Atkin's 8th Edition Self-test 9.7

Dr. Kevin Range


Date: 2007-02-08

The following is Self-test 9.7 from Atikin's 8th Edition, page 295, done correctly. Thanks to Jared for catching my misapplication of the Hermite polynomial recursion relations.

What went wrong before?

Before we start, let's examine where things went wrong. The recursion relation for the Hermite polynomials is

$\displaystyle y H_v = v H_{v-1} + \frac{1}{2} H_{v+1}$ (1)

The is indubitably correct. However, when faced with a term like $ y H_{v-1}$ I incorrectly said

$\displaystyle y H_{v-1} = v H_{v-2} + \frac{1}{2} H_{v}$ (2)

which, as Jared pointed out, is not correct. The correct relationship in this case is

$\displaystyle y H_{v-1} = \mathbf{(v-1)} H_{v-2} + \frac{1}{2} H_{v}$ (3)

Other recursion relations can be generated in the analogous way. With this knowledge in hand, let's proceed to calculate $ \left<x^2\right>$ for the quantum harmonic oscillator.

$ \left<x^2\right>$ for the quantum harmonic oscillator


$\displaystyle \left<x^2\right>$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty} N_v H_v\left(\frac{x}{\alpha}\right) \mat... ...H_v\left(\frac{x}{\alpha}\right) \mathrm{e}^{-\frac{x^2}{2 \alpha}} \mathrm{d}x$ (4)
  $\displaystyle =$ $\displaystyle N_v^2 \int_{-\infty}^{\infty} H_v\left(\frac{x}{\alpha}\right) x^2 H_v\left(\frac{x}{\alpha}\right) \mathrm{e}^{-\frac{x^2}{\alpha}} \mathrm{d}x$ (5)

Substitute


$\displaystyle y$ $\displaystyle =$ $\displaystyle \frac{x}{\alpha}$ (6)
$\displaystyle x$ $\displaystyle =$ $\displaystyle \alpha y$ (7)
$\displaystyle \mathrm{d}x$ $\displaystyle =$ $\displaystyle \alpha \mathrm{d}y$ (8)

$\displaystyle \left<x^2\right> = N_v^2 \alpha^3 \int_{-\infty}^{\infty} H_v\left(y\right) y^2 H_v\left(y\right) \mathrm{e}^{-y^2} \mathrm{d}y$ (9)

Now we use the recursion relations


$\displaystyle y^2 H_{v}$ $\displaystyle =$ $\displaystyle y v H_{v-1} + \frac{y}{2} H_{v+1}$ (10)
$\displaystyle y v H_{v-1}$ $\displaystyle =$ $\displaystyle v (v-1) H_{v-2} + \frac{v}{2} H_{v}$ (11)
$\displaystyle \frac{y}{2} H_{v+1}$ $\displaystyle =$ $\displaystyle \frac{v+1}{2} H_v + \frac{1}{4} H_{v+2}$ (12)

$\displaystyle \left<x^2\right> = N_v^2 \alpha^3 \int_{-\infty}^{\infty} H_v \le... ...+ \frac{v+1}{2} H_v + \frac{1}{4} H_{v+2} \right] \mathrm{e}^{-y^2} \mathrm{d}y$ (13)

Terms involving Hermite polynomials of different $ v$ will be zero due to orthogonality.


$\displaystyle \left<x^2\right>$ $\displaystyle =$ $\displaystyle N_v^2 \alpha^3 \int_{-\infty}^{\infty} H_v \left[ \frac{v}{2} H_{v} + \frac{v+1}{2} H_v \right] \mathrm{e}^{-y^2} \mathrm{d}y$ (14)
  $\displaystyle =$ $\displaystyle N_v^2 \alpha^3 \int_{-\infty}^{\infty} H_v \left[ \left(v + \frac... ...{-y^2} \mathrm{d} y \; \fbox{There's the $v+\frac{1}{2}$ we were looking for!}$ (15)
  $\displaystyle =$ $\displaystyle N_v^2 \alpha^3 \left(v + \frac{1}{2} \right) \int_{-\infty}^{\infty} H_v H_{v} \mathrm{e}^{-y^2} \mathrm{d} y$ (16)
  $\displaystyle =$ $\displaystyle \frac{1}{\alpha \sqrt{\pi} 2^v v!} \alpha^3 \left(v + \frac{1}{2} \right) \sqrt{\pi} 2^v v!$ (17)
  $\displaystyle =$ $\displaystyle \boxed{\alpha^2 \left(v + \frac{1}{2} \right)}$ (18)

Now that you know how to correctly apply the Hermite polynomial recursion relations multiple times, calculating $ \left<x^3\right>$ and $ \left<x^4\right>$ should be simple.

About this document ...


Atkin's 8th Edition Self-test 9.7

This document was generated using the LaTeX2HTML translator Version 2002-2-1 (1.70)

Copyright © 1993, 1994, 1995, 1996, Nikos Drakos, Computer Based Learning Unit, University of Leeds.
Copyright © 1997, 1998, 1999, Ross Moore, Mathematics Department, Macquarie University, Sydney.

The command line arguments were:
latex2html -local_icons -split 0 -no_navigation 9_7.tex

The translation was initiated by Kevin Range on 2007-02-08

Kevin Range 2007-02-08