= Solution
For any nonzero constant $C$, differentiating $\psi_0=Ce^{-x^2/2}$ gives $\psi_0''=(x^2-1)\psi_0$. Hence
$$
\boxed{(-d^2/dx^2+x^2)\psi_0=\psi_0,\qquad\lambda_0=1.}
$$
It is indeed the lowest <eigenvalue>, since the factorization $H=(-d/dx+x)(d/dx+x)+1$ makes $\langle\psi,H\psi\rangle\geq\|\psi\|^2$, with equality for this Gaussian. The variational bound exceeds the exact value by $\sqrt{10/7}-1\simeq0.195229$, or about \b[$19.52\%$].
The PDF's prefactor $1/\sqrt{2\pi}$ does not normalize this <wavefunction>: its squared norm is $1/(2\sqrt\pi)$. The normalized <ground state> has prefactor $\pi^{-1/4}$. This normalization error does not affect the eigenvalue, because the equation is homogeneous in the wavefunction.
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