A particle of mass m and charge e confined to motion in the x direction oscillates in a one-dimensional harmonic potential with angular frequency ω.
(a) Show, using perturbation theory, that the effect of an applied uniform electric field E in the x direction is to lower all the energy levels by e^{2}\left|E\right|^{2} /2m\omega ^{2}.
(b) Compare this with the classical result.
(c) Use perturbation theory to calculate the new ground-state wave function.
(a) The solution follows that already given in this chapter.
(b) The new energy levels of the oscillator are to second order
E=\hbar \omega \left(n+\frac{1}{2} \right)-\frac{e^{2}\left|E\right|^{2} }{2\kappa }which is the same as the exact result. Physically, the particle oscillates at the same frequency, ω, as the unperturbed case, but it is displaced a distance of e\left|E\right|/m\omega ^{2}, and the new energy levels are shifted by -e ^{2}\left|E\right|^{2}/ 2m\omega ^{2}.
(c) The ground-state wave function of the unperturbed harmonic oscillator is
\psi _{0}(x)=\left(\frac{m\omega }{\pi \hbar } \right)^{1/4} e^{-x^{2}m\omega /2\hbar }After the perturbation, it is
\psi _{0}(x)=\left(\frac{m\omega }{\pi \hbar } \right)^{1/4} e^{-\frac{m\omega }{2\hbar }\left(x-\frac{e\left|E\right| }{m\omega^{2} } \right)^{2} }The result using second-order perturbation theory is