 Solution

Part (a) implies $[\hat{A},\hat{B}]=2m\hslash \omega$. Since $\hat{H}=\hat{B}\hat{A}/(2m)+\hslash \omega /2$, it follows thatFor an eigenfunction $\varphi$ of the Hamiltonian with eigenvalue $E$,Thus $\hat{A}$ is a ladder operator lowering the energy by $\hslash \omega$, provided $\hat{A}\varphi \ne 0$. The zero function is not an eigenfunction; in particular the ground state is annihilated by this operator. This qualification is essential for the later argument.