= Solution
For the <quantum harmonic oscillator>, put $\ell_0=\sqrt{\hbar/(m\omega)}$ and $\xi=x/\ell_0$. The normalized <energy eigenstates> are
$$
\boxed{E_n=\hbar\omega(n+\tfrac12)},\qquad
\psi_n(x)=\frac{H_n(\xi)e^{-\xi^2/2}}{\pi^{1/4}\sqrt{2^nn!\ell_0}},\qquad n=0,1,\ldots,
$$
where $H_n$ is a <Hermite polynomial>. Successive <energies> are equally spaced; the <ground state> retains <zero-point energy> $\hbar\omega/2$. The $n$th <wavefunction> has $n$ <nodes of a wavefunction> and <parity> $(-1)^n$. In particular, the first three shapes are proportional to $e^{-\xi^2/2}$, $\xi e^{-\xi^2/2}$, and $(2\xi^2-1)e^{-\xi^2/2}$. The <classical turning points> satisfy $x=\pm\ell_0\sqrt{2n+1}$; the <wavefunctions> extend beyond them and decay, rather than ending abruptly.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2015/paper-1-oscillator.png]
{title=Quantum harmonic oscillator energies, signed wavefunctions, nodes and classical turning points}
The <wavefunctions> are plotted with arbitrary vertical scale about their <energy> levels. Their signed excursions are amplitudes, not additional <energies> or <probability> densities.
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