= Solution
The <Gaussian integral> gives $\int_{\mathbb R^3}e^{-\gamma|y|^2}\,d^3y=(\pi/\gamma)^{3/2}$. Thus the normalized <ground state> is, up to an overall phase,
$$
\boxed{\psi_0(r)=\left(\frac{m\omega}{\pi\hbar}\right)^{3/4}
\exp\left[-\frac{m\omega}{2\hbar}\left|r-\frac{q\mathcal E}{m\omega^2}\right|^2\right].}
$$
Its <probability> density is even in each shifted coordinate, so all three expectations $\langle y_i\rangle$ vanish by oddness. Therefore
$$
\boxed{\langle r\rangle=\frac{q\mathcal E}{m\omega^2}.}
$$
The mean position is the minimum of the displaced potential, rather than the original coordinate origin.
Back to article page
