# Quantum harmonic oscillator inner product

1. Jul 7, 2015

### Dazed&Confused

1. The problem statement, all variables and given/known data
Using the equations that are defined in the 'relevant equations' box, show that

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

2. Relevant 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 $$b = \left ( \frac{\hbar}{m \omega} \right)^{1/2}.$$

3. 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: Jul 7, 2015
2. Jul 7, 2015

### Dazed&Confused

Ok if I write it like this $$b^2 A_n A_{n'}( \delta_{n',n+1} [\pi^{1/2} 2^n (n+1)!] + \delta_{n',n-1} [\pi^{1/2}2^{n-1}(n-1)!]).$$
The product $A_nA_{n'}$ is $$\frac{1}{b \pi^{1/2}} \frac{1}{ 2^{n'/2} 2^{n/2} (n!)^{1/2} (n'!)^{1/2 } }.$$

If you replace $n'$ with $n+1$ in the first term and $n'$ with $n-1$ in the second, you get

$$\frac{b}{\sqrt{2}} [ \delta_{n',n+1} (n+1)^{1/2} + \delta_{n',n-1} n^{1/2}],$$ which is the correct answer. This makes sense to me, but is it correct?

Last edited: Jul 7, 2015