How do I simplify <n|n> for Hermetian Operators?

[Biest]

Hi,

So for an harmonic oscillator we need to to find the average value for $$x^4$$, so $$<n|x^4|n>$$. We split it up to $$\sum_m |<n|x^2|n>|^2$$ and recognize that only m = n+2, m=n and m = n-2 can be used. We find that

m=n

$$\frac{\hbar}{2m\omega}<n|\hat{A}\hat{A^\dagger}|n>$$

m= n+2 $$\frac{\hbar}{2m\omega}<n+2|\hat{A^\dagger}\hat{A^\dagger}|n>$$

m = n-2

$$\frac{\hbar}{2m\omega}<n-2|\hat{A}\hat{A}|n>$$

So we can reduce it all to

$$<n|x^4|n> = \frac{1}{4} \hbar^2 \omega^2 (2n+1)^2 + \frac{1}{2} (\frac{\hbar}{m \omega})^2 <n|n>$$How I simplify the $$<n|n>$$.

Thanks.

Cheers,

Biest

 

[quantumdude]

The basis states are orthonormal, so $<n|m>=\delta_{mn}$.