I just calculated the commutators for the number operator and the lowering/raising operators of the harmonic oscillator and they are non zero. Is there any significance to that? I got $[\hat{N},\hat{a}]=-\hat{a}$ and $[\hat{N},\hat{a^\dagger}]=\hat{a}^\dagger$.
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Take the case of $[\hat{N},\hat{a^\dagger}]=\hat{a}^\dagger$. Operate it on an eigen state of $\hat{N}$, ie $\hat{N}|n>=n|n>$.
So
$(\hat{N}\hat{a^\dagger}-\hat{a^\dagger}\hat{N})|n>=\hat{a^\dagger}|n>$
$\hat{N}\hat{a^\dagger}|n>=\hat{a^\dagger}|n>+\hat{a^\dagger}\hat{N}|n>=\hat{a^\dagger}|n>+\hat{a^\dagger}n|n>$
$\hat{N}\hat{a^\dagger}|n>=(n+1)\hat{a^\dagger}|n>$
Which means that $\hat{a^\dagger}|n>$ is again an eigen state of $\hat{N}$ but now with an eigen value increased by one. So $\hat{a^\dagger}$ acting on $|n>$ takes it to a state of higher eigen value with respect to $\hat{N}$ say $|n+1>$.
Similarly one can show that $\hat{a}$ maps $|n>$ to a state of lower eigen value. That is $\hat{N}\hat{a}|n>=(n-1)\hat{a}|n>$
