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

Set $r_0=q\mathcal{E}/(m{\omega }^2)$ and $y=r-r_0$. Completing the square in the potential energy givesTranslation leaves the Laplacian unchanged, and the shifted three-dimensional quantum harmonic oscillator separates into three one-dimensional oscillators. Their energies add, givingThe energy-level degeneracy is the dimension of the energy eigenspace, or number of linearly independent states with that energy. Counting nonnegative integer triples of total $N$ by stars and bars givesThe constant electric field displaces the oscillator and shifts all levels by the same amount, preserving their degeneracies.

Put $\gamma =m\omega /\hslash$. Substitution of $\psi =F(y)e^{-\gamma y^2/2}$ in the one-dimensional Schrödinger equation with energy $(n+1/2)\hslash \omega$ givesFor degrees zero, one and two, coefficient comparison yields, up to independent nonzero constants,For the shifted coordinates $y_i=x_i-(r_0{)}_i$, let $G(y)=e^{-\gamma |y{|}^2/2}$. A basis of wave functions for the first three energy levels is:

These are products of the one-dimensional polynomial solutions. Arbitrary linear combinations within each listed eigenspace are also stationary states. Their counts $1,3,6$ agree with part (a), so the degeneracies have all been included.

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.