 Solution

Direct differentiation gives ${\psi }^{\prime \prime }/\psi =x^2/a^4-1/a^2$. Inserting this into the Schrödinger equation and separately matching the $x^2$ and constant terms yieldsFor $V_0=m{\omega }_0^2/2$, $a^2=\hslash /(m{\omega }_0)$ and $E=\hslash {\omega }_0/2$. The normalizable Gaussian has no nodes, as expected for the nondegenerate one-dimensional ground state of this confining potential.

The normalization integral is $|C{|}^{2}a\sqrt{\pi }=1$. Choose its irrelevant global phase so thatFor the oscillator ground-state density in a comparison wavelength, choose a free comparison particle with the same total energy. It has $p=\sqrt{2mE}=\hslash /a$, hence comparison wavelength $\lambda =h/p=2\pi a$. ConsequentlyThe bound density is localized on scale $a=\lambda /(2\pi )$, with standard deviation $\lambda /(2\pi \sqrt{2})$, rather than being spatially uniform like the density of a single free plane wave. The comparison does not assign constant free-particle momentum to the bound state; its total energy also includes potential energy.