# PlanetPhysics/Hermite Polynomials

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

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

since this is a polynomial having the highest degree term ${\displaystyle (2z)^{n}}$ and satisfying the Hermite equation.\, The first six Hermite polynomials are

${\displaystyle H_{0}(z)\equiv 1,}$\\ ${\displaystyle H_{1}(z)\equiv 2z,}$\\ ${\displaystyle H_{2}(z)\equiv 4z^{2}-2,}$\\ ${\displaystyle H_{3}(z)\equiv 8z^{3}-12z,}$\\ ${\displaystyle H_{4}(z)\equiv 16z^{4}-48z^{2}+12,}$\\ ${\displaystyle H_{5}(z)\equiv 32z^{5}-160z^{3}+120z,}$

and the general polynomial form is

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

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

${\displaystyle {\begin{matrix}H'_{n}(z)=2nH_{n-1}(z).\end{matrix}}}$

We shall now show that the Hermite polynomials form an orthogonal set on the interval\, ${\displaystyle (-\infty ,\,\infty )}$\, with the weight factor ${\displaystyle e^{-x^{2}}}$.\, Let\, ${\displaystyle m;\, using (1) and integrating by parts we get ${\displaystyle (-1)^{n}\int _{-\infty }^{\infty }H_{m}(x)H_{n}(x)e^{-x^{2}}\,dx=\int _{-\infty }^{\infty }H_{m}(x){\frac {d^{n}e^{-x^{2}}}{dx^{n}}}\,dx=}$ $\displaystyle = \sijoitus{-\infty}{\quad\infty}H_m(x)\frac{d^{n-1}e^{-x^2}}{dx^{n-1}} -\int_{-\infty}^\infty H'_m(x)\frac{d^{n-1}e^{-x^2}}{dx^{n-1}}\, dx.$ The substitution portion here equals to zero because ${\displaystyle e^{-x^{2}}}$ and its derivatives vanish at ${\displaystyle \pm \infty }$.\, 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 ${\displaystyle \int _{-\infty }^{\infty }H_{m}(x)H_{n}(x)e^{-x^{2}}\,dx=2^{m}(-1)^{m+n}m!\int _{-\infty }^{\infty }H_{0}(x){\frac {d^{n-m}e^{-x^{2}}}{dx^{n-m}}}\,dx=}$ $\displaystyle = 2^m(-1)^{m+n}m!\sijoitus{-\infty}{\quad\infty}\frac{d^{n-m-1}e^{-x^2}}{dx^{n-m-1}} = 0,$ whereas in the case\, ${\displaystyle m=n}$\, the result ${\displaystyle \int _{-\infty }^{\infty }(H_{n}(x))^{2}e^{-x^{2}}\,dx=2^{n}(-1)^{2n}n!\int _{-\infty }^{\infty }e^{-x^{2}}\,dx=2^{n}n!{\sqrt {\pi }}}$ (see the area under Gaussian curve). The results mean that the functions\, ${\displaystyle x\mapsto {\frac {H_{n}(x)}{\sqrt {2^{n}n!{\sqrt {\pi }}}}}e^{-{\frac {x^{2}}{2}}}}$\, form an orthonormal set on\, ${\displaystyle (-\infty ,\,\infty )}$.\\

The Hermite polynomials are used in the quantum mechanical treatment of a harmonic oscillator, the wave functions of which have the form ${\displaystyle \xi \mapsto \Psi _{n}(\xi )=C_{n}H_{n}(\xi )e^{-{\frac {\xi ^{2}}{2}}}.}$