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I know how to derive below equations found on wikipedia and have done it myselt:
\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.
\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.