= Solution
For the <quantum harmonic oscillator>, the <time-independent Schrödinger equation> is
$$
-\frac{\hbar^2}{2m}\psi''+\frac12m\omega^2x^2\psi=E\psi,
$$
with $m,\omega,\hbar>0$. The given dimensionless coordinate yields $\psi_{\xi\xi}+(2E/(\hbar\omega)-\xi^2)\psi=0$. Substituting the Gaussian factor gives
$$
f''-2\xi f'+\left(\frac{2E}{\hbar\omega}-1\right)f=0.
$$
By the supplied normalizability fact, $f$ is a nonzero polynomial. If its degree is $n$, comparison of the leading coefficient forces $2E/(\hbar\omega)-1=2n$. Conversely, at this value the coefficient recurrence
$$
a_{j+2}=\frac{2j-2n}{(j+2)(j+1)}a_j
$$
produces a degree-$n$ polynomial with parity $n$: start with a nonzero constant for even $n$ or a nonzero linear coefficient for odd $n$, and the recurrence terminates at degree $n$. This is a multiple of the <Hermite polynomial>. Its product with the Gaussian is square integrable and can be normalized. Therefore all and only the allowed energies are
$$
\boxed{E_n=\left(n+\frac12\right)\hbar\omega,\qquad n=0,1,2,\ldots.}
$$
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