= Solution
Differentiate the Gaussian before substituting into the <Hamiltonian (quantum mechanics)>:
$$
\Psi_0'=-\alpha x\Psi_0,\qquad
\Psi_0''=(\alpha^2x^2-\alpha)\Psi_0.
$$
Thus
$$
\hat H\Psi_0=\left[\frac{\hbar^2\alpha}{2m}
+\left(\frac{m\omega^2}2-\frac{\hbar^2\alpha^2}{2m}\right)x^2\right]\Psi_0.
$$
An <energy eigenvalue> must be independent of $x$, so the coefficient of $x^2$ vanishes. Normalizability requires $\alpha>0$, giving
$$
\boxed{\alpha=\frac{m\omega}{\hbar},\qquad E_0=\frac12\hbar\omega.}
$$
The constant term is the <quantum harmonic oscillator>'s zero-point energy; a Gaussian width $\alpha^{-1/2}$ balances its kinetic and potential contributions.
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