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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?

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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

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