# I Degenerate Perturbation Theory

1. Sep 28, 2016

### bananabandana

I'm struggling to understand degenerate perturbation theory. It's clear that in this case the 'normal' approximation method fails completely seeing as you get a divide by zero.

I follow the example for a two state system given in e.g D.J Griffiths "Introduction to Quantum Mechanics"

However, I don't understand the reasoning of how we go from that to say that we should generally "diagonalize the perturbation in the subspace of degenerate states" - how are we sure from the get go that the only significant contributions to the perturbed energy should come from the states which are degenerate in the original energy? Haven't we just completely excluded any others from consideration by this method?

2. Sep 28, 2016

### Orodruin

Staff Emeritus
Look at it this way:

Since the states are degenerate in the unperturbed system, there is no preferred basis of eigenvectors in the degenerate subsystem. You can therefore chose to work in the basis which diagonalises the perturbation in that subsystem. This will give you the zeroth order eigenstates.

Since the perturbation is now diagonal in the subsystem, you will have no problems applying first order perturbation theory to this set of states - the perturbation is diagonal in the degenerate subspace, indicating that the matrix elements (the ones that are divided by zero otherwise) are equal to zero. You therefore will find that the correction to these states appear at first order in perturbation theory and therefore are suppressed by the energy difference between the degenerate states and all other states. The first order energy shifts are, as usual, given by the expectation of the perturbation. In this case it just means the elements of the diagonalised restricted (to the degenerate subspace) perturbation.

If you will, consider it a move of the part of the perturbation that is in the degenerate subspace to the main Hamiltonian and consider the rest of the perturbation as the actual perturbation.