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Given the relativistic correction $$ H_1' = - \frac{p^4}{8m^3 c^2} $$ to the Hamiltonian (i.e. a perturbation), what does it mean when $[H_1', \mathbf{L}] = 0$? The book I'm reading says this implies that the degenerate states belonging to the level $E_n^{(0)}$ are not connected to first order by $H_1'$. Does this mean that we don't have to use degenerate perturbation theory?

Edit: Also, Griffiths uses nondegenerate perturbation theory to compute the first order energy correction. His explanation doesn't make any sense to me. He says:

You might have noticed that I used non degenerate perturbation theory in this calculation, even though the hydrogen atom is highly degenerate. But the perturbation is spherically symmetric, so it commutes with $L^2$ and $L_z$. Moreover, the eigenfunctions of these operators (taken together) have distinct eigenvalues for the $n^2$ states with a given $E_n$. Luckily then, the wave functions $\psi_{nlm}$ are 'good' states for this problem, so as it happens the use of non degenerate perturbation theory was legitimate.

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    $\begingroup$ whether you use degenerate perturbation theory or not is independent of the perturbation itself, but about the degeneracy of the unperturbed states. $\endgroup$ – AccidentalFourierTransform May 2 '16 at 20:45
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Let's think about a system that has a two-fold degeneracy for some given energy level. That is, two states $ \psi_{a} $ and $ \psi_{b} $, both of which correspond to energy $ E_{0} $. An example would be a spin-1/2 particle with a Hamiltonian that is spin-independent.

Now imagine that when we apply a perturbation, H', to the system, the degeneracy breaks into two distinct energy levels, $ E_{1} $ and $ E_{2}$.

The subtlety is that these two distinct energy levels do not necessarily correspond to the two degenerate states (it is not necessarily the case that $ H'\psi_{a} = E_{1}\psi_{a} $ and $ H'\psi_{b} = E_{2}\psi_{b} $). It is possible that the two distinct perturbed energies correspond to linear combinations of the two degenerate states, i.e. $ H'(\alpha \psi_{a} + \beta \psi_{b}) = E_{1}(\alpha \psi_{a} + \beta \psi_{b}) $, and similarly for some other linear combination (orthogonal to the first).

This subtlety is the reason that in degenerate perturbation theory, the first-order corrections to the energy require calculation of off-the-diagonal elements, i.e. things that look like $ \langle \psi_{a} | H' | \psi_{b} \rangle $. This is annoying, because in first order perturbation theory, we only needed to calculate one inner product. In degenerate perturbation theory, we have to compute a whole matrix of inner products to calculate the correction to the energy.

Since it is tedious to compute inner products, it'd be nice to know a trick to find out if the off-the-diagonal elements of the perturbing Hamiltonian are 0. The trick is given and proven on pg. 259-260 of Griffiths.

In the case of your question, the original Hamiltonian is spherically symmetric (Coulomb potential has no angular dependence). Also, the the perturbing Hamiltonian is spherically symmetric. If you look at the form of the angular momentum operator, you will note that it only involves things with $ \theta $ and $ \phi$ (derivatives and cosines and stuff). The nice thing about that, is that if I have a Hamiltonian that is purely radial (depends only on r), than it definitely commutes with L.

All Griffiths is doing is showing that = the conditions of the theorem on pg. 259 are satisfied, which shows that degenerate perturbation theory collapses to non-degenerate perturbation theory.

I'm not familiar with the idea of states being "connected" by a Hamiltonian. I would speculate that two states are connected by a Hamiltonian if the expected value of one state is nonzero given that it is initially in the other state.

Hope this helps!

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This perturbation operator is a scalar (in the sense that it doesn't change upon rotations). This is the meaning of the fact that it commutes with the angular momentum operators (or that it is spherically symmetrical, as you say).

Since it is a scalar, it cannot connect two states with different angular momentum values or different projections of the angular momentum (its matrix element between any two such states must be zero). It will not lift the angular momentum degeneracy and therefore degenerate perturbation theory will lead to the same result as non-degenerate one -- you can pick whatever basis you want to work with.

The Wigner-Eckart theorem allows us to relate the symmetries of a perturbation to the non-vanishing matrix elements that are possible.

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Since the energy levels of the hydrogen atom are degenerate (the energy depends only on the quantum number $n$ so the energy is the same for all values of $l$ and $m$), we might think we need to apply degenerate perturbation theory. However, we can recall a theorem that we proved earlier which states that if we can find an operator that commutes with the unperturbed hamiltonian and the perturbation, then the eigenvectors of that operator can be used as the ’special’ states and we can get away with using the non-degenerate theory[David J Griffith's Introduction to QM]. In this case, the stationary states $\psi_{nlm}$ are eigenstates of the angular momentum operators $L^2$ and $L_z$ and these two operators commute with $p^2$ and $p^4$, so the $\psi_{nlm}$ functions are already the special states and we can just apply non-degenerate theory using these functions directly [Ref:http://physicspages.com/pdf/Griffiths%20QM/Griffiths%20Problems%2006.12.pdf].

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