Questions: Spin-Orbit Coupling

Question 1

Spin-orbit coupling is added to the hydrogen Hamiltonian. Which set of quantum numbers correctly describes the new good quantum numbers for the electron?

• A. n, l, m_l, m_s — the same as the unperturbed hydrogen atom
• B. n, l, s, j, m_j — because H_SO commutes with J² and J_z but not with L_z or S_z individually
• C. n, j, m_j only — because spin-orbit coupling destroys all other quantum numbers
• D. n, l, m_l, j — a hybrid set that mixes the old and new bases

Answer: (B)

Spin-orbit coupling H_SO ∝ L·S does not commute with L_z or S_z individually, so m_l and m_s are no longer conserved. However, L·S is rotationally invariant (it is a scalar), so it commutes with J² = (L + S)² and with J_z = L_z + S_z. The quantum numbers n (from the radial equation), l (from L²), s = 1/2 (unchanged), j (from J²), and m_j (from J_z) all remain good. The key change is replacing the pair (m_l, m_s) with the pair (j, m_j).

Question 2

Why is the identity L·S = (1/2)(J² − L² − S²) essential for computing the energy correction from spin-orbit coupling?

• A. It converts L·S into a form involving only position operators, making it easier to compute expectation values
• B. It expresses L·S in terms of J², L², and S², whose eigenvalues ℏ²j(j+1), ℏ²l(l+1), ℏ²s(s+1) are known in the |n,l,s,j,m_j⟩ basis, giving a direct formula for ΔE_SO
• C. It eliminates the coupling between spin and orbital motion, reducing the problem to two independent angular momenta
• D. It is only needed for states with l > 1; for l = 0 and l = 1 states, L·S can be computed directly

Answer: (B)

The identity L·S = (1/2)(J² − L² − S²) is powerful because J², L², and S² all commute with H_SO and with each other in the coupled basis |n,l,s,j,m_j⟩. Their eigenvalues are ℏ²j(j+1), ℏ²l(l+1), ℏ²s(s+1) respectively. Therefore ⟨L·S⟩ = (ℏ²/2)[j(j+1) − l(l+1) − s(s+1)], which directly gives the perturbative energy correction without any complicated matrix computation. Without this identity, computing ⟨L·S⟩ would require evaluating off-diagonal matrix elements between m_l and m_s states.

Question 3

The spin-orbit energy correction ΔE_SO takes different values for j = l + 1/2 and j = l − 1/2 states with the same n and l, splitting what was previously a degenerate level.

• T. True
• F. False

Answer: True

For j = l + 1/2: j(j+1) − l(l+1) − s(s+1) = (l + 1/2)(l + 3/2) − l(l+1) − 3/4 = l. For j = l − 1/2: the analogous calculation gives −(l+1). Since l ≠ −(l+1) for l > 0, the two j values give different energy shifts, breaking the degeneracy. For example, the hydrogen 2p level splits into 2p₁/₂ (j = 1/2) and 2p₃/₂ (j = 3/2) with different energies. This fine structure splitting is one of the key experimental consequences of spin-orbit coupling.

Question 4

In the electron's rest frame, the orbiting proton creates an electric field at the electron's location, and it is the interaction of this electric field with the electron's charge that produces spin-orbit coupling.

• T. True
• F. False

Answer: False

The electric field produced by the orbiting proton exerts a Coulomb force on the electron's charge — this is the dominant (unperturbed) Hamiltonian, not the spin-orbit coupling. Spin-orbit coupling arises from the *magnetic* field that the orbiting proton produces in the electron's rest frame. A moving charge creates a magnetic field, and the proton's orbital motion (as seen from the electron's frame) generates a magnetic field B at the electron's location. This magnetic field then interacts with the electron's *spin magnetic moment* μ_S, giving H_SO ∝ μ_S · B ∝ L·S.

Question 5

Explain why L_z and S_z are no longer individually conserved when spin-orbit coupling is present, and what quantity is conserved instead.

Model answer: Spin-orbit coupling adds H_SO ∝ L·S to the Hamiltonian. This term does not commute with L_z or S_z individually: [L·S, L_z] = [L_xS_x + L_yS_y + L_zS_z, L_z] ≠ 0, because L_x and L_y do not commute with L_z. Physically, H_SO mixes states with different (m_l, m_s) pairs that have the same m_j = m_l + m_s, so individual m_l and m_s fluctuate while their sum is preserved. What is conserved is the total angular momentum: J² and J_z commute with H_SO because L·S is rotationally invariant. The good quantum numbers are therefore j and m_j rather than m_l and m_s.

The semiclassical picture is that L and S each precess around the fixed total angular momentum vector J. Neither L_z nor S_z is individually stable — they oscillate as L and S precess — but J_z = L_z + S_z remains constant. This precession picture explains why (l, s, j, m_j) are the right quantum numbers: j characterizes the magnitude of J (which is fixed by the energy eigenstate), and m_j characterizes its projection on the z-axis (which is conserved). The individual projections m_l and m_s average to well-defined values only in special cases.