Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For separable potential, say $x^4+y^4$, its symmetry are degenerate. Is that a generic case to every separable potential? I will explain my question:

The potential $x^4+y^4$ has $A_1, B_1, A_2, B_2, E$ symmetries. However, all of them are degenerate. Quantum mechanically, the energy level are all degenerate and one can consider, for example, only the $A_1$ symmetry, why?

share|cite|improve this question
I don't think the notation you use is standard. What are these symmetries? What does "separable" mean? – yohBS Feb 5 '12 at 21:57
I think by separable he mean the potential will lead to a solution of the form $\psi(x,y) = f(x)g(y)$. – Evan Sosenko Sep 17 '12 at 19:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.