= Solution
Direct differentiation gives $\psi''/\psi=x^2/a^4-1/a^2$. Inserting this into the <Schrödinger equation> and separately matching the $x^2$ and constant terms yields
$$
\boxed{a^4=\frac{\hbar^2}{2mV_0},\qquad a^2=\frac{\hbar}{\sqrt{2mV_0}},\qquad E=\frac{\hbar^2}{2ma^2}.}
$$
For $V_0=m\omega_0^2/2$, \b[$a^2=\hbar/(m\omega_0)$ and $E=\hbar\omega_0/2$]. The <normalizable> <Gaussian> has no <nodes>, as expected for the nondegenerate one-dimensional <ground state> of this confining potential.
The normalization integral is $|C|^2a\sqrt\pi=1$. Choose its irrelevant <global phase> so that
$$
\boxed{C=(\pi a^2)^{-1/4},\qquad dP=\frac1{\sqrt\pi a}e^{-x^2/a^2}\,dx.}
$$
For the <oscillator ground-state density in a comparison wavelength>, choose a free comparison particle with the same total <energy>. It has $p=\sqrt{2mE}=\hbar/a$, hence comparison <wavelength> $\lambda=h/p=2\pi a$. Consequently
$$
\boxed{dP=\frac{2\sqrt\pi}{\lambda}\exp\!\left(-\frac{4\pi^2x^2}{\lambda^2}\right)dx.}
$$
The bound density is localized on scale $a=\lambda/(2\pi)$, with <standard deviation> $\lambda/(2\pi\sqrt2)$, 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>.
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