- #1

thecourtholio

- 19

- 1

## Homework Statement

Prove that ##\psi_n## in Eq. 2.85 is properly normalized by substituting generating functions in place of the Hermite polynomials that appear in the normalization integral, then equating the resulting Taylor series that you obtain on the two sides of your equation. As a by-product, you should also find that ## \psi_m ## and ## \psi_n ## are orthogonal when m ≠ n.

## Homework Equations

Eq 2.85: $$ \psi_n(x) = \bigg(\frac{m\omega}{\pi\hbar}\bigg)^{1/4} \frac{1}{\sqrt{2^nn!}}H_n(\xi) e^{-\xi^2/2} $$

where $$\xi \equiv \sqrt{\frac{m\omega}{\hbar}}x $$

and ## H_n(\xi) ## are the Hermite polynomials.

## The Attempt at a Solution

Ok so the main thing throwing me off here is where it says to equate the two taylor series obtained because I don't really see how I'm supposed to obtain a series on both sides.

First off, when I go to normalize I use

$$ \int_{-\infty}^{\infty} |\psi_n|^2d\xi = 1$$

and equating all the constants in front of the ## H_n(\xi) ## term to ## A## the integral becomes

$$ \int_{-\infty}^{\infty} A^2 H_n^2 e^{-\xi^2} d\xi = 1 $$

Then, substituting the generating function (using the form ## e^{2\xi z-z^2} ##) in, this becomes

$$ \int_{-\infty}^{\infty} A^2 e^{-(\xi-z)^2} = 1 $$

But this is a simple gaussian integral right? So then its just

$$ \int_{-\infty}^{\infty} A^2 e^{-(\xi-z)^2} = \sqrt{\pi}A^2 $$

But I feel like all this is wrong because I don't see how a Taylor series comes into play. Am I doing the normalization all wrong or is there just something I'm not seeing?

Any advice is much appreciated!