How Does Spin-Orbit Coupling Affect Energy Levels in a Hydrogen Atom?

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Homework Help Overview

The discussion revolves around the effects of spin-orbit coupling on the energy levels of an electron in a hydrogen atom, specifically in the n = 2, l = 1 state. Participants are exploring the implications of the spin-orbit interaction on energy level shifts.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the degeneracy of the n, l state and the implications of the spin-orbit interaction. There are attempts to evaluate the matrix elements of the perturbation and to understand the total angular momentum states. Questions arise regarding the total number of energy shifts and the nature of degeneracy in the context of the spin-orbit coupling.

Discussion Status

The discussion is active, with participants providing insights into the evaluation of angular momentum states and the resulting energy levels. Some guidance has been offered regarding the evaluation of the spin-orbit interaction, but there is no explicit consensus on the total number of energy shifts or the interpretation of degeneracy.

Contextual Notes

Participants are working within the framework of degenerate perturbation theory and are considering the implications of rotational invariance on the energy levels. There is an ongoing exploration of the assumptions related to the angular momentum states and their contributions to the energy shifts.

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Homework Statement



An electron in a hydrogen atom is in the n = 2, l = 1 state. It experiences a spin-orbit interaction H' = \alpha \mathbf{L} \cdot \mathbf{S}. Calculate the energy level shifts due to the spin-orbit interaction.

Homework Equations



Degenerate perturbation theory.

The Attempt at a Solution



This n,l state is triply degenerate due to the three possible values of m = -1,0,1.

The unperturbed Hamiltonian is just what goes in the Schrödinger equation right? In which case the eigenfunctions of the unperturbed hamiltonian are just the spherical harmonics Y_{lm} multiplied by strictly radial functions. So I put

\psi^{(0)} = \alpha Y_{10} + \beta Y_{1-1} + \gamma Y_{11}

So I then write down the matrix \langle Y_{1,i} |H'| Y_{1,j} \rangle and find the eigenvalues.

Am I getting warm?
 
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You have to evaluate L.S for each angular state (J,L,S) by using
J^2=L^2+S^2+2L\cdot S.
You don't need the explicit wave functions.
 
Hi Meir Achuz,

Thanks for your reply.

So for n = 2, l = 1 we have two possibilities for the total angular momentum corresponding to j = 1/2 and j=3/2 right?

But we also have some degeneracy coming from the possible values of m = -1,0,1.

Does this mean there will be a total of 6 energy shifts?
 
L^2 and S^2 are known.
There are two energy levels.
One for J^2=(3/2)(5/2), and one for J^2=(1/2)(3/2).
There are 6 states, but because of rotational invariance, the 4 J=3/2 states are still, degenerate, as are the l2 J=1\/2 states.
 

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