Proof for commutator relation $[\hat{H},\hat{a}] = - \hbar \omega \hat{a}$ I know how to derive below equations found on wikipedia and have done it myselt too: 
\begin{align}
\hat{H} &= \hbar \omega \left(\hat{a}^\dagger\hat{a} + \frac{1}{2}\right)\\
\hat{H} &= \hbar \omega \left(\hat{a}\hat{a}^\dagger - \frac{1}{2}\right)\\
\end{align} 
where $\hat{a}=\tfrac{1}{\sqrt{2}} \left(\hat{P} - i \hat{X}\right)$ is a annihilation operator and $\hat{a}^\dagger=\tfrac{1}{\sqrt{2}} \left(\hat{P} + i \hat{X}\right)$ a creation operator. Let me write also that:
\begin{align}
\hat{P}&= \frac{1}{p_0}\hat{p} = -\frac{i\hbar}{\sqrt{\hbar m \omega}} \frac{d}{dx}\\
\hat{X}&=\frac{1}{x_0} \hat{x}=\sqrt{\frac{m\omega}{\hbar}}x
\end{align}
In order to continue i need a proof that operators $\hat{a}$ and $\hat{a}^\dagger$ give a following commutator with hamiltonian $\hat{H}$: 
\begin{align}
\left[\hat{H},\hat{a} \right] &= -\hbar\omega \, \hat{a}\\
\left[\hat{H},\hat{a}^\dagger \right] &= +\hbar\omega \, \hat{a}^\dagger
\end{align}
These statements can be found on wikipedia as well as here, but nowhere it is proven that the above relations for commutator really hold. I tried to derive $\left[\hat{H},\hat{a} \right]$ and my result was: 
$$
\left[\hat{H},\hat{a} \right] \psi = -i \sqrt{\frac{\omega \hbar^3}{4m}}\psi
$$
You should know that this this is 3rd commutator that i have ever calculated so it probably is wrong, but here is a photo of my attempt on paper. I would appreciate if anyone has any link to a proof of the commutator relations (one will do) or could post a proof here.
 A: Start with your $\hat{H} = \hbar \omega \left( \hat{a}^\dagger\hat{a} + \frac{1}{2} \right)$. I will omit hat notation from this point. The commutator then reads as
\begin{equation}
\left[ H, a \right] = \hbar \omega \left[ \left( \hat{a}^\dagger\hat{a} + \frac{1}{2} \right) a - a \left( \hat{a}^\dagger\hat{a} + \frac{1}{2} \right) \right] = \hbar \omega \left( a^\dagger a a - a a^\dagger a \right) ,
\end{equation}
which is nothing but
\begin{equation}
\left[ H, a \right] =  \hbar \omega (a^\dagger a - a a^\dagger)a = \hbar \omega \left[ a^\dagger, a \right]a,
\end{equation}
but we know that
\begin{equation}
\left[a^\dagger, a \right] = -1 ,
\end{equation}
therefore
\begin{equation}
\left[ H, a \right] = -\hbar \omega a,
\end{equation}
QED.
Proof of the second relation is done in the same way.
A: On the Wikipedia page you link to there is a derivation of the commutation relation between $\hat{a}$ and $\hat{a}^{\dagger}$,
$$ [\hat{a},\hat{a}^{\dagger}] = 1.$$
This directly leads to (use the relation $[AB,C]=[A,C]B+A[B,C]$)
$$[\hat{a}^{\dagger}\hat{a},\hat{a}] = -\hat{a} , 
\qquad
[\hat{a}^{\dagger}\hat{a},\hat{a}^{\dagger}] = +\hat{a}^{\dagger}.$$
Up to a constant this is the same as $[\hat{H},\hat{a}]$ and $[\hat{H},\hat{a}^{\dagger}]$.
