TITLE: How can I construct a rotationally symmetric energy eigenstate for a particle on a ring?
QUESTION [0 upvotes]: The energy eigenstates for a particle on a ring are $\psi_m(\phi) \propto e^{im\phi}$, which are two-fold degenerate. A general energy eigenstate is any linear combination of these, which I can write as $a\psi_m + b\psi_{-m}$. These eigenstates are in general not rotationally symmetric. How can I construct a rotationally symmetric energy eigenstate for an arbitrary energy?

REPLY [4 votes]: You cannot. If it $\psi(\theta)$ is independent of the angle $\theta$, then it is an eigenstate of the $L= -i\partial_\theta$  with eigenvalue $0$, and so an eigenstate of $H=L^2= -\partial^2_\theta$ with eigenvalue $E=0$.
{"set_name": "stack_exchange", "score": 0, "question_id": 625651}