0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $L_z$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.