 Solution

For any nonzero constant $C$, differentiating ${\psi }_0=C{e}^{-x^2/2}$ gives ${\psi }_0^{\prime \prime }=(x^2-1){\psi }_0$. HenceIt is indeed the lowest eigenvalue, since the factorization $H=(-d/dx+x)(d/dx+x)+1$ makes $⟨\psi ,H\psi ⟩\ge \Vert \psi {\Vert }^2$, with equality for this Gaussian. The variational bound exceeds the exact value by $\sqrt{10/7}-1\simeq 0.195229$, or about $19.52\%$.

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.