Rotating harmonic oscillator

Question

The Hamiltonian operator $$H=\frac{{\bf p}^2}{2m} +\frac{m\omega^2}{2}{\bf r}^2-\Omega L_z$$ with $L_z=xp_y-yp_x$, can be written as $$H=\hbar\left(\omega+\Omega\right)\alpha^\dagger\alpha+\hbar\left(\omega-\Omega\right)\beta^\dagger\beta+\hbar\omega$$ where the ladder operators $\alpha$ and $\beta$ are given by $$ \alpha=\frac{1}{\sqrt{2}}(a_x+ia_y)$$ $$ \beta=\frac{1}{\sqrt{2}}(a_x-ia_y). $$ and $a_x$, $a_y$ are the standard ladder operators for the harmonic oscillator. That's a nice way to see that the energy for this problem doesn't have a lower bound (when $\Omega>\omega$).

QUESTION: Does anyone have a clear picture of what kind of excitations do the operators $\alpha^\dagger\alpha$ and $\beta^\dagger\beta$ represent?

Attempt answer: would it be correct to say that they represent clockwise and anticlockwise rotations?

Issue with the attempt: it doesn't make sense to have the lower bound on one of them and not on the other.

Second attempt: anything to do with maxons and rotons?

Answer 1

For the sake of simplicity lets assume $\Omega > 0$.

I believe that your initial attempt is correct and that $\alpha$ and $\beta$ can be thought of as states where the particle is "moving in a circle around the $z$ axis". Lets look at the term you added to the standard harmonic Hamiltonian $$ -\Omega L_z\;. $$ This term says that orbiting around the $z$ axis in a negative direction requires energy, but orbiting in the positive direction reduces your energy. If this effect is larger than the energies associated with the harmonic oscillator, then you can reduce your energy as much as you like by orbiting in the appropriate direction. In other words this term does exactly what your "Issue with the attempt" was struggling to explain.

This explains why, as @secavra states in the comments, $L_z \propto \beta^\dagger\beta - \alpha^\dagger\alpha$, i.e. the angular momentum is given by the number of excitations for circling one way minus the number for circling in the opposite direction. It can also be seen by looking at the representations in terms of $a_x$ and $a_y$; $\alpha$ and $\beta$ represent harmonic motion in the $x$ and $y$ directions phase shifted by $\pm \frac{\pi}{2}$, which classically would be circular motion.