PlanetPhysics/Hermite Polynomials

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

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

since this is a polynomial having the highest degree term $(2z)^{n}$ 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.

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

We shall now show that the Hermite polynomials form an orthogonal set on the interval\, $(-\infty ,\,\infty )$\, with the weight factor $e^{-x^{2}}$.\, Let\, $m<n$;\, using (1) and integrating by parts we get $(-1)^{n}\int _{-\infty }^{\infty }H_{m}(x)H_{n}(x)e^{-x^{2}}\,dx=\int _{-\infty }^{\infty }H_{m}(x){\frac {d^{n}e^{-x^{2}}}{dx^{n}}}\,dx=$ 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 $\pm \infty$.\, 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 $\int _{-\infty }^{\infty }H_{m}(x)H_{n}(x)e^{-x^{2}}\,dx=2^{m}(-1)^{m+n}m!\int _{-\infty }^{\infty }H_{0}(x){\frac {d^{n-m}e^{-x^{2}}}{dx^{n-m}}}\,dx=$ 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 $\int _{-\infty }^{\infty }(H_{n}(x))^{2}e^{-x^{2}}\,dx=2^{n}(-1)^{2n}n!\int _{-\infty }^{\infty }e^{-x^{2}}\,dx=2^{n}n!{\sqrt {\pi }}$ (see the area under Gaussian curve). The results mean that the functions\, $x\mapsto {\frac {H_{n}(x)}{\sqrt {2^{n}n!{\sqrt {\pi }}}}}e^{-{\frac {x^{2}}{2}}}$\, 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 $\xi \mapsto \Psi _{n}(\xi )=C_{n}H_{n}(\xi )e^{-{\frac {\xi ^{2}}{2}}}.$