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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.

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

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

$\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.

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