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 $$\begin{matrix} H_n(z) := (-1)^ne^{z^2}\frac{d^n}{dz^n}e^{-z^2}, \end{matrix}$$ 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 $$(-1)^n\int_{-\infty}^\infty H_m(x)H_n(x)e^{-x^2}\,dx = \int_{-\infty}^\infty H_m(x)\frac{d^ne^{-x^2}}{dx^n}\,dx =$$ $$= \sijoitus{-\infty}{\quad\infty}H_m(x)\frac{d^{n-1}e^{-x^2}}{dx^{n-1}} -\int_{-\infty}^\infty H'_m(x)\frac{d^{n-1}e^{-x^2}}{dx^{n-1}}\, dx.$$ 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 $$\int_{-\infty}^\infty H_m(x)H_n(x)e^{-x^2}\,dx = 2^m(-1)^{m+n}m!\int_{-\infty}^\infty H_0(x)\frac{d^{n-m}e^{-x^2}}{dx^{n-m}}\,dx =$$ $$= 2^m(-1)^{m+n}m!\sijoitus{-\infty}{\quad\infty}\frac{d^{n-m-1}e^{-x^2}}{dx^{n-m-1}} = 0,$$ whereas 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 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 $$\xi \mapsto \Psi_n(\xi) = C_nH_n(\xi)e^{-\frac{\xi^2}{2}}.$$