I know how to derive below equations found on wikipedia and have done it myselt:(adsbygoogle = window.adsbygoogle || []).push({});

\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.

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# Proof for commutator $[\hat{H},\hat{a}] = - \hbar \omega \hat{a}

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