Questions: Spin-Orbit Coupling and Fine Structure

Question 1

An electron in the 2p state of hydrogen (orbital quantum number l = 1) has spin s = 1/2. How many distinct energy levels does spin-orbit coupling produce from this single 2p manifold, and what are the corresponding total angular momentum quantum numbers?

• A. One energy level — spin-orbit coupling is too small in hydrogen to resolve the 2p states
• B. Two energy levels — j = l + s = 3/2 and j = l − s = 1/2 have different expectation values of L·S and therefore different energies
• C. Three energy levels — corresponding to mⱼ = −1, 0, +1 projections of the total angular momentum
• D. Six energy levels — each of the six 2p states (mₗ = −1, 0, +1 and mₛ = ±1/2) splits independently

Answer: (B)

Spin-orbit coupling couples L and S into J = L + S, making j the good quantum number. For l = 1 and s = 1/2, the possible total angular momenta are j = 3/2 and j = 1/2, giving two distinct energy levels (the 2P₃/₂ and 2P₁/₂ states). The energy split depends on ⟨L·S⟩, which equals [j(j+1) − l(l+1) − s(s+1)]/2 — different for j = 3/2 and j = 1/2. Options C and D confuse magnetic sublevels (mⱼ values, which are degenerate in the absence of a magnetic field) with the spin-orbit split itself.

Question 2

A spectroscopist studying lead (Z = 82) finds that the orbital angular momentum quantum number L of individual electrons cannot be assigned well-defined values. What is the physical reason?

• A. Lead has too many electrons to track individual quantum numbers — the many-body problem prevents meaningful assignment
• B. The relativistic speeds of electrons in heavy atoms destroy orbital angular momentum conservation
• C. In j-j coupling (valid for heavy atoms), strong spin-orbit interaction so thoroughly mixes L and S for each electron that only j per electron is a good quantum number, not l and s individually
• D. The Pauli exclusion principle prevents electrons in the same shell from having well-defined orbital quantum numbers

Answer: (C)

The coupling regime depends on the relative strength of spin-orbit coupling versus inter-electron interactions. In light atoms (L-S coupling), inter-electron forces dominate and the individual orbital angular momenta couple into a total L, while spins couple into total S; then L and S couple weakly via spin-orbit. In heavy atoms (j-j coupling), spin-orbit coupling ∝ Z⁴ dominates for each electron, so l and s couple into j per electron before inter-electron effects mix them. The result is that L and S are no longer good quantum numbers — each electron is better described by its own j. This isn't a measurement problem; it reflects the actual eigenstates of the Hamiltonian.

Question 3

The physical origin of spin-orbit coupling can be understood by transforming to the electron's rest frame, where the orbiting nucleus creates a magnetic field that interacts with the electron's spin magnetic moment.

• T. True
• F. False

Answer: True

This is the correct physical picture. In the lab frame, the electron orbits the nucleus (a stationary positive charge). In the electron's own rest frame, the nucleus appears to orbit it, creating a circulating current and therefore a magnetic field. The electron's intrinsic spin magnetic moment sits inside this field, and the interaction energy depends on whether the spin is aligned or anti-aligned with the orbital angular momentum. The Hamiltonian H_SO = ξ(r)L·S captures exactly this coupling, with ξ(r) > 0 encoding the radial dependence of the effective magnetic field.

Question 4

In the L-S (Russell-Saunders) coupling scheme, spin and orbital angular momenta combine as described above; this scheme is valid for heavy atoms where spin-orbit coupling is large, and breaks down in light atoms where it is weak.

• T. True
• F. False

Answer: False

This is exactly backwards. L-S coupling is valid for *light* atoms (low Z), where spin-orbit interaction is weak compared to inter-electron Coulomb repulsion. In that regime, orbital angular momenta couple to form total L, spins couple to form total S, and then L and S combine weakly via spin-orbit. For *heavy* atoms (high Z), the spin-orbit term ∝ Z⁴ dominates, and L-S coupling breaks down in favor of j-j coupling. The confusion often arises because the sodium D-line doublet (a light atom) is the paradigm example of spin-orbit splitting, which seems to suggest strong coupling, but the splitting is still a small perturbation in Na compared to heavy-atom behavior.

Question 5

Why does spin-orbit coupling make J = L + S the 'good quantum number' rather than L and S separately? Explain both the physical picture and the mathematical consequence using the L·S interaction.

Model answer: Physically: spin-orbit coupling means the electron's spin magnetic moment interacts with the magnetic field generated by its own orbital motion. This interaction causes spin and orbital angular momenta to precess around their sum J — neither L nor S points in a fixed direction, so neither has a conserved z-component. Only the total angular momentum J = L + S is conserved. Mathematically: the interaction Hamiltonian H_SO = ξ(r)L·S does not commute with Lz or Sz individually, but it does commute with J², L², and S². Using J² = L² + S² + 2L·S gives L·S = (J² − L² − S²)/2, which is simultaneously diagonal in the basis |n, l, s, j, mⱼ⟩. So j (and mⱼ) labels the energy eigenstates; mₗ and mₛ individually do not.

This is the general principle: when two angular momenta interact, their sum is the conserved quantity and the good quantum number. The coupling 'locks' L and S into precessing around J together, rather than precessing independently around a fixed axis. In spectroscopic notation, the label ²S+1L_J (e.g., ²P₃/₂) encodes exactly this: the superscript gives the spin multiplicity 2S+1, the letter encodes L (S, P, D, F...), and the subscript gives J. This notation only makes sense because J is the conserved quantity in the presence of spin-orbit coupling.