= Solution
The <commutators> imply
$$
H(a|n\rangle)=(E_n-\hbar\omega)a|n\rangle,\qquad H(a^\dagger|n\rangle)=(E_n+\hbar\omega)a^\dagger|n\rangle.
$$
Thus any nonzero resulting vector is the adjacent <energy eigenstate>. Since $a^\dagger a|n\rangle=n|n\rangle$ and $aa^\dagger=a^\dagger a+1$, their squared norms are
$$
\|a|n\rangle\|^2=n,\qquad \|a^\dagger|n\rangle\|^2=n+1.
$$
Choosing consistent phases for the <eigenstates> yields
$$
\boxed{a|n\rangle=\sqrt n\,|n-1\rangle\quad(n\geq1),\qquad a^\dagger|n\rangle=\sqrt{n+1}\,|n+1\rangle,\qquad a|0\rangle=0.}
$$
The vanishing last expression terminates the lowering sequence at the <ground state>, rather than introducing a negative-energy oscillator state.
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