On the linear dispersion–linear potential quantum oscillator

With the quantum theory, as it was called at the time, nearing it first centennial anniversary, it is a rare opportunity to study an one-dimensional ideal oscillator which has not been solved long ago. The motion of a (non-relativistic) quantum particle with a linear dispersion, $\epsilon_{p}=v\cdot|p|$, where $p=\hbar k$ is the momentum and $v$ is a parameter, in a linearly confining potential $V(x)=F\cdot|x|$, where $x$ is the position and the constant force $F$ again a parameter, however, appears to provide an example. While the problem may look trivial at first, it is not. The usual method of quantization by replacing either $p\rightarrow-i\hbar\frac{\partial}{\partial x}$ or $x\rightarrow i\hbar\frac{\partial}{\partial p}$ cannot be applied directly, as one cannot sensibly define the absolute value of a differential operator.

The problem is not just of academic interest, but even of relevance to a recent experiment lake-09np50 ; greiter09np5 . Spinons, the fractionally quantized and elementary excitations in antiferromagnetic spin chains, are well known to disperse linearly at low energies, with $v$ proportional to the antiferromagnetic exchange constant $J$ along the chains Giamarchi04 . Spinons carry the spin of an electron but no charge. Since the antiparticle for a spinon is just another spinon with its spin reversed, the spectrum has only a positive energy branch. An one couples two chains antiferromagnetically Dagotto-96s618 , the coupling $J_{\perp}$ will induce a linear confinement potential between pairs of spinons, as the rungs between two spinons become effectively decorrelated Shelton-96prb8521 ; greiter02prb054505 . To a very first approximation, the energy gap in the spin ladder is hence given by the ground state energy of the bi-linear oscillator

which we study in this Article. The ground state is symmetric under one-dimensional parity $x\rightarrow-x$ and corresponds to a spinon pair in the triplet channel, while the first excited state is antisymmetric under $x\rightarrow-x$ corresponds to the lowest singlet excitation in the spin ladder. It the context of this problem, it is hence desirable to know what the lowest eigenvalues of (1) are. From dimensional considerations, it is immediately clear that they must scale like $\sqrt{\hbar\;\!vF}$.

Even though the usual method of quantization can not be applied directly, the problem can still be approached quasi-classically. Applying the Bohr-Sommerfeld quantization condition Landau3

where we are supposed to integrate over the entire classical orbit, results with $p(x)=\frac{E_{n}-F|x|}{v}$ in

Carrying out the integration yields

We expect this to constitute a reasonable approximation for the higher energy levels, but probably not for the low lying ones. Indeed, this is what we will find as we solve the problem numerically below.

Before proceeding with the numerical solution, let us rewrite the eigenvalue equation $H\psi(x)=E\psi(x)$ as a differential (and integral) equation in position space. For convenience, we consider the dimensionless Hamiltonian

which is obtained from (1) by rescaling

Let us denote the eigenvalues of (5) by $\lambda$ and the eigenfunctions by $\phi(x)$. With

we may write

where

is the sign function and

where $\mathcal{P}$ denotes the principal part, is the Fourier transform thereof. The eigenfunctions $\phi(x)$ with eigenvalues $\lambda$ of (5) are hence the solutions of

While (11) provides a clear mathematical formulation of the problem, we are not aware of any method to solve it analytically, nor consider it a viable starting point for numerical work.

To solve (5) numerically, we exactly diagonalize a finite Hamiltonian matrix we obtain through discretization of position space with a suitably chosen cutoff.

Let this discrete Hilbert space consist of $N$ sites, with the positions

where $i=1,2,\ldots N$ and $a$ is the lattice constant. The cutoff $|x_{c}|={Na}/{2}$ in real space implies a cutoff

for the potential energy in (5), which must be chosen significantly larger than the largest eigenvalue $\lambda_{n}$ we wish to evaluate reliably. (From (4), we expect $\lambda_{n}$ to be of order $\sqrt{\pi\left(n+\frac{1}{2}\right)}$.) On the other hand, the classically allowed part of the Hilbert space will contain only of the order of ${N}/{\lambda_{c}}$ sites for the ground state, which implies that we must further require $\lambda_{c}\ll N$.

The lattice provides us simultaneously with a cutoff in momentum space, $-\pi\leq ak\leq\pi$. We may hence expand $|k|$ in a Fourier series,

with

as one may easily verify through integration by parts. We proceed by writing (5) in second quantized notation,

where

Since

we obtain

with

where we have substituted $\frac{2\lambda_{c}}{N}$ for $a$.

Numerical diagonalization of $h_{ij}$ yields the eigenvalues $\lambda_{n}$ and eigenfunctions $\phi_{n}(x_{i})$ of (5), and hence the eigenvalues and eigenfunctions

of (1). The results for $N=20001$, $\lambda_{c}=20$ are listed in Table 1 and Figures 1 and 2. (We have chosen an odd number for $N$, because this means that the position $x=0$, where the potential $|x|$ is not differentiable, coincides with a lattice point. Including this point improves the convergence of the eigenvalues and functions for $n$ even.) From Table 1, we see that the quasi-classically obtained eigenvalues converge towards the numerically obtained values as $n$ is increased.

The eigenfunctions obtained numerically can be approximated by

for $n=0,1$ and by

for $n=2,3$, with parameters $a_{n}$, $b_{n}$, $c_{n}$, and $d_{n}$ listed in Table 2. Comparisons of these fits to the numerically obtained eigenfunctions are shown in Figure 3. The fits are not as good an approximation as Figure 3 may suggest, however, as they fall off as $\exp(-a|x|)$ while the true eigenfunctions $\phi_{n}(x)$ fall off as $1/x^{3}$ for $n$ even and as $1/x^{4}$ for $n$ odd as $x\rightarrow\infty$.

This asymptotic behavior of the eigenfunctions can be understood physically through second order perturbation theory. If we consider a small region around a point $x\gg\lambda$ (i.e., very far away from the classically allowed region for the eigenstate with energy $\lambda$), the amplitude there will be governed by scattering into this region from the classically allowed region, which contains almost the entire amplitude of the state. From (20), this scattering is proportional to

where $\lambda_{t}$ is a cutoff to insure that we include the tail immediately surrounding the classically allowed region in the integral (from Figs. 1 and 2, we see that $\lambda_{t}=3$ would be a reasonable choice). With the potential energy in the region we consider given by $|x|$, the amplitude for finding the particle there will be proportional to $1/x^{3}$ for $n$ even and as $1/x^{4}$ for $n$ odd.

The numerical work reported here indicates that, within the limits of accuracy, the solutions are differentiable at $x=0$, i.e., the expansion of $\phi_{n}(x)$ around $x=0$ does not contain a term proportional to $|x|$ for $n$ even or $x\,|x|$ for $n$ odd. Unfortunately, we have not been able to reach a conclusion regarding higher terms, and cannot tell whether there are terms proportional to $x^{2}|x|$ for $n$ even or $x^{3}|x|$ for $n$ odd.

It would be highly desirable to identify the exact eigenvalues and functions of (5). Unfortunately, we have as of yet not even succeeded in obtaining those for the ground state. A few thoughts on this problem, however, are possibly worth mentioning.

We have succeeded in solving the bi-linear oscillator $H=v|p|+F|x|$ both quasi-classically and numerically. In an attempt to solve it analytically as well, we have derived a differential and integral equation, and obtained the asymptotic behavior for large $x$. We further formulated several conditions the solutions must satisfy. The problem of obtaining an analytical solution, however, is still open.