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For a one-dimensional quantum harmonic oscillator, let $b=\sqrt{\hslash /(m\omega )}$ and $\xi =x/b$. The stationary Schrödinger equation gives the displayed curvature relation. Inflections occur at classical turning points $\xi =\pm \sqrt{2n+1}$ and at simple wavefunction wavefunction nodes, provided the curvature changes sign. Thus the first excited state has an inflection at its central wavefunction node as well as at both turning points. Restricting attention to where $V=E$ misses wavefunction node inflections.

For the nonrelativistic point-Coulomb hydrogen Hamiltonian, the lowest radial state at fixed orbital angular momentum $\ell$ has principal quantum number $n=\ell +1$ and no positive-radius wavefunction nodes. Substituting the displayed radial shape cancels the centrifugal inverse-square term and fixes $a_0=4\pi {ϵ}_0{\hslash }^2/(m{e}^2)$. With $b=a_0(\ell +1)$ and unit-normalized spherical harmonic, $A=(2/b{)}^{\ell +3/2}/\sqrt{(2\ell +2)!}$. Excited radial states require additional polynomial factors.

An operational-amplifier configuration applying the signal to the non-inverting input and returning a fraction of output to the inverting input. In stable unsaturated ideal operation, zero input current and equal input voltages give the displayed voltage gain. The feedback resistor $R_f$ connects output to the inverting input and $R_g$ connects that input to the reference wavefunction node.

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.

Let $V_B$ be the non-inverting input voltage. The feedback divider gives $V_B=V_{\mathrm{out}}/K$. With $s=i\omega$, current conservation at this wavefunction node gives $(V_A-V_B)/R_2=s{C}_2{V}_B$, because the amplifier input draws no current. Hence

A point or surface at which a wavefunction vanishes. In one dimension a simple node of a real stationary wavefunction changes its sign. A node is not generally the same as a classical turning point; for a smooth finite potential, the stationary Schrödinger equation also makes the second derivative vanish at a wavefunction node.