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Given two harmonic oscillators with frequencies $\Omega$ and $\Omega'$, the eigenstates themselves are exactly known. Let's call them $\Psi_n$ and $\Psi'_n$.

Is there a compact expression for the basis transformation that relates the $\Psi_n$ to the $\Psi'_n$?

For the ground state, I think I have an idea of how to do that: The matrix element $\langle \Psi'_0 | \Psi_0\rangle$ is easy to calculate as it's just a Gaussian integral, and for the matrix elements of the type $\langle \Psi'_n | \Psi_0 \rangle$ I can use $$\langle \Psi'_n | a | \Psi_0 \rangle = 0$$ and then use the fact that the operator $a$ must be related to operator $a'$ for the other oscillator via a unitary transformation of the type $a = u a' + v a'^\dagger$.

That gives me a recursive relation for the coefficients of a relatively simple form that I can solve.

But I'm not sure if that's the most elegant way, and how I'd generalized that to get all the matrix elements.

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1 Answer 1

up vote 5 down vote accepted

Yes, of course, they're the squeezed states.

If your change involves a simple change of the frequency only, you may simply "rescale" all the energy eigenstates by $x\to Cx$ and $p\to p/C$. This is achieved by the operation $$|n\rangle \to |n'\rangle = \exp[\frac {i\ln(C)}{2\hbar}(xp+px)] |n\rangle$$ I inserted the Hermitian part of the operator $xp$ (multiplied by a pure imaginary parameter) so that the normalization of the vectors in the Hilbert space was preserved. The operator $(xp+px)/2\hbar$ may also be written as $(aa^\dagger + a^\dagger a)/2$ in terms of the raising and lowering operator.

Because the width in $x$ scale like $\sqrt{\hbar / m\omega}$, rescaling $\omega$ by the factor of $\Omega'/\Omega$ means rescaling $x$ by $C=\sqrt{\Omega/\Omega'}$ (thanks, Ron) which is the value you should substitute. Now, just be careful about the passive vs active transformation etc. but there is a simple compact dictionary between the eigenstates of both oscillators. You may manipulate with the formula above in various ways, using properties of the exponential such as its Taylor expansion and the CBH identity.

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C should be $\sqrt{\Omega/\Omega'}$ (typo) but +1. You can also insert the creation annihilation operators and normal order to make the answer complete and explicit, but it's probably in the link. – Ron Maimon Oct 31 '12 at 6:01
You're right! For others, the fix was done thanks to Ron. – Luboš Motl Oct 31 '12 at 7:21

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