Suppose there is a free particle on a circle with radius r.

The energy spectrum is then

$$E_n = \frac{n^2\hbar^2}{2mr^2} \,.$$

Thus, when $n \neq 0$, then the spectrum of energies is degenerate for $n = \pm1, \pm2, \pm3, ...$

How can I construct a hermitian operator that distinguishes the degenerate states physically? I know that the degenerate states have to be eigenfunctions of said operator, but with different eigenvalues for the degenerate states.


A free particle on a circle of fixed radius r is known as a 2D rigid rotor.

As for the classical case, the particle's energy, and therefore its Hamiltonian, can be expressed in terms of the angular momentum ${\bf L}$ and the moment of inertia $I = mr^2$ as $$ H = \frac{{\bf L}^2}{2I} = \frac{{\bf L}^2}{2mr^2} $$ If the particle's trajectory is constrained to the $xy$-plane, the only surviving angular momentum component is $L_z$, so $$ H = \frac{L_z^2}{2mr^2} = - \frac{\hbar^2}{2mr^2}\frac{\partial^2 }{\partial \phi^2} $$ To understand what is happening and what the physical interpretation of the degeneracy is, look at the eigenvalues of $L_z$.

  • $\begingroup$ So, in other words, the eigenvalues for $\large L_z$, call them $\ell$ are distinct whenever the energy level, $n$, is degenerate. Thus, for example, if $ n = \pm1, \pm2,... , \ell = 0, 1, 2,...$ ? $\endgroup$ – Hugo Bethancourt Oct 23 '15 at 15:23
  • 1
    $\begingroup$ No, not at all. The eigenvalues of $L_z$ are simply $n\hbar$, just solve its eigenvalue equation. Then the eigenvalues for $H$ are $(n\hbar)^2/(2mr^2) = \hbar^2 n^2/(2mr^2)$. The $l$ you are mentioning labels eigenvalues of ${\bf L}^2 = L_x^2 +L_y^2 +L_z^2$ in 3D. The eigenvalues of ${\bf L}^2$ are in fact $\hbar l(l+1)$, for $l = 0, 1, 2,...$. In that case, we also have $[{\bf L}^2, L_z] = 0$, from which follows that for every given $l$ there are $2l+1$ degenerate eigenfunctions of ${\bf L}^2$, $\psi_{ln}$, labeled by different eigenvalues of $L_z$, $n = 0, \pm 1, \pm 2, ..., \pm l$. $\endgroup$ – udrv Oct 23 '15 at 15:52

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