# Normalizing Hermite Polynomials

1. May 5, 2017

### rmiller70015

1. The problem statement, all variables and given/known data
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.

2. Relevant 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) $H'_n(x)=2nH_{n-1}(x)$

3. 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.

2. May 5, 2017

### I like Serena

Hi rmiller,

Let's take a closer look at the IBP:
Shouldn't that be:
$$(-1)^n\left[H_n(x)e^{-x^2}|^{\infty}_{-\infty}-\int^{\infty}_{-\infty}2nH_{n-1}(x)\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}dx\right] \\ =(-1)^n \cdot \left[ -\int^{\infty}_{-\infty}2nH_{n-1}(x)\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}dx \right] \\ =(-1)^{n-1} \int^{\infty}_{-\infty}2nH_{n-1}(x)\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}dx$$
?

3. May 8, 2017

### 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.