The infinite-fold degeneracy of an oscillator when becoming a free particle

Question

Considering a one-dimensional case, and if it has the following relation: $$(1) : [H,a]= \pm \omega a$$

then there are evenly spaced spectrum lines.

If we sink $\omega \rightarrow 0$ , or, accordingly, the potential becomes less and less curved, and eventually - flat. then we have:

$$(2):[H,a]=0$$

the spectrum lines will theoretically get narrower and narrower, and eventually overlap, and hence degeneracy. However, it seems very strange to me:

Based on the argument, then one eigenvalue corresponds to infinite energy eigenstates for a one-dimension "oscillator" (now technically a free particle).

Really? Infinite? I think for a a free particle with certain eigenenergy, it only has two eigenstates, as it either moving right or left (no infinity, at least for the one-dimensional situation)?

I think my intuition is not quite right, but I am not sure where. (I am not even very sure if the whole argument works)

Answer 1

This question is a good reminder that we can't define a limit just by specifying what goes to zero. We also need to specify what remains fixed.

The harmonic-oscillator Hamiltonian can be written either as $$ \newcommand{\da}{a^\dagger} H=\omega \da a \tag{3} $$ or as $$ H= p^2+\omega^2 x^2. \tag{4} $$ They are related to each other by \begin{align} a = \frac{p-i\omega x}{\sqrt{2\omega}}. \tag{5} \end{align} Equation (3) says that if we take the limit $\omega\to 0$ with $a$ held fixed, we get $H=0$, which gives equation (2) in the question. But equation (4) says that if we take the limit $\omega\to 0$ with $x$ and $p$ held fixed, we get $H=p^2$, which gives the words shown in the question ("the potential becomes less and less curved" and "...a free particle with certain eigenenergy... only has two eigenstates, as it either moving right or left"). The paradox is resolved by taking care to distinguish between these two different limits.