# 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.

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.
• Or just take $\dagger$ in your last equation :). – user2820579 May 26 at 18:43
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 $[\hat{H},\hat{a}^{\dagger}]$.