Degenerate Perturbation Theory: Operator commuting with Ho and H'

[meanyack]

Homework Statement
Question is:

Prove the following:
Let A be a Hermitian operator that commutes with H₀ and perturbation H'. If two degenerate states have distinct eigenvalues for A, then the matrix element of perturbation between them is zero!

The real problem is I don't understand problem completely.
I know if [A, H'] = 0, then it is called good states and one can find a unique wave functions for this. Can anyone help me about this?

 

[NateD]

Homework Equations A is a Hermitian operator, so its eigenvalues are real.[A, H0] = 0, [A, H'] = 0The Attempt at a SolutionLet A be an Hermitian operator that commutes with H0 and H'. Let λ1 and λ2 be the eigenvalues of A for two degenerate states. Since A is Hermitian, its eigenvalues are real. Now assume that λ1 ≠ λ2.We will show that the matrix element of the perturbation between the two states is zero. Since A commutes with H', it follows that [A, H'] = 0. This implies that the wave functions for the two degenerate states are good states, i.e. they are eigenstates of the combined Hamiltonian (H0 + H'). Since the wave functions are good states, the matrix element of the perturbation between them must be zero. Therefore, if two degenerate states have distinct eigenvalues for A, then the matrix element of the perturbation between them is zero.