# Quantum harmonic oscillator as the potential becomes zero

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.

• As $\omega \rightarrow 0$ the separation of the energy levels tends to 0, giving you the continuous spectrum of a free particle. Can you explain your issue a little more? Commented Jul 11, 2023 at 8:42
• I get that intuitively it becomes continuous, however I wasn't able to find a more mathematical way to make $E_n\to \hbar^2 k^2/2m$ in this limit. Commented Jul 11, 2023 at 8:51
• I suppose the reason is simply that you (more or less implicitly) assumed a non-zero (perhaps even positive) $\omega$ in the derivation of the eigenvalues and eigenvectors of the QHO. Also note that for $\omega=0$, all the harmonic oscillator eigenstates become the zero vector, which by definition is not an eingenstate of any operator. Commented Jul 11, 2023 at 9:02
• Ok thanks a lot, I think I got it. Commented Jul 11, 2023 at 9:38
• Correction to my previous comment: I meant the normalized vectors. Commented Jul 11, 2023 at 11:36

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 $$\infty\cdot 0~=~0$$.

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.