Eigenstates in hexagonal well I am currently solving the 2D Schrödinger Equation Numerically for a hexagonal well. The modulus squared of the first four eigenstates I find are shown on the attached picture (please comment if they look correct). 
Now my question is: Should there not be a 6-fold degeneracy due to the 6-fold symmetry of the hexagon? My intuition tells me this but on the other hand it sounds weird since in general there would be an N-fold degeneracy for a symmetric polygon of N sides, which is obviously absurd. Where does my logic fail? 
 A: Since the Hamiltonian is invariant under a unitary representstion of the symmetry group, you can conclude that every eigenspace of the Hamiltonian is invariant and  supports a representation of that group. However nothing implies that the dimension of the eigenspace is equal to the number of elements of the group. For instance it could be one-dimensional because the only (up factors)  eigenvector  is invariant under the action of the group.
More generally, the action of the symmetry group on an eigenvector can always be defined as a  linear combination of a finite set of eigenvectors with the same eigenvalue. Think of the hydrogen atom, for a given  value of the energy we have a finite dimensional eigenspage even if the symmetry group $SO(3)$ contains a continuous infinity of elements. This infinity corresponds to the infinite number  of possible  linear combinations of a finite number of eigenvectors forming the base of the given eigenspace of the Hamiltonian.
A: Look at the paper by Cureton and Kuttler (Journal of Sound and Vibration 220, 83-98 (1999))
in which the real counterpart of the Schroedinger Equation is considered. There EIGHT classes os symmetries appear which I'm sure are respected when going to the complex version.
