Degenerate eigenstates of irreducible representations ↩

aliases: []
shorthands: {"GL": "{\text{GL}(V)}"}
created: 2021-11-21 13:59:29
modified: 2022-01-19 10:53:26

Proof

Let be a fully decomposable or irreducible representation of the group and the invariant subspace of corresponding to the irreducible representation of .

If we take any vector and consider its orbit by (that is ), it will span some subspace of . In other words: . Where is less than , we would have found a non-trivial invariant subspace of of dimension less than , which is impossible by definition. QED

Remark

Therefore any vector (including those that describe the eigenstate of a system) in an irreducible subspace, will have linearly independent degenerate "copies" in the same subspace generated by the symmetry. These copies have to span at least as many dimensions inside as the irreducible representation has, i.e. you will find at least as many linearly independent degenerate modes as the number of dimensions of the irreducible representation.