Special Functions – Orthogonal polynomials – Hermite polynomials

The Hermite polynomials are orthogonal on the interval \((-\infty,\infty)=\mathbb{R}\) with respect to the normal distribution \(w(x)=e^{-x^2}\). They can be defined by means of their Rodrigues formula:

\[H_n(x)=\frac{(-1)^n}{w(x)}D^nw(x)=(-1)^ne^{x^2}D^ne^{-x^2},\quad n=0,1,2,\ldots.\tag1\]

Since \(D^{n+1}=D\,D^n\), we obtain

\begin{align*} D^{n+1}w(x)&=D\left[D^nw(x)\right]=(-1)^nD\left[w(x)H_n(x)\right]=(-1)^n\left[w'(x)H_n(x)+w(x)H_n'(x)\right]\\[2.5mm] &=(-1)^{n+1}w(x)\left[2xH_n(x)-H_n'(x)\right],\quad n=0,1,2,\ldots, \end{align*}

which implies that

\[H_{n+1}(x)=2xH_n(x)-H_n'(x),\quad n=0,1,2,\ldots.\tag2\]

The definition (1) implies that \(H_0(x)=1\). Then (2) implies by induction that \(H_n(x)\) is a polynomial of degree \(n\). Further we have that \(H_{2n}(x)\) is even and \(H_{2n+1}(x)\) is odd and that the leading coefficient of the polynomial \(H_n(x)\) equals \(k_n=2^n\).

The Hermite polynomials satisfy the orthogonality relation

\[\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-x^2}H_m(x)H_n(x)\,dx=2^n\,n!\,\delta_{mn},\quad m,n\in\{0,1,2,\ldots\}.\tag3\]

To prove this we use the definition (1) to obtain

\[\int_{-\infty}^{\infty}e^{-x^2}H_m(x)H_n(x)\,dx=(-1)^n\int_{-\infty}^{\infty}H_m(x)D^ne^{-x^2}\,dx.\]

Now we use integration by parts \(n\) times to conclude that the integral vanishes for \(m < n\).

For \(m=n\) we have using integration by parts

\[\int_{-\infty}^{\infty}e^{-x^2}H_n(x)H_n(x)\,dx=(-1)^n\int_{-\infty}^{\infty}H_n(x)D^ne^{-x^2}\,dx =\int_{-\infty}^{\infty}D^nH_n(x)\cdot e^{-x^2}\,dx=k_n\,n!\,\int_{-\infty}^{\infty}e^{-x^2}\,dx=2^n\,n!\,\sqrt{\pi}.\]

This proves the orthogonality relation (3).

In order to find the three-term recurrence relation we start with

\[w(x)=e^{-x^2}\quad\Longrightarrow\quad w'(x)=-2xw(x).\]

Then we have by using Leibniz' rule

\[D^{n+1}w(x)=D^nw'(x)=D^n\left[-2xw(x)\right]=-2xD^nw(x)-2nD^{n-1}w(x),\]

which implies that

\[H_{n+1}(x)=2xH_n(x)-2nH_{n-1}(x),\quad n=1,2,3,\ldots.\tag4\]

Combining (2) and (4) we find that

\[H_n'(x)=2nH_{n-1}(x),\quad n=1,2,3,\ldots.\tag5\]

Differentiation of (2) gives

\[H_{n+1}'(x)=2xH_n'(x)+2H_n(x)-H_n''(x),\quad n=0,1,2,\ldots.\]

Now we use (5) to conclude that

\[2(n+1)H_n(x)=2xH_n'(x)+2H_n(x)-H_n''(x),\quad n=0,1,2,\ldots,\]

which implies that \(H_n(x)\) satisfies the second-order linear differential equation

\[y''(x)-2xy'(x)+2ny(x)=0,\quad n\in\{0,1,2,\ldots\}.\]

Finally we will prove the generating function

\[e^{2xt-t^2}=\sum_{n=0}^{\infty}\frac{H_n(x)}{n!}\,t^n.\tag6\]

We start with

\[f(t)=e^{-(x-t)^2}=e^{-x^2}\cdot e^{2xt-t^2}.\]

The Taylor series for \(f(t)\) is

\[f(t)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}\,t^n\]

with, by using the substitution \(x-t=u\),

\[f^{(n)}(0)=\left[\frac{d^n}{dt^n}\,e^{-(x-t)^2}\right]_{t=0} =(-1)^n\left[\frac{d^n}{du^n}\,e^{-u^2}\right]_{u=x}=(-1)^nD^ne^{-x^2}=e^{-x^2}H_n(x),\quad n=0,1,2,\ldots.\]

Hence we have

\[e^{-x^2}\cdot e^{2xt-t^2}=e^{-(x-t)^2}=f(t)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}\,t^n =e^{-x^2}\sum_{n=0}^{\infty}\frac{H_n(x)}{n!}\,t^n.\]

This proves the generating function (6).

The Hermite polynomials \(H_2(x)\), \(H_3(x)\) and \(H_4(x)\).

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Last modified on May 22, 2021