= Solution
Let $c=\sqrt{m\omega/(2\hbar)}$ and $d=1/\sqrt{2m\hbar\omega}$, so $a=cx+idp$ and $a^\dagger=cx-idp$. The <canonical commutation relation> $[x,p]=i\hbar$ gives
$$
a^\dagger a=c^2x^2+d^2p^2+icd[x,p]=\frac{m\omega x^2}{2\hbar}+\frac{p^2}{2m\hbar\omega}-\frac12.
$$
Hence $\boxed{H=\hbar\omega(a^\dagger a+1/2)}$. Similarly,
$$
[a,a^\dagger]=-2icd[x,p]=\boxed{1}.
$$
Using the product rule for a <commutator>, $[AB,C]=A[B,C]+[A,C]B$, gives
$$
\boxed{[H,a]=-\hbar\omega a,\qquad [H,a^\dagger]=\hbar\omega a^\dagger.}
$$
Thus $a^\dagger a$ is the <number operator>, while $a$ and $a^\dagger$ are the <annihilation operator> and <creation operator>.
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