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?