# Normalizing Hermite Polynomials

## Homework Statement

Evaluate the normalization integral in (22.15). Hint: Use (22.12) for one of the $H_n(x)$ factors, integrate by parts, and use (22.17a); then use your result repeatedly.

## Homework Equations

(22.15) ##\int_{-\infty}^{\infty}e^{-x^2}H_n(x)H_m(x)dx = \sqrt{\pi}2^nn!## when ##n=m##
(22.12) ##H_n(x)=(-1)^ne^{x^2}\frac{d^n}{dx^n}e^{-x^2}##
(22.17a) ##H'_n(x)=2nH_{n-1}(x)##

## The Attempt at a Solution

I started out by writing my integral from (22.15) as:
##\int_{-\infty}^{\infty}e^{-x^2}[H_n(x)]^2dx##
Expanding one of the ##H_n(x)## terms gives:
##\int_{-\infty}^{\infty}e^{-x^2}H_n(x)e^{x^2}(-1)^n\frac{d^n}{dx^n}e^{-x^2}dx##
##=(-1)^n\int_{-\infty}^{\infty}H_n(x)\frac{d^n}{dx^n}e^{-x^2}dx##
Integration by parts and using equation 22.17a gives:
##(-1)^n[H_n(x)e^{-x^2}|^{\infty}_{-\infty}-\int^{\infty}_{-\infty}2nH_{n-1}(x)\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}dx##
The first term goes to zero and leaves:
##(-1)^n\int^{\infty}_{-\infty}2nH_{n-1}(x)\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}dx##
IBP n-1 more times gives:
##(-1)^n2^nn!\int^{\infty}_{-\infty}H_0(x)e^{-x^2}dx=(-1)^n2^nn!\int^{\infty}_{-\infty}e^{-x^2}dx = (-1)^n2^nn!\sqrt{\pi}##

This is correct except I have this extra factor of ##(-1)^n## that shouldn't be there, I'm not sure how to get rid of it.

I like Serena
Homework Helper
Hi rmiller,

Let's take a closer look at the IBP:
##(-1)^n[H_n(x)e^{-x^2}|^{\infty}_{-\infty}-\int^{\infty}_{-\infty}2nH_{n-1}(x)\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}dx##

Shouldn't that be:
$$(-1)^n\left[H_n(x)e^{-x^2}|^{\infty}_{-\infty}-\int^{\infty}_{-\infty}2nH_{n-1}(x)\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}dx\right] \\ =(-1)^n \cdot \left[ -\int^{\infty}_{-\infty}2nH_{n-1}(x)\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}dx \right] \\ =(-1)^{n-1} \int^{\infty}_{-\infty}2nH_{n-1}(x)\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}dx$$
?

rmiller70015
For future people, you get a factor of -2 each time you do ibp and in the end you're left with (-1)^n (-2)^n when you combine them it gives 2^n.