 Solution

The time-independent Schrödinger equation isPut $\alpha =m\omega /\hslash$. For $\psi =e^{-\alpha x^2/2}$, differentiation gives ${\psi }^{\prime \prime }=({\alpha }^2{x}^2-\alpha )\psi$. Substitution cancels the terms in $x^2$ and leaves $E_0={\hslash }^2\alpha /(2m)=\hslash \omega /2$.

The Gaussian integral gives $\int e^{-\alpha x^2}dx=\sqrt{\pi /\alpha }$. Hence a normalized wave function isMultiplication by any constant of modulus one gives the same quantum state. The position probability density is $|{\psi }_0{|}^2=\sqrt{\alpha /\pi }{e}^{-\alpha x^2}$. Its first moment vanishes by oddness. Integration by parts gives $\int x^2{e}^{-\alpha x^2}dx=(2\alpha {)}^{-1}\int e^{-\alpha x^2}dx$, with the boundary term zero. Therefore