# Energy eigenvalues of a Q.H.Oscillator with $[\hat{H},\hat{a}] = -\hbar \omega \hat{a}$ and $[\hat{H},\hat{a}^\dagger] = \hbar \omega \hat{a}^\dagger$

I just finished deriving the commutators:

\begin{align} [\hat{H}, \hat{a}] &= -\hbar \omega \hat{a}\\ [\hat{H}, \hat{a}^\dagger] &= \hbar \omega \hat{a}^\dagger\\ \end{align}

On the Wikipedia it is said that these commutators can be used to find energy eigenstates of Quant. harm. oscillator, but explanation is a bit too fast there. Anyway i strive to be able to derive the equation $W_n = \hbar \omega \left(n + \tfrac{1}{2}\right)$ in full, but first i need to clarify why theese two relations hold:

\begin{align} \hat{H}\hat{a} \psi_n &= (W_n - \hbar \omega) \hat{a} \psi_n\\ \hat{H}\hat{a}^\dagger \psi_n &= (W_n + \hbar \omega) \hat{a}^\dagger \psi_n \end{align}

I can't see any commutators in above relations, so how do the commutators i just calculated help us to get and solve these two relations?

I am sorry for asking such a basic questions. I am a self-taught and a real freshman to commutators algebra.

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We start from $\hat{H}\hat{a}\psi_n$. Using the commutator $[\hat{H},\hat{a}] = \hat{H}\hat{a}-\hat{a}\hat{H} = -\hbar\omega\hat{a}$, we can write $\hat{H}\hat{a}\psi_n = (\hat{a}\hat{H}-\hbar\omega\hat{a})\psi_n$. Because we have $\hat{H}\psi_n = W_n\psi_n$, we get $(\hat{a}\hat{H}-\hbar\omega\hat{a})\psi_n = (\hat{a}W_n-\hbar\omega\hat{a})\psi_n = (W_n-\hbar\omega)\hat{a}\psi_n$ (note that we can change the order of the annihilation operator and c-numbers $W_n$ and $\hbar\omega$). Therefore, we have $\hat{H}\hat{a}\psi_n = (W_n-\hbar\omega)\hat{a}\psi_n$ and we conclude that $\hat{a}\psi_n$ is an eigenstate of the Hamiltonian with eigenvalue $W_n-\hbar\omega$.
I don't quite understand this statement: ˝Because we have $\hat{H}\psi_n = W_n\psi_n$ we get $(W_n-\hbar\omega)\hat{a}\psi_n$˝. –  71GA May 7 '13 at 8:44
Ok i understand now that you swapped opperator $\hat{H}$ with its eigenvalue $W_n$. All i know from this is that i get an expectation value for energy $\langle W \rangle$ like this: $$\langle W \rangle = \int \limits_{-\infty}^{\infty} \overline{\Psi}\, \left(- \frac{\hbar^2}{2m} \frac{d^2}{d \, x^2} + W_p\right) \Psi \, d x$$ and $$\hat{H} = - \frac{\hbar^2}{2m} \frac{d^2}{d \, x^2} + W_p$$ But i am weak on eigenvalues, eigenvectors and Hilbert space... So how could i connect what i already know to confirm that $\hat{H} = W_n$? –  71GA May 7 '13 at 9:26
You can't say $\hat{H} = W_n$ but you can say $\hat{H}\psi_n = W_n\psi_n$ if $\psi_n$ is an eigenstate of $\hat{H}$ with eigenvalue $W_n$. And that is your starting point, together with the commutator, to find what $\hat{H}\hat{a}\psi_n$ is. –  Ondřej Černotík May 7 '13 at 9:29