Orthogonality of Hermite polynomials
This is a problem I encountered in one of my math classes. I enjoyed the problem and figured I’d document the results.
Problem
The nth degree Hermite polynomial, \(H_{n}(x)\), is defined as
\[H_{n}(x) = (-1)^n e^{x^2}\frac{d^n}{dx^n}e^{-x^2}\]In this post, I will prove that
\[\int_{-\infty}^{\infty}{H_n(x)H_m(x)e^{x^2}dx=2^{n}n!\sqrt{\pi}\delta_{nm}}\]1. Recurrence relationship
We will first take the derivative of the Hermite polynomial
\[\frac{d}{dx}H_n(x) = \frac{d}{dx}\left[(-1)^n e^{x^2}\frac{d^n}{dx^n}e^{-x^2}\right]\] \[= (-1)^n \left[ 2xe^{x^2}\frac{d^n}{dx^n}e^{-x^2} + e^{x^2}\frac{d^{n+1}}{dx^{n+1}}e^{-x^2} \right]\] \[= (-1)^n \left[ 2xe^{x^2}\frac{d^n}{dx^n}e^{-x^2} + e^{x^2}\frac{d^{n}}{dx^{n}}(-2xe^{-x^2}) \right]\] \[= (-1)^n \left[ 2xe^{x^2}\frac{d^n}{dx^n}e^{-x^2} - 2e^{x^2}\frac{d^{n}}{dx^{n}}(xe^{-x^2}) \right]\]Focusing for a moment on the second term in this result, we will expand the nth derivative using the General Leibniz rule:
\[\frac{d^{n}}{dx^{n}}(xe^{-x^2}) = \sum_{k=0}^{n} {n \choose k} x^{(n-k)} (e^{-x^2})^{(k)}\] \[= x \frac{d^n}{dx^n}e^{-x^2} + n\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}\]Substituting this result into the above equation, we get
\[= (-1)^n \left[ 2xe^{x^2}\frac{d^n}{dx^n}e^{-x^2} - 2xe^{x^2}\frac{d^n}{dx^n}e^{-x^2} -2ne^{x^2}\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}\right]\] \[= (-1)^n \left[-2ne^{x^2}\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}\right]\] \[= (2n)\left((-1)^{n-1} e^{x^2}\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}\right)\] \[\boxed{= 2nH_{n-1}(x)}\]2. Orthogonality
With this recurrence relationship derived, we return to the original problem. We will first consider the case for \(n\neq m\). Without loss of generality, we will say that \(m > n\). A direct integration follows:
\[\int_{-\infty}^{\infty}H_n(x)H_m(x)e^{-x^2}dx\]Substituting in the definition of the Hermite polynomial:
\[= (-1)^m\int_{-\infty}^{\infty}H_n(x)\frac{d^m}{dx^m}e^{-x^2}dx\]Using integration by parts, we get
\[= (-1)^m\left[H_n(x)\frac{d^{m-1}}{dx^{m-1}}e^{-x^2}\right]_{-\infty}^{\infty} -(-1)^m\int_{-\infty}^{\infty}H'_n(x)\frac{d^{m-1}}{dx^{m-1}}e^{-x^2}dx\]The left term is a polynomial term multiplied by an \(e^{-x^2}\). Regardless of the polynomial, the exponential term will ultimately dominate the expression as \(x \to \pm \infty\). Thus, this term is zero.
\[= -(-1)^m \int_{-\infty}^{\infty}H'_n(x)\frac{d^{m-1}}{dx^{m-1}}e^{-x^2}dx\]Repeating the integration by parts \(m\) times, we get
\[= (-1)^{2m} \int_{-\infty}^{\infty}\left(\frac{d^m}{dx^m}H_n(x)\right)e^{-x^2}dx\] \[= \int_{-\infty}^{\infty}\left(\frac{d^m}{dx^m}H_n(x)\right)e^{-x^2}dx\]Because \(m>n\), and \(H_n(x)\) is an nth order polynomial, the mth order derivative of \(H_n(x)\) is zero. Thus, the integral is zero.
We now consider the case of \(n=m\). In this case, the problem proceeds in the same way, with \(m\) repeated integration by parts.
\[= \int_{-\infty}^{\infty}\left(\frac{d^m}{dx^m}H_m(x)\right)e^{-x^2}dx\]In this case, however, the derivative is not equal to zero. To derive the value of this derivative, we use the recurence relationship derived above.
\[= \int_{-\infty}^{\infty}2^mm!H_0(x)e^{-x^2}dx\] \[= 2^mm!H_0(x)\int_{-\infty}^{\infty}e^{-x^2}dx\]The \(0\)th Hermite polynomial is \(1\), and the integral to the right evaluates to \(\sqrt{\pi}\). Thus, the final result
\[= 2^mm!\sqrt{\pi}\] \[\blacksquare\]