Quantum harmonic oscillator as the potential becomes zero

Question

I'm having a question concerning the quantum harmonic oscillator: If, for instance, $\omega\to 0$ the Hamiltonian $$ \hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega^2 x^2\tag{1} $$ becomes that of a free particle, but the energy eigenvalues don't. Why is there such a distinction between the two cases? I get that intuitively it becomes continuous, however I wasn't able to find a more mathematical way to make $$E_n\to \hbar^2 k^2/2m\tag{2}$$ in this limit.

Answer 1

1. One idea is to put the QHO$^1$ $$H~=~\frac{p^2}{2m} + \frac{m\omega^2x^2}{2}+\infty \cdot\theta(\frac{L}{2}-|x|) $$ in a box of length $L$.

2. On the one hand, well above a threshold energy $$ E_{\rm threshold}~=~\frac{m\omega^2L^2}{8} $$ the energy eigenvalues behaves like a particle in a box $$E_n~=~\frac{(\hbar k_n)^2}{2m}, \qquad k_n~=~\frac{n\pi}{L}, \qquad n~\in~\mathbb{N},$$ which should be compared with OP's eq. (2).

3. On the other hand, well below the threshold energy $E_{\rm threshold}$ the energy eigenvalues behaves like a QHO.

4. Now remove the harmonic potential $\omega\to 0$ and remove the box $L\to \infty$ in such a way that $\omega L\to 0$ so that $E_{\rm threshold}\to 0$.

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$^1$ Here we assume that $\infty\cdot 0~=~0$.

Answer 2

Picking up on @Tobias's argument, the natural unit of length for this problem is $\lambda=\sqrt{\frac{\hbar}{m\omega}}$, and so depends on $\omega$. Since all solutions are functions of the dimensionless quantity $x/\lambda$, you immediately see you are in trouble if you let $\omega\to 0$ as any solution of the form $e^{-(x/\lambda)^2/2}H_n(x/\lambda)$ behaves badly. In other words, taking the $\omega\to 0$ limit is delicate business and may not be feasible at all.