Why is Spin-Orbit Coupling Important in Understanding Quantum Systems?

[oddiseas]

In a quantum system we have that the angular momentum is quantised in integer values. Then to my understanding we have like an extra momentum that is assocaited with the spin of the electron which is also found to be quantised but in half integer values and this can add or subtract from the "total" momentum.
Say we have a p orbital. There are three possible orientations for the orbit -1,0,1.
In the lectures we were shown that this would separate into 3 distinct spectral lines, one which stays the same, one that becomes 2p(3/2) and one that becomes 2p(1/2) each with distinct energy.
1)Can this add or subtract in "any" scenario or only when there is a magnetic field?

in addition finding the z component for the J, (the total momentum) i get values of -3/2
-1/2,1/2,3/2, for one electron with spin up in a 2p orbital.And two more if i include the spin down electron, so i am starting to get confused.This gives me six states, so why in the lecture are we told we will get three distinct lines?

In addition solving the bohr model i get quantised angular momentum L=nh, this says nothing about the z component of the momentum, why are we always concerned with the z component of the momentum and not say the x or y component?

 

[DrClaude]

1)Can this add or subtract in "any" scenario or only when there is a magnetic field?

This is spin-orbit coupling and is present in the absence of external magnetic fields. It can be seen as the interaction between the magnetic moment due to orbital angular momentum and that due to spin.

In the lectures we were shown that this would separate into 3 distinct spectral lines, one which stays the same, one that becomes 2p(3/2) and one that becomes 2p(1/2) each with distinct energy.

This doesn't really make sense.

in addition finding the z component for the J, (the total momentum) i get values of -3/2
-1/2,1/2,3/2, for one electron with spin up in a 2p orbital.And two more if i include the spin down electron, so i am starting to get confused.This gives me six states, so why in the lecture are we told we will get three distinct lines?

The six states coming from 3 $m_l$ states x 2 $m_s$ states get recombined into 4 $M_J$ states for P_3/2 and two for P_1/2, so it is still 6 states. The P_3/2 are degenerate, and os are the P_1/2, such that the number of spectral lines here should be two.

In addition solving the bohr model i get quantised angular momentum L=nh, this says nothing about the z component of the momentum, why are we always concerned with the z component of the momentum and not say the x or y component?

By convention, we choose to single out z. It is an arbitrary choice. But the fact that $J_x$, $J_y$, and $J_z$ donut commute amongst themselves means that we can only have a state that is an eigenstate of a single one of those three.