n}e^{-x^{2}}}{dx^{n}% }\,dx$ $\displaystyle\;=\;\operatornamewithlimits{\Big{/}}_{\!\!\!-\infty}^{\,\quad% \infty}\!H_{m}(x)\frac{d^{n-1}e^{-x^{2}}}{dx^{n-1}}-\int_{-\infty}^{\infty}H^{% \prime}_{m}(x)\frac{d^{n-1}e^{-x^{2}}}{dx^{n-1}}\,dx.$

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

 $\displaystyle\int_{-\infty}^{\infty}H_{m}(x)H_{n}(x)e^{-x^{2}}\,dx$ $\displaystyle\;=\;2^{m}(-1)^{m+n}m!\int_{-\infty}^{\infty}H_{0}(x)\frac{d^{n-m% }e^{-x^{2}}}{dx^{n-m}}\,dx$ $\displaystyle\;=\;2^{m}(-1)^{m+n}m!\!\operatornamewithlimits{\Big{/}}_{\!\!\!-% \infty}^{\,\quad\infty}\frac{d^{n-m-1}e^{-x^{2}}}{dx^{n-m-1}}\;=\;0,$

whereas in the case  $m=n$  the result

 $\int_{-\infty}^{\infty}(H_{n}(x))^{2}e^{-x^{2}}\,dx\;=\;2^{n}(-1)^{2n}n!\int_{% -\infty}^{\infty}e^{-x^{2}}\,dx\;=\;2^{n}n!\sqrt{\pi}$

(see area under Gaussian curve). The results that the functions  $x\mapsto\frac{H_{n}(x)}{\sqrt{2^{n}n!\sqrt{\pi}}}e^{-\frac{x^{2}}{2}}$  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

 $\xi\;\,\mapsto\,\;\Psi_{n}(\xi)\;=\;C_{n}H_{n}(\xi)e^{-\frac{\xi^{2}}{2}}.$
 Title Hermite polynomials Canonical name HermitePolynomials Date of creation 2013-03-22 15:16:25 Last modified on 2013-03-22 15:16:25 Owner pahio (2872) Last modified by pahio (2872) Numerical id 28 Author pahio (2872) Entry type Definition Classification msc 33E30 Classification msc 33B99 Classification msc 26C05 Classification msc 26A09 Classification msc 12D99 Related topic SubstitutionNotation Related topic AppellSequence Related topic LaguerrePolynomial