# Is $H = (-\Delta)^{\alpha/2} + (X^2)^{\beta/2}$, $\frac{1}{\alpha} + \frac{1}{\beta} = 1$, a Quantum Harmonic Oscillator ?

David Coulette, Raphael Ducatez

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David Coulette, Raphael Ducatez. Is $H = (-\Delta)^{\alpha/2} + (X^2)^{\beta/2}$, $\frac{1}{\alpha} + \frac{1}{\beta} = 1$, a Quantum Harmonic Oscillator?. 2025. hal-05303065

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# Is $H = (-\Delta)^{\alpha/2} + (X^2)^{\beta/2}$, $\frac{1}{\alpha} + \frac{1}{\beta} = 1$, a Quantum Harmonic Oscillator ?

David Coulette, Raphael Ducatez

October 8, 2025

## Abstract

We observe numerically the discrepancy between the spectrum of operators obtained by the quantification of classical isochrone systems and the spectrum of the quantum harmonic oscillator $\omega(N + \frac{1}{2})$. We are particularly interested in the operator $(-\Delta)^{\alpha/2} + (X^2)^{\beta/2}$ with $\frac{1}{\alpha} + \frac{1}{\beta} = 1$.

## 1 Introduction

Let us start with a very natural question which is as old as quantum mechanics but still far from being well understood: “What properties can be deduced, from a classical system to its associated quantum model?” The difficulty (and the interest) here is that we are dealing with two mathematical formalisms that are radically different one to another : trajectories of a finite set of particles and PDE’s on the one hand, wave functions and operators on the other. It is all the most surprising that such a link even exist. We should, of course, mention Ehrenfest theorem, Weyls law and the Semi-classical Analysis in general. For a more recent example, we could add the relation between chaotic behaviours of the classical dynamics and Quantum Ergodic theorems. However almost all the existing results are stated in the limit $\hbar \to 0$ and give a correspondance between classical and quantum quantities with an error that is small in this limite. On the other hand, *exact results* for the finite regime $\hbar \asymp 1$ are very rare and seem very difficult to obtain in general.

In these notes we are interested in what we will called generalized harmonic oscillators that are systems with isochrone trajectories. We will focus on two particular models :

1. The fractional Laplacian quantum harmonic oscillator on $\mathbb{R}$ (see [4])

$$H_{\alpha, \beta} = (-\Delta)^{\frac{\alpha}{2}} + (X^2)^{\frac{\beta}{2}}, \quad \frac{1}{\alpha} + \frac{1}{\beta} = 1$$

2. The two frequencies asymmetric harmonic oscillator on $\mathbb{R}$ (see [1])

$$G_{\omega_1, \omega_2} = -\Delta + \omega_1^2 X^2 1_{X \ge 0} + \omega_2^2 X^2 1_{X < 0}$$
We would like to know if the spectrum of these operators is given by $\omega(\mathbb{N} + \frac{1}{2})$, $\omega \in \mathbb{R}$ as for the standard quantum harmonic oscillators. The main interest of the second model is that it gives a counter example of our main question. It is also a simple enough model so that one is able to give a computation. The first model is by far the most intriguing one. In this case, we have very little ideas how to start a rigorous proof and the numerics are not very convincing. Therefore our main question is still open. On the other hand, the fractional Laplacian is nowadays, a very popular operator and any results about $H_{\alpha,\beta}$ would then be very interesting and useful. Because we, unfortunately, have few rigorous results to present, the main goal of these notes is to advertize the model and the question.

## 2 Generalized Harmonic Oscillator, from classical to quantum ?

In these notes, we will only consider simple one particle, one dimensional systems. We denote $H_0(x,p) = \frac{1}{2}\omega(p^2 + x^2)$ the usual classical harmonic oscillator. For a classical Hamiltonian $H(x,p)$ and denote $x(t), p(t)$ the solution of the Hamilton equations

$$ \frac{d}{dt}x(t) = \partial_p H(x(t), p(t)), \quad \frac{d}{dt}p(t) = -\partial_x H(x(t), p(t)), \quad x(0) = x_0, \quad p(0) = p_0 $$

and denote $\phi_t : \mathbb{R}^2 \to \mathbb{R}^2$ the map $\phi_t(x_0, p_0) = (x(t), p(t))$.

**Definition 1.** We call $H(x,p)$ a Generalized Classical Harmonic Oscillator<fnref n="1" /> if one of the following properties is satisfied

1. There exists a change of variable $(x, p) \to (\tilde{x}, \tilde{p})$ so that one recovers $H_0$

2. All trajectories are periodic with same period $T = \frac{2\pi}{\omega}$,

3. $\phi_T = I_{\mathbb{R}^2}$, the identity on $\mathbb{R}^2$.

Up to some small technical hypothesis and maybe non-relevant constants the above conditions are equivalent. We now denote $H_0 = \frac{1}{2}\omega(-\Delta + X^2)$ the usual quantum harmonic oscillator. For a quantum operator $\hat{H}$ that act on wave functions on $\mathbb{R}$ we write $u(t, x)$ the solution of the Schrodinger equation

$$ i\hbar\partial_t u(t, x) = \hat{H}u(t, x), \quad u(t, x) = e^{-it\hat{H}/\hbar}u(0, x) $$

and $E_0 \le E_1 \le \dots$ the eigenvalues of $\hat{H}$.

**Definition 2.** We call $\hat{H}$ a Generalized Quantum Harmonic Oscillator if one of the following properties is satisfied

<footnote n="1">¹This is a local appellation. Unfortunately, “general harmonic oscillator” could mean already many things in the literature...</footnote>
1. There exists a unitary operator U so that $UHU^* = H_0$,

2. For all $i \in \mathbb{N}$, $E_{i+1} - E_i = \bar{\hbar}\omega$,

3. $e^{iTH/\hbar} = I$ the identity.

Here again up to some technical hypothesis, added constants or a global phase these properties are almost equivalent. The correspondence between the lists of the two definitions is transparent. Then it is natural to ask the following.

**Question 3.** Is the quantification of a generalized classical harmonic oscillator a generalized quantum harmonic oscillator?

We start by an easy example.

**Example 4.** Let $f$ a smooth strictly increasing real function on $\mathbb{R}$. The systems defined as

$$H_f(x, p) = \left( \frac{1}{f'(x)} p \right)^2 + f(x)^2 \quad H_f = \left( \frac{1}{f'(X)} \partial_x \right)^2 + f(X)^2$$

are generalized classical and quantum harmonic oscillators.

*Proof.* This is a simple change of variable $y = f(x)$ and $q = \frac{1}{f'(x)}p$. Because of $\frac{1}{f'(x)}\partial_y = \partial_x$ we have the Poisson bracket and the commutator that satisfy

$$\left\{ \frac{1}{f'(x)} p, f(x) \right\} = 1 \quad \text{and,} \quad \left[ \frac{1}{f'(X)} \partial_x, f(X) \right] = 1.$$

For example $H_{\log}(x, p) = (xp)^2 + (\log(x))^2$ is then an Harmonic oscillator. □

Unfortunatly for a general Hamiltonian, Question 3 doesn't fully make sense because quantification is not a very well defined process. Indeed the standard procedure $p \to i\hbar\partial_x$ to too naive (For a typical example, we have $xp = px$ but $x\partial_x f(x) \neq \partial_x[xf(x)]$) and one can then define many different quantifications which are not equivalent. In the following we will only consider Hamiltonian of the form $H(x, p) = F(p) + V(x)$ and the quantification $H = F(i\partial_x) + V(X)$. Before going to particular examples, we mention another heuristic for Question 3

**Remark 5.** (Bohr-Sommerfield quantification [3, (48.2)] The level spacings are expected to satisfy

$$h = \int_{E_i \le H(x,p) \le E_{i+1}} dx dp = \int_{E_i}^{E_{i+1}} T(E) dE$$

where $T(E)$ is the period of the classical trajectory with energy $E$. For a isochrone classical system, $T(E) = T$ constant, this gives

$$E_{i+1} - E_i = \frac{h}{T} = \hbar\omega.$$

Unfortunatly this is only an heuristic and can't be used for a rigorous computation.
# 3 Fractional Laplacian Quantum Harmonic Oscillator

## 3.1 The model

We consider the following classical Hamiltonian

$$H(x, p) = |p|^{\alpha} + |x|^{\beta}, \quad \begin{cases} \frac{d}{dt}x(t) = \alpha \operatorname{sign}(p)|p|^{\alpha-1} \\ \frac{d}{dt}p(t) = -\beta \operatorname{sign}(x)|x|^{\beta-1} \end{cases} .$$

which is of course, the usual harmonic oscillator in the case $\alpha = \beta = 2$. It is a simple exercise to compute the period at a fixed energy $E > 0$,

$$\frac{T_{\alpha, \beta}(E)}{4} = \frac{E^{\frac{1}{\alpha} - \frac{1}{\beta} - 1} \Gamma(\frac{1}{\alpha}) \Gamma(\frac{1}{\beta})}{\alpha \beta \Gamma(\frac{1}{\alpha} + \frac{1}{\beta})}$$

with $\Gamma$ the standard Gamma function and in the case $\frac{1}{\alpha} + \frac{1}{\beta} = 1$ all trajectories are periodic with frequency

$$\omega_{\alpha, \beta} = \frac{2\pi}{T_{\alpha, \beta}} = \frac{\pi \alpha \beta}{2 \Gamma(\frac{1}{\alpha}) \Gamma(\frac{1}{\beta})} = \frac{\alpha^2 \sin(\frac{\pi}{\alpha})}{2(\alpha - 1)}.$$

We then have a Generalized Classical Harmonic Oscillator and we consider the associated quantum operator, replacing $p \to i\hbar\partial_x$. Up to a scaling, we can assume $\hbar = 1$ and we simply have $|p|^{\alpha} \to (-\Delta)^{\alpha/2}$ the so called fractional Laplacian. This operator is usually defined through the Fourier transform

$$\langle (-\Delta)^{\alpha/2} f(p) \rangle = |p|^{\alpha} \hat{f}(p).$$

or with a long-range kernel

$$\left[(-\Delta)^{\alpha/2} f\right](x) = \frac{2^{\alpha} \Gamma(\frac{1+\alpha}{2})}{\sqrt{\pi} \Gamma(-\frac{\alpha}{2})} \int_{\mathbb{R}} \frac{f(x+y) - f(x)}{|y|^{1+\alpha}} dy.$$

This can also be seen as the generator of the $\alpha$-stable Levy process. Because of the non-locality, the fractional Laplacian is notoriously difficult to deal with. We now ask our main question.

**Question 6.** Are the eigenvalues of $H_{\alpha, \beta} = (-\Delta)^{\alpha/2} + (X^2)^{\beta/2}$ with $\frac{1}{\alpha} + \frac{1}{\beta} = 1$ given by $\omega_{\alpha, \beta}(\mathbb{N} + \frac{1}{2})$?

If Question 6 were true and if one could compute its eigenvectors and the associated creation-annihilation operator, it would give a complete new and powerful tools to study others operator constructed with the Fractional Laplacian.

**Remark 7.** Here are a few simple observations

* For the limiting case $\alpha = 1, \beta = \infty$, we obtain the operator $H_{1,\infty} = |\partial_x|$ restricted on $[-1, 1]$ (an infinite well) with Dirichlet boundary condition. Here, the eigenstates are just $\cos((k + \frac{1}{2})\pi x)$ with energy $E_k = (k + \frac{1}{2})\pi$.
* These operators are symmetric by Fourier transform : $\widehat{H}_{\alpha,\beta} = H_{\beta,\alpha}$.

* Because $H_{\alpha,\beta}$ is homogenuous, Viriel Theorem states that for any $\psi$ eigen-state of $H_{\alpha,\beta}$ we have

$$\alpha \langle (-\Delta)^{\alpha/2} \rangle_{\psi} = \beta \langle (X^2)^{\beta/2} \rangle_{\psi}$$

* Using a Feynman-Kac formula approach, we have that Question 6 is equivalent to

$$\mathrm{Tr}(e^{-\gamma H_{\alpha,\beta}}) = q(\gamma) \int_{\mathbb{R}} \mathbb{E}[e^{\int_{0}^{\gamma} |x+W_{t}^{(\alpha)}|^{\beta} dt} | W_{\gamma}^{(\alpha)} = 0] dx = \frac{2}{\sinh(\gamma \omega_{\alpha,\beta}/2)}$$

for all $\gamma > 0$ where $W_t^{(\alpha)}$ is the $\alpha$-stable Levy process starting 0, $q(\gamma)$ the probability density of going back to 0 at time $\gamma$ and $\mathbb{E}$ the expectation on all such processes conditionnally on going back to 0.

## 3.2 Numerical simulation

In Figure 1 we numerically compute (with finite element method) the first eigenvalues of $H_{\alpha,\beta}$. The result seems to match pretty well the conjecture. We should stress that for example $H_{\frac{4}{3},4} = (-\Delta)^{2/3} + X^4$ is a very different operator than $-\Delta + X^2$. So such a plot is very surprising. If we zoom in, we observe small deviations ($\approx 10^{-3}$) from the prediction. But because we make some errors 1) in computing $-\Delta$ with the finite difference method giving an errort term of order $O(\frac{1}{a})$ with $a$ the size of discretization step and 2) in considering only a finite domain which cannot be neglected because the kernel is long range and giving an errort term of order $O(\frac{1}{L^{1+\alpha}})$ with $L$ the size of the domain. It is difficult to see if it is a real discrepancy or just due to the numerics.

In Figure 2, we compute again the first eigenvalues of $H_{\alpha,\beta}$ but in the regime $\alpha, \beta$ close to two. Here we use $\psi_n$ the eigenvectors of the usual Quantum Harmonic Oscillator (ie: the Hermite functions $\psi_n(x) = H_n(x)e^{-x^2/2}$). We compute $a_{n,m}(s) = \langle \psi_n, |X|^s \psi_m \rangle [2, (124)]$ to construct an $N \times N$ matrix $M$ with entries $M_{n,m} = a_{n,m}(\alpha) + (i)^{m-n} a_{n,m}(\beta)$ and diagonalize it. The error between the eigenvalues of this matrix and the prediction is shown in Figure 2, Again the error is non zero but too small to rule out numerical mistakes.
<figure><img src="image_2.png" /><figcaption>Figure 1: The ten first energy levels of $\frac{1}{\omega_{\alpha,\beta}}H_{\alpha,\beta}$ with $\frac{1}{\alpha} + \frac{1}{\beta} = 1$ for some value of $\beta \in [2, 5]$.</figcaption></figure>

<figure><img src="image_3.png" /><figcaption>Figure 2: The log error of the ten first eigenvalue of $H_{\alpha,\beta}$, $\beta - 2 \in \{10^{-4}, 10^{-3}, 10^{-2}, 10^{-1}\}$.</figcaption></figure>
# 4 Two frequencies non symmetric harmonic oscillator

## 4.1 The model

Here we consider the following classical Hamiltonian

$$G(x, p) = \frac{1}{2}p^2 + V(x) \quad \text{with} \quad V(x) = \begin{cases} \frac{1}{2}\omega_{-}^2 x^2 & \text{if } x \le 0 \\ \frac{1}{2}\omega_{+}^2 x^2 & \text{if } x > 0 \end{cases}$$

which is isochrone with uniform frequency $\omega = \frac{2\omega_+\omega_-}{\omega_+ + \omega_-}$. We denote $a = \frac{\omega_+ - \omega_-}{\omega_+ + \omega_-} \in (-1, 1)$ an asymmetric parameter. We therefore introduce the associated quantum operator

$$G_{\omega_1, \omega_2} = -\frac{1}{2}\Delta + V(X)$$

and ask the same question

**Question 8.** Are the eigenvalues of $G_{\omega_1, \omega_2}$ given by $\omega(\mathbb{N} + \frac{1}{2})$?

For this question we can answer by the negative. However we still have the following proposition.

**Proposition 9.** The eigenvalues $E$ of the asymmetric harmonic oscillator are given by

$$\frac{E}{\omega} = n + \frac{1}{2} + \epsilon_n, \quad n \in \mathbb{N}, \quad \epsilon_n \in [0, 1[, \quad \lim_{n \to \infty} \epsilon_n = 0.$$

*Proof.* We denote

$$\alpha_+ = 1 + a, \quad \alpha_- = 1 - a, \quad s = \frac{1+a}{1-a}, \quad \mathcal{E} = \frac{E}{2\omega}$$

$$\nu_- + \frac{1}{2} = \frac{E}{\omega_-} = 2\alpha_+\mathcal{E}, \quad \nu_+ + \frac{1}{2} = \frac{E}{\omega_-} = 2\alpha_-\mathcal{E},$$

We also write

$$A_+ = A(a) = \frac{1}{\sqrt{\alpha_+ \Gamma(\frac{3}{4} - \alpha_- \mathcal{E}) \Gamma(\frac{1}{4} - \alpha_+ \mathcal{E})}} \\ A_- = A(-a) = \frac{1}{\sqrt{\alpha_- \Gamma(\frac{3}{4} - \alpha_+ \mathcal{E}) \Gamma(\frac{1}{4} - \alpha_- \mathcal{E})}}$$

We use the formula of [1, (11)]. $E$ is an eigenvalue of the asymmetric harmonic oscillator if

$$A_+ + A_- = 0. \qquad (1)$$
Because $\sin(\pi z)\Gamma(1-z)\Gamma(z) = \pi$, we have

$$ 
\begin{aligned}
A_+ &= \frac{\pi^2}{\sqrt{\alpha_+}} \Gamma\left(\alpha_-\mathcal{E} - \frac{1}{4}\right) \Gamma\left(\alpha_+\mathcal{E} - \frac{3}{4}\right) \sin\left(\pi\left(\frac{3}{4} - \alpha_-\mathcal{E}\right)\right) \sin\left(\pi\left(\frac{1}{4} - \alpha_+\mathcal{E}\right)\right) \\
&= \frac{\pi^2}{2\sqrt{\alpha_+}} \Gamma\left(\alpha_-\mathcal{E} - \frac{1}{4}\right) \Gamma\left(\alpha_+\mathcal{E} - \frac{3}{4}\right) (\cos(2\pi\mathcal{E}) - \sin(2\pi a\mathcal{E})) \\
&=: F_+(\cos(2\pi\mathcal{E}) - \sin(2\pi a\mathcal{E}))
\end{aligned}
$$

Similarly

$$ 
\begin{aligned}
A_- &= \frac{\pi^2}{2\sqrt{\alpha_-}} \Gamma\left(\alpha_-\mathcal{E} - \frac{3}{4}\right) \Gamma\left(\alpha_+\mathcal{E} - \frac{1}{4}\right) (\cos(2\pi\mathcal{E}) + \sin(2\pi a\mathcal{E})) \\
&=: F_-(\cos(2\pi\mathcal{E}) + \sin(2\pi a\mathcal{E}))
\end{aligned}
$$

Then using the equality $A_+ + A_- = 0$ we obtain

$$ \cos(2\pi\mathcal{E}) = \left(\frac{F_- - F_+}{F_+ + F_-}\right) \sin(2\pi a\mathcal{E}). $$

We have

$$ \lim_{z \to \infty} \frac{\Gamma\left(z + \frac{1}{2}\right)}{\Gamma(z)\sqrt{z}} = 1 \quad , \quad \text{then} \quad \lim_{\mathcal{E} \to \infty} \frac{F_- - F_+}{F_+ + F_-} = 0. $$

and we finally get

$$ 2\mathcal{E} = n + \frac{1}{2} + o_{n \to \infty}(1). $$

Moreover with

$$ g(\mathcal{E}) := \cos(2\pi\mathcal{E}) - \left(\frac{F_- - F_+}{F_+ + F_-}\right) \sin(2\pi a\mathcal{E}) $$

$$ \frac{1}{2\pi} g'(\mathcal{E}) := -\sin(2\pi\mathcal{E}) - a \left(\frac{F_- - F_+}{F_+ + F_-}\right) \cos(2\pi a\mathcal{E}) - \frac{\sin(2\pi a\mathcal{E})}{2\pi} \frac{d}{d\mathcal{E}} \left(\frac{F_- - F_+}{F_+ + F_-}\right) $$

The solution of $g'(\mathcal{E}) = 0$ should give only solutions such that $2\mathcal{E} = n + o_{n \to \infty}(1)$. This implies that for each intervals, there is exactly one eigenvalue $E \in (\omega n, \omega(n+1))$. $\square$

Therefore Question 3 cannot hold for any generalized harmonic oscillator. However, notice that the error is still small and get smaller for larger eigenvalues. We can then propose a weaker formulation of Question 3.

**Conjecture 10.** (weak version of Question 3) The quantification of a generalized classical harmonic oscillator is of the form $H + B$ with $H$ a generalized quantum harmonic oscillator and $B$ "small".

## 4.2 Numerical simulations

In Figure 3 we compute the energy levels $E$ that are the solutions of the equation (1). This time the numerics are clear enough that the answer to Question 8 is "no (but not very far)".
<figure><img src="image_4.png" /><figcaption>Figure 3: The error between $(i + \frac{1}{2})$ and the ten first eigenvalues of $G_{\omega_1, \omega_2}$, $\omega = 1$ for different value of $a \in [0, 1)$.</figcaption></figure>

References

[1] Goce Chadzitaskos and Jiri Patera. Asymmetric harmonic oscillator. arXiv preprint arXiv:2204.06415, 2022.

[2] Richard Herrmann. Solutions of the fractional schrödinger equation via diagonalization-a plea for the harmonic oscillator basis part 1: the one dimensional case. arXiv preprint arXiv:1805.03019, 2018.

[3] Lev Davidovich Landau and Evgenii Mikhailovich Lifshitz. *Quantum mechanics: non-relativistic theory*, volume 3. Elsevier, 2013.

[4] Nick Laskin. Fractional schrödinger equation. *Physical Review E*, 66(5):056108, 2002.