= Solution
From $H=a_+a_-/(2m)+\hbar\omega/2$ and the <commutator product rule>,
$$
[H,a_+]=\frac1{2m}a_+[a_-,a_+]=\hbar\omega a_+,\qquad [H,a_-]=\frac1{2m}[a_+,a_-]a_-=-\hbar\omega a_-.
$$
Consequently, if $H\psi=E\psi$,
$$
\boxed{H(a_\pm\psi)=(E\pm\hbar\omega)(a_\pm\psi).}
$$
Thus the <creation operator> raises the <energy eigenvalue> by $\hbar\omega$, and the <annihilation operator> lowers it by $\hbar\omega$, \b[provided the resulting vector is nonzero]. The zero vector is not an <eigenstate>; in particular the <ground state> will be annihilated by $a_-$. The displayed operator identity holds even in that case.
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