Solution
= Solution
For the Gaussian, $\phi_0''=(x^2/\sigma^4-1/\sigma^2)\phi_0$. Using $\hbar/(m\omega)=\sigma^2$, the $x^2$ terms cancel and leave $E_0=\hbar\omega/2$. For $\phi_1\propto x\phi_0$, <differentiation> gives $\phi_1''=(x^2/\sigma^4-3/\sigma^2)\phi_1$, and the same cancellation gives $\boxed{E_1=3\hbar\omega/2}$. The <Normalization constant>[normalization constants] do not affect these <Eigenvalue equation>[eigenvalue equations].