Past exam of the options II course of the University of Oxford / 2013 / J2 / c / Solution

The Gaussian integral gives ${\int }_{{\mathbb{R}}^3}{e}^{-\gamma |y{|}^2}{d}^{3}y=(\pi /\gamma {)}^{3/2}$. Thus the normalized ground state is, up to an overall phase,Its probability density is even in each shifted coordinate, so all three expectations $⟨y_i⟩$ vanish by oddness. ThereforeThe mean position is the minimum of the displaced potential, rather than the original coordinate origin.