 Solution

Let $\omega =\sqrt{k/m}$, $b=\sqrt{\hslash /(m\omega )}$ and $\xi =x/b$. With ${\psi }_0={\pi }^{-1/4}{b}^{-1/2}{e}^{-{\xi }^2/2}$, the next two normalized harmonic-oscillator wavefunctions are ${\psi }_1=\sqrt{2}\xi {\psi }_0$ and ${\psi }_2=(2{\xi }^2-1){\psi }_0/\sqrt{2}$. Their energies are $E_n=(n+1/2)\hslash \omega$, with respectively zero, one and two wavefunction nodes, alternating even and odd parity. The Schrödinger equation givesFor harmonic-oscillator wavefunction inflections, the inflections include both classical turning points and simple wavefunction wavefunction nodes. In dimensionless coordinates the complete sets areThe curvature changes sign at every listed point; none of these states has a wavefunction node coinciding with its turning points.