0
$\begingroup$

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)?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ +1 for the parity argument, which is the sort of things I'm looking for. If you make the appropriate change of variable, you can transform the Hamiltonian into that of another SHO, with an energy shift precisely equal to the second order pertubation you would have computed in the first Hamiltonian. In that sense, the exact solution is given by the second-order perturbation (and contributions from all other orders vanish). $\endgroup$ – Demosthene Mar 19 '17 at 15:19
  • $\begingroup$ @Demosthene of course! How silly of me! My original answer had a long discussion of the shifted oscillator but it didn't seem relevant. I see your point now. $\endgroup$ – ZeroTheHero Mar 19 '17 at 15:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.