The quantum harmonic oscillator on a circle -- fragmentation of the algebraic method

Abstract

A quantum particle on a circle in a quadratic potential exhibits a spectrum that is not harmonic, despite having all algebraic properties of the quantum harmonic oscillator. This raises the question where the usual algebraic argument -- implying integer gaps -- fails. The answer is illuminating and covers a surprisingly rich range of physical phenomena for such a simple model.

Figures (8)

• Figure 1: Geometry of our setup and the choice of position $x$
• Figure 2: Even eigenvalues $E_2k$ (orange) and odd eigenvalues $E_{2k+1}$(blue) vs $\ell=\pi r$. One has $E_n(\ell)\to n+\frac{1}{2}$ as $\ell\to\infty$ (spectrum of the harmonic oscillator on the line), and $E_n(\ell)\sim \lceil \frac{n}{2} \rceil^2 \frac{\pi^2}{2\ell^2}$ as $\ell\to 0$ (spectrum of a free particle on a circle).
• Figure 3: Eigenfunctions $u_n(x)$ vs $x$ for $r=1$ and $n=0,1,2,3,4,5$. We plot the potential $x^2/2$ and set the baseline of $u_n$ to be the corresponding eigenvalue $E_n$. Notice the periodicity. For small energies we have that $u_n$ approximate the Hermite eigenfunctions of the harmonic oscillator on the line, while for large energies they approximate the trigonometric eigenfunctions of the free particle on the circle.
• Figure 4: Venn diagram of the domains of the Hamiltonians $H=\frac{1}{2}(p^2+q^2)$, $H_N=a^\dag a +\frac{1}{2}$ and $H_A= aa^\dag - \frac{1}{2}$. Shown are also the eigenfunctions $u_n$, $v_n$, and $w_n$ in their respective domains. In particular, odd eigenfunctions are in the domain of the commutator and are shared by the three Hamiltonians, $u_{2k+1}=v_{2k+1}=w_{k+1}$. On the other hand, each Hamiltonian has its own even eigenfunctions ---and in particular a ground state--- characterized by a different behavior at the antipodal point $A$.
• Figure 5: The lowest energy levels of $H_N$ and $H_A$ for $\ell=\pi r = 2$. Shown are the remnants of the usual ladder structure: starting on odd eigenfunctions, we obtain new eigenstates of $H_N$ (purple) by applying $a^\dag$ and of $H_A$ (orange) by applying $a$. This works only once for each odd level, which correspond to joint eigenstates of both Hamiltonians (black). Applying the opposite operator brings the state back. The blue ground states are annihilated by $a$ for $H_N$ and by $a^\dag$ for $H_A$, and have eigenvalues $1/2$ and $-1/2$ respectively.