Commutators of $A$, $A^\dagger$ and $H$

We will use the commutator between $A$ and $A^\dagger$ to solve the HO problem. The operators are defined to be

$$\begin{eqnarray*} A &=&\left(\sqrt{m\omega\over2\hbar}x+i{p\over\sqrt{2m\hbar\om... ...t{m\omega\over2\hbar}x-i{p\over\sqrt{2m\hbar\omega}}\right) .\\ \end{eqnarray*}$$

The commutator is

$$\begin{eqnarray*}[A,A^\dagger]&=&{m\omega\over2\hbar}[x,x]+{1\over 2m\hbar\omega... ...\\ &=&{i\over 2\hbar}(-[x,p]+[p,x])={i\over \hbar}[p,x]=1 .\\ \end{eqnarray*}$$

Lets use this simple commutator

$$\bgroup\color{black} [A,A^\dagger]=1 \egroup$$

to compute commutators with the Hamiltonian. This is easy if $H$ is written in terms of $A$ and $A^\dagger$.

$$\begin{eqnarray*}[H,A]&=&\hbar\omega[A^\dagger A,A]=\hbar\omega[A^\dagger,A]A=-\... ...er]=\hbar\omega A^\dagger[A,A^\dagger]=\hbar\omega A^\dagger \\ \end{eqnarray*}$$