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.


2 Answers 2


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.

  • $\begingroup$ Or just take $\dagger$ in your last equation :). $\endgroup$ May 26, 2020 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 $[\hat{H},\hat{a}^{\dagger}]$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.