# Griffiths's take on degenerate perturbation theory [duplicate]

I've searched around here and it seems I'm not alone with my frustration towards Griffith's explanation of degenerate perturbation theory. Right from the start, Griffiths claims that some perturbation $H'$ will split a degeneracy (assume 2-fold degeneracy) and if we 'turn off' the perturbation then the 'upper' and 'lower' state will be some mutually orthogonal linear combinations of the degenerate states $\psi_A$ and $\psi_B$. Where did this claim come from? Is this because the linear sum $c_1\psi_A+c_2\psi_B$ is also an eigenstate of the unperturbed Hamiltonian $H^0$?

My intuition so far is that we resort to linear sums because we can now have distinct wave functions to use in the first order correction $E_1^1=\langle \psi_A^0 |H'|\psi_A^0 \rangle$ or $E_1^1=\langle \psi_B^0 |H'|\psi_B^0 \rangle$; I guess it was unclear which wave function we must choose? Is this why Griffiths goes on to solve $H\psi^0=E^0\psi^0$ where $H=H'+H^0$ and $\psi^0=c_1\psi_A+c_2\psi_B$?

I found some more formal mathematical reasoning behind this. I can't understand the explanations as I don't have a very advanced understanding of linear algebra, but from what I get we are trying to change basis? Can I get some clarification on the mechanism and reasoning behind this?

• I'm pretty sure this post is a duplicate of this other one. Could you read the link and let me know if it answers your question? Commented Apr 9, 2018 at 4:36
• or closely related to physics.stackexchange.com/q/314947/36194 Commented Apr 9, 2018 at 5:12
• Commented Apr 9, 2018 at 13:45