= Solution
On a suitable dense domain of smooth rapidly decreasing <wave functions>, the canonical <commutator> is $[x,p]=i\hbar$. Expanding the two <ladder operators> gives
$$
[a,a^\dagger]=\frac12\left[\beta x+\frac{ip}{\beta\hbar},\beta x-\frac{ip}{\beta\hbar}\right]
=\frac12(1+1)=1.
$$
Their sum and difference give
$$
\boxed{x=\frac{a+a^\dagger}{\sqrt2\beta},\qquad
p=\frac{\beta\hbar}{i\sqrt2}(a-a^\dagger).}
$$
Direct multiplication, retaining the operator order, gives
$$
a^\dagger a=\frac12\left[\beta^2x^2+\frac{p^2}{\beta^2\hbar^2}+\frac{i[x,p]}\hbar\right]
=\frac12\left[\beta^2x^2+\frac{p^2}{\beta^2\hbar^2}-1\right].
$$
Using $\beta^2=m\omega/\hbar$ identifies the <Hamiltonian operator>:
$$
\boxed{H=\hbar\omega(a^\dagger a+\tfrac12).}
$$
For any admissible <wave function>, adjointness gives
$$
\langle\Psi,H\Psi\rangle=\hbar\omega\left(\|a\Psi\|^2+\tfrac12\|\Psi\|^2\right)
\geq\tfrac12\hbar\omega\|\Psi\|^2.
$$
For the conventional normalized state $\|\Psi\|=1$, this is the stated lower bound $E\geq\hbar\omega/2$. For an unnormalized <function> the <norm> factor must be retained; rescaling a <wave function> cannot leave its unnormalized <energy> <integral> bounded below by a fixed positive number.
Equality holds exactly when $a\Psi_0=0$. In position representation this equation is $(\beta x+\beta^{-1}\partial_x)\Psi_0=0$, whose normalized square-integrable solution is
$$
\boxed{\Psi_0(x)=\left(\frac{\beta^2}{\pi}\right)^{1/4}e^{-\beta^2x^2/2},\qquad E_0=\tfrac12\hbar\omega.}
$$
This construction establishes existence of a state attaining the bound, and hence proves that it is the <ground state>.
The <commutator> $[a,a^\dagger]=1$ implies $[H,a^\dagger]=\hbar\omega a^\dagger$. Repeatedly commuting through the product gives
$$
H(a^\dagger)^n\Psi_0=(a^\dagger)^nH\Psi_0+n\hbar\omega(a^\dagger)^n\Psi_0
=\hbar\omega(n+\tfrac12)(a^\dagger)^n\Psi_0.
$$
These <vectors> are nonzero: the identity $a(a^\dagger)^n\Psi_0=n(a^\dagger)^{n-1}\Psi_0$ gives their <norm> squared $n!$ by induction. Therefore
$$
\boxed{\Psi_n=\frac{(a^\dagger)^n}{\sqrt{n!}}\Psi_0,\qquad E_n=(n+\tfrac12)\hbar\omega.}
$$
For time-dependent stationary states multiply each spatial eigenfunction by $e^{-iE_nt/\hbar}$.
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