Consider a charged particle placed in a simple one-dimensional quantum harmonic oscillator. If we turn on an electric field $\mathbf{F}=F\mathbf{x}$, it will experience a pertubation $V_1(x)=-qFx$.
Without doing the actual computations (standard material for a course on QM), can we argue that the first-order perturbation to the energy levels is zero, and that the second-order perturbation is exact?
Furthermore, can we generalize these arguments so as to encompass a wider class of potentials? or are they restricted to the QHO only (like the property of having an evenly spaced eigenspectrum)?
Best Answer
Yes. The first order correction to the energy is the average value of the perturbation in the unperturbed states. For any odd power of $x$ this average is $0$ by parity.
I don't know what you mean by "second order perturbation is exact". It will contain terms of the type $\vert \langle n+1 \vert \hat x \vert n\rangle \vert^2 $ and $\vert \langle n-1 \vert \hat x \vert n\rangle\vert^2$ which is quadratic in $x$ but (unless I'm mistaken) the fourth order correction will contain terms quadratic in $x$ which can have non-trivial overlaps.