Suppose I have a harmonic oscillator with frequency $\Omega_1$ and another one with frequency $\Omega_2$. Is there a simple relationship between the eigenstates of the two? Especially, how would the ground-state of one of them be expressed in terms of eigenstates of the other one?

An application of my question would be a harmonic oscillator whose frequency can be controlled. Suppose then I start out in the ground state and then suddenly change the frequency. I'd expect that I'm then not in a ground state of the (new) oscillator any more, and I'd be interested in the time evolution of my state. For that, I need to do a basis transform of my groundstate.

The problem seems basic enough to me that there should be previous work done on it. A brute force solution would probably be to perform integrals over the eigenstates in real space, but I have hope that an algebraic solution in terms of creation and destruction operators exists.


They are related by rescaling x linearly, so they are just more compressed/expanded but have the same shape.

$$ \psi'_n (x) = \psi_n(\sqrt{\Omega_1\over \Omega_2} x)$$

The reason is dimensional analysis--- the scale of x is determined from the physical length scale, which is the decay-constant of the ground state Gaussian:

$$ \Delta X = \sqrt{\hbar \over m \Omega} $$

mass times frequency is momentum over length, and $\hbar$ is momentum times length, so the ratio inside the square root has units of length squared. Don't bother with the formulas you find in books, set m and $\Omega$ to 1, and use the dimensional analysis scaling laws to find the rest of the solutions. There is only one harmonic oscillator, up to choice of units.

  • $\begingroup$ Ah I see, that's swell. Good one Ron! $\endgroup$ – Dylan O. Sabulsky Oct 31 '12 at 1:40
  • $\begingroup$ Thanks. But I'm not completely happy yet :) What I'm look for is this: Given an eigenfunction $\Psi_n$ of the one oscillator, what are the $a_i$ so that $\Psi_n = \sum_i a_i \Psi'_i$? I think I'm on the way of figuring this out for the ground state $\endgroup$ – Lagerbaer Oct 31 '12 at 2:48
  • $\begingroup$ @Lagerbaer: That is a completely different question. It is also exactly known, but it is best asked separately. It is unrelated to this simpler problem. $\endgroup$ – Ron Maimon Oct 31 '12 at 3:02

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