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

Question

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.

Answer 1

The commutators in the above expressions are sued to change the order of the Hamiltonian and annihilation or creation operators. I'll show you the first one in some detail, the second one should not give you problems afterwards.

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 $W_n-\hbar\omega$.