 Solution

For the quantum harmonic oscillator, put ${\ell }_0=\sqrt{\hslash /(m\omega )}$ and $\xi =x/{\ell }_0$. The normalized energy eigenstates arewhere $H_n$ is a Hermite polynomial. Successive energies are equally spaced; the ground state retains zero-point energy $\hslash \omega /2$. 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.

The wavefunctions are plotted with arbitrary vertical scale about their energy levels. Their signed excursions are amplitudes, not additional energies or probability densities.