2
$\begingroup$

So I am looking at the problem of the (charged) harmonic oscillator in a weak electric field - the problem that defines e.g. the polarizibility of the oscillator.

Let the fieldless Hamiltonian be:

$$H=\frac{p^2}{2m}+\frac12m\omega^2x^2$$

so that the ground state wavefunction can be written as:

$$\psi_0=\left(\frac{m\omega}\pi\right)^\frac14\exp{\left(-\frac12m\omega x^2\right)}$$

Let the perturbation be $-qE_fx$. The exact solution of the perturbed problem can be found by rearranging the perturbed Hamiltionian:

$$ H=\frac{p^2}{2m}+\frac12m\omega^2x^2-qE_fx=\frac{p^2}{2m}+\frac12m\omega^2\left(x-\frac{qE_f}{m\omega^2}\right)^2-\frac{q^2E_f^2}{2m\omega^2} $$

This Hamiltionian has the ground state:

$$ \psi_0^E=\left(\frac{m\omega}\pi\right)^\frac14\exp{\left(-\frac12m\omega\left(x-\frac{qE_f}{m\omega^2}\right)^2\right)} $$

So far this is all clear. What I wanted to do is to express the wavefunction $\psi_0^E$ using the Rayleigh-Schroedinger perturbation theory, as in using the original energy eigenstates as a basis. For example, writing up the expression for the first order correction:

$$ \Psi_n^{(1)} = -\sum_{m \neq n} \frac{\langle \Psi_m^{(0)}| W | \Psi_n^{(0)} \rangle}{E_m^{0}-E_n^{0}} | \Psi_m^{(0)} \rangle ~~~~~$$

Now my problem is the following: the exact wavefunction of the perturbed problem is basically the same function as the unperturbed problem, but the origin is shifted by the field. With this perturbation expression all I can do is express a linear combination of functions that are solutions of the fieldless problems, so centered around the origin. There is no way for me to express a shift of the wavefunction using the original energy eigenstates as a basis, as none of them contain any shift whatsoever. So I conclude that Rayleigh-Schroedinger perturbation fails in this case.

My question is: how to state the failure in a more rigorous way? In which cases can the original eigenfunctions not be used as a basis set for expansion? Is it general for problems where the symmetry is broken by perturbation? Also: what approximate methods could I use to solve problems like this?

$\endgroup$
2
  • 1
    $\begingroup$ Try working out the first order correction (only one term in the summation survives), and compare to the first term in the Taylor expansion of your shifted eigenfunction around $qE_f = 0$ $\endgroup$ – Kasper Oct 8 '19 at 14:06
  • $\begingroup$ @Ezze - I think you got confused because you didnt realize that, e.g. the first excited state is odd in $x$, $\approx x e^{-x^2/2}$. And subsequent levels are polynomials in $x$. Surely there is no obstacle here to expand the shifted wavefunction. $\endgroup$ – Kostas Oct 8 '19 at 14:25

Your Answer

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

Browse other questions tagged or ask your own question.