What Are the Selection Rules for the Hydrogen Atom in LS Coupling?

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SUMMARY

The selection rules for the hydrogen atom in LS coupling dictate specific transitions based on angular momentum. In this context, the rules are defined as follows: ΔJ = ±1 (except J=0 to J=0), Δmj = ±1 or 0, and ΔS = 0 due to the electron's spin-1/2 nature. Additionally, transitions from spin up to spin down allow for Δms = ±1, which can lead to Δmj = ±2 under certain conditions. However, the overall selection rules maintain that ΔJ must change by ±1, ensuring that transitions adhere to these constraints.

PREREQUISITES
  • Understanding of LS coupling in quantum mechanics
  • Familiarity with angular momentum quantum numbers (l, m_l, J, m_j)
  • Knowledge of electron spin and its implications in quantum transitions
  • Basic grasp of selection rules in atomic physics
NEXT STEPS
  • Study the implications of spin-orbit coupling on selection rules
  • Explore the derivation of angular momentum addition in quantum mechanics
  • Investigate the role of selection rules in spectroscopy
  • Learn about the differences between LS coupling and jj coupling
USEFUL FOR

Students of quantum mechanics, physicists specializing in atomic structure, and researchers focusing on atomic transitions and spectroscopy will benefit from this discussion.

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I have a couple of questions about selection rules for the hydrogen atom.

When we talk about these rules in an undergraduate context we are usually assuming LS coupling where we ignore spin orbit coupling so orbital and spin angular momentum are essentially independent. Is that correct?

Subsuming the basic l,m_l rules in this case we have
<br /> \Delta J=\pm 1, \text{except J=0} \rightarrow J=0.<br /> \\<br /> \Delta m_j=\pm 1,0<br /> \\<br /> \Delta S=0 \text{ since the electron is always spin half}<br /> \\<br /> \text{but } \Delta m_s=\pm 1\text{ since we can transition from spin up to spin down?}<br />
If we can indeed transition from spin up to spin down couldn't we possibly have \Delta m_j=\pm -2 from (m_l=1, up) to (m_l=0,down) then in violation of the above rule?
 
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Yes, that is possible in LS coupling. However, the selection rules dictate that the total angular momentum should change by $\Delta J=\pm 1$, so the only way for this transition to be allowed is if $\Delta m_j=0$ or $\pm 2$, with $\Delta m_l=\pm 1$ and $\Delta m_s=\pm 1$.
 

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