- #1

Dazed&Confused

- 191

- 3

## Homework Statement

Using the equations that are defined in the 'relevant equations' box, show that

$$\langle n' | X | n \rangle = \left ( \frac{\hbar}{2m \omega} \right )^{1/2} [ \delta_{n', n+1} (n+1)^{1/2} + \delta_{n',n-1}n^{1/2}]$$

## Homework Equations

$$\psi_n(x) = \left ( \frac{m \omega}{\pi \hbar 2^{2n} (n!)^2} \right )^{1/4} \text{exp} \left ( \frac{-m \omega x^2}{2 \hbar} \right )H_n \left [ \left ( \frac{m \omega}{\hbar} \right )^{1/2} x \right ] $$

where ##H_n## are the Hermite polynomials.

The questions asks you to use the useful relations:

$$ H_n^{'}(y) = 2nH_{n-1}$$ $$H_{n+1}(y) = 2yH_n -2nH_{n-1}.$$

I think that the following is also needed:

$$\int_{-\infty}^{\infty} H_n(y)H_{n'}(y) e^{-y^2} dy =\delta_{nn'}(\pi^{1/2}2^n n!).$$

Here ##yb = x## where $$b = \left ( \frac{\hbar}{m \omega} \right)^{1/2}.$$

## The Attempt at a Solution

Since $$| n \rangle = \int_{-\infty}^{\infty} | x' \rangle \langle x'| n \rangle dx'= \int_{-\infty}^{\infty} | x' \rangle \psi_n(x') dx',$$ $$ \langle n' | X | n \rangle = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \langle x | x' \rangle x' \psi_{n'}(x) \psi_n(x') dx' dx.$$

This becomes $$\int_{-\infty}^{\infty} x \psi_{n'}(x) \psi_{n} (x) dx. $$

Substituting into ##y##, we have $$b^2 A_n A_{n'} \int_{-\infty}^{\infty} ye^{-y^2} H_n(y) H_{n'}(y) dy. $$

Using the relations, this becomes $$b^2 A_n A_{n'} \int_{-\infty}^{\infty} \frac12 H_{n+1} H_{n'} e^{-y^2} + n H_{n-1} H_{n'}e^{-y^2} \ dy.$$

I'm unsure how to write these integrals. In particular, which number to choose as the ##n## corresponding to the useful integral.

Last edited: