= Solution
On a smooth <wavefunction> $f$, the <position operator> and <momentum operator> give
$$
[X,P]f=-i\hbar xf'+i\hbar(xf)'=i\hbar f.
$$
Thus the <canonical commutation relation> is $\boxed{[X,P]=i\hbar I}$. Since $X$ and $P$ are <self-adjoint operators> and $m,\omega$ are real, the <adjoint operator> of $a_-$ is
$$
\boxed{a_+=P+im\omega X.}
$$
Expanding the products in the stated order is essential:
$$
\begin{aligned}
a_+a_-&=P^2+m^2\omega^2X^2+im\omega[X,P]=2mH-m\hbar\omega,\\
a_-a_+&=P^2+m^2\omega^2X^2-im\omega[X,P]=2mH+m\hbar\omega.
\end{aligned}
$$
Subtracting gives $\boxed{[a_-,a_+]=2m\hbar\omega I}$. These identities initially hold on smooth rapidly decaying functions, which contain the <quantum harmonic oscillator> <energy eigenstates> used below.
The factorization also proves a lower bound: for a normalized <energy eigenstate> $H\psi=E\psi$,
$$
0\leq\|a_-\psi\|^2=\langle\psi,a_+a_-\psi\rangle=2m\left(E-\frac12\hbar\omega\right).
$$
Therefore $E\geq E_0:=\hbar\omega/2$. The <commutator product rule>, applied to $H=a_+a_-/(2m)+E_0$, gives
$$
[H,a_+]=\hbar\omega a_+,\qquad [H,a_-]=-\hbar\omega a_-.
$$
So a nonzero application of the <creation operator> $a_+$ raises the <energy eigenvalue> by $\hbar\omega$, while the <annihilation operator> $a_-$ lowers it by the same amount.
Begin with any <energy eigenstate>, whose existence is assumed. Repeated lowering must terminate: otherwise it would eventually produce a nonzero <energy eigenstate> below $E_0$, contradicting the lower bound. More explicitly, if $(E-E_0)/(\hbar\omega)=N+r$ with integer $N\geq0$ and $0\leq r<1$, the first $N$ lowering steps are nonzero. If $r>0$, one further step is also nonzero by its positive squared <norm> and has energy $E_0+(r-1)\hbar\omega<E_0$, an impossibility. Thus $r=0$ and the last state satisfies $a_-\psi_0=0$. This proves
$$
\boxed{E_n=\left(n+\frac12\right)\hbar\omega,\qquad n=0,1,2,\ldots.}
$$
Every level exists as well: $\|a_+\psi_n\|^2=2m(E_n+\hbar\omega/2)=2m\hbar\omega(n+1)>0$, so repeated raising from $\psi_0$ never vanishes. The same factorization bounds the <energy expectation> of any normalized state in the <one-dimensional harmonic oscillator form domain> below by $E_0$, proving that the terminated state really is a <ground state>.
To find its <probability density>, solve the first-order equation
$$
(P-im\omega X)\psi_0=-i\left(\hbar\psi_0'+m\omega x\psi_0\right)=0.
$$
Its solutions are $\psi_0=C_0\exp[-m\omega x^2/(2\hbar)]$. With $\alpha=m\omega/\hbar>0$, the supplied <Gaussian integral> gives
$$
1=|C_0|^2\int_{-\infty}^{\infty}e^{-\alpha x^2}\,dx=|C_0|^2\sqrt{\frac\pi\alpha}.
$$
Taking $C_0=(\alpha/\pi)^{1/4}$ fixes an irrelevant <global phase>. Consequently
$$
\boxed{|\psi_0(x)|^2=\sqrt{\frac{m\omega}{\pi\hbar}}\exp\left(-\frac{m\omega x^2}{\hbar}\right).}
$$
The <ground-state energy> is positive even though the potential minimum is zero: it is the oscillator's <zero-point energy>.
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