 Solution

For a normalized wavefunction, the expectation value of a function of position isprovided the integral exists. For an unnormalized state, divide this expression by $\int |\psi {|}^{2}dx$. Put $b=m\omega /\hslash >0$. The ground state has probability density $\sqrt{b/\pi }{e}^{-b{x}^2}$, which is even. ThusThe zero mean position ${\mathbb{E}}_{{\psi }_0}(x)=0$ expresses symmetry, whereas this positive expectation value measures the mean distance from the centre.