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:


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)

  • $\begingroup$ related: physics.stackexchange.com/q/576110 $\endgroup$
    – Shing
    May 2, 2021 at 16:05
  • 1
    $\begingroup$ "as it either moving right or left" — if you start from the oscillator, then your options are rather "is either even or odd", and it's the states you'd get in the limit. As for your construction, the problem here is that the initial and final Hamiltonians' spectra are qualitatively different: discrete vs continuous. I'd view all the final levels as "infinitesimally different", although some of them would actually be equal (due to double degeneracy created by uncovered additional symmetry). Any finite level of the oscilltr will be zero in the limit; this seems to be a poor proces to get free. $\endgroup$
    – Ruslan
    May 2, 2021 at 17:01

1 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.

  • $\begingroup$ Either equation (3) or (4) should include a constant term, because the operators don't commute with each other. But that doesn't change the message, and editing the answer would bump it back to the home page, so I'm posting this minor correction as a comment instead. $\endgroup$ May 2, 2021 at 17:38
  • $\begingroup$ The second limit ($x$ and $p$ held fixed) makes a lot of physical sense to me : just turn off the harmonic confinement and get back the free particle. Then the ladder operator diverges, as there are no such thing when the particle is free. Is there a similar interpretation for the other limit ($a$ held fixed) ? $\endgroup$ May 3, 2021 at 10:04
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    $\begingroup$ @SolubleFish For just one harmonic oscillator by itself, I don't know of any physically useful interpretation of the first limit. But a free quantum field can be viewed as a system of harmonic oscillators, one for each momentum $\vec p$, and in that case we have $\omega=\sqrt{\vec p^2+m^2}$, where $m$ is the mass of a single particle and $\omega$ is its energy. In the massless case $m=0$, $\omega$ goes to zero as $\vec p\to 0$. An arbitrarily large number of massless particles costs arbitrarily little total energy in the zero-momentum limit. $\endgroup$ May 3, 2021 at 13:12

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