# Momentum in quantum harmonic oscillator with step up and step down operators [closed]

I'm hitting a wall in my understanding of the momentum operator in a quantum harmonic oscillator. I've showed that $p = (a^\dagger - a)\sqrt{\frac{m w \hbar}{2}}i$ where $a^\dagger$ and $a$ are the step up and step down operators, and $i$ is the imaginary number.

I'm trying to calculate $<m|p|n>$ using this momentum operator, where |m> and |n> are just different states of the system.

My work so far is:

\begin{align} <m|p|n> &= <m|(a^\dagger - a)\sqrt{\frac{m w \hbar}{2}}i|n> \\ &=<m|a^\dagger\sqrt{\frac{m w \hbar}{2}}i|n> - <m|a\sqrt{\frac{m w \hbar}{2}}i|n> \\ &=\sqrt{\frac{m w \hbar}{2}}i (<m|a^\dagger|n> - <m|a|n>) \end{align}

And this is where I'm stuck. I'm not sure what I can do from here.

## closed as off-topic by ACuriousMind♦, Kyle Kanos, user36790, Gert, Sebastian RieseNov 25 '15 at 20:23

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• What happens if $m = n \pm 1$? What happens otherwise? – By Symmetry Nov 23 '15 at 3:57
• I believe if $m = n \pm 1$, then the inner product is 1 and 0 otherwise since they are othonormal – TheStrangeQuark Nov 23 '15 at 4:00

$\bf{a^{\dagger}}$$|n>=\sqrt{n+1}|n+1> \bf{a}$$|n>=\sqrt{n}|n-1>$
Taking into account $<n|m>=\delta_{n,m}$ , you get the answer.