# Degenerate pertubation theory when the first order fails

1. Jan 11, 2010

### znbhckcs

The basic algorithm of degenerate perturbation theory is quite simple:
1.Write the perturbed Hamiltonian as a matrix in the degenerate subspace.
2.Diagonalize it.
3.The eigenstates are the 'correct' states to which the system will go as the perturbation ->0.

But what to do if the first order does not break the degeneracy?
For instance, if the perturbation matrix elements are all 0 in the degenerate subspace.

It seems unlikely to me that there is nothing else to be done in that case....

2. Jan 11, 2010

### chafelix

Actually when you diagonalize you solve the problem EXACTLY to all orders(in the framework of the set of states you use). It's when you have both degenerate and nondegenerate that order makes sense. To your question, if in the degerenate subspace the perturbation is 0, then to get a nonzero result you need more states.
Example: Take a static electric field (z-axis) and the n=2 level of atomic hydrogen.
210 and 200 get mixed, but 211 and 21-1 are not affected. So to affect them, you need to
include more states, e.g. n=3, when they an mix with 321 and 32-1 for example.
This is not degenerate anymore, of course