= Solution
Part (a) implies $[\hat A,\hat B]=2m\hbar\omega$. Since $\hat H=\hat B\hat A/(2m)+\hbar\omega/2$, it follows that
$$
[\hat H,\hat A]=\frac1{2m}[\hat B\hat A,\hat A]=-\hbar\omega\hat A.
$$
For an <eigenfunction> $\phi$ of the <Hamiltonian> with <eigenvalue> $E$,
$$
\boxed{\hat H(\hat A\phi)=(E-\hbar\omega)\hat A\phi.}
$$
Thus $\hat A$ is a <ladder operator> lowering the energy by $\hbar\omega$, \b[provided $\hat A\phi\ne0$]. 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.
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