Spin–orbit coupling

The previous section ended on a cliffhanger: the six states of a p electron regroup into a $j=\tfrac32$ family and a $j=\tfrac12$ family, but nothing in the Hamiltonian tells them apart — same energy, so the regrouping is a mere paper exercise.

Yet experiment says the grouping is real. Take the easiest observation there is: the spectrum of a sodium lamp (the yellow glow of old street lights). Under a good enough spectrometer the famous yellow line is actually two lines, 589.0 nm and 589.6 nm, split by about $2.1\times10^{-3}\ \text{eV}$ — roughly one part in a thousand of the level energy itself. Hydrogen’s lines carry the same kind of finer hairlines (about $4.5\times10^{-5}\ \text{eV}$), hence the name fine structure. A doublet means the p level really has split in two — one line each for the $j=\tfrac32$ and $j=\tfrac12$ families.

What force divided the estate between the two families? The answer is already sitting in the toolbox of the past few sections: a magnetic moment, a magnetic field, $\vec\mu\cdot\vec B$ — except that this time the field is not applied by an experimenter. The atom makes it itself.

Take the electron’s seat

The textbook picture: the nucleus sits at the centre, the electron circles it. Now switch to the electron’s seat: the nucleus circles the electron. The nucleus is positively charged, a circling positive charge is a loop of current, and a current makes a magnetic field — so in the electron’s own frame it is bathed in a field perpendicular to the orbital plane, i.e. along the orbital angular momentum $\vec L$ (faster circling and a tighter loop mean a stronger field; quantitatively $\vec B\propto\vec L$).

And the electron happens to carry a magnetic moment $\vec\mu_s=-\frac{g_e\mu_B}{\hbar}\hat{\vec S}$ (the protagonist of section 5.5). A moment in a field has energy $-\vec\mu_s\cdot\vec B$. Combine the two proportionalities:

$$\hat H_{\text{SO}}=\xi(r)\,\hat{\vec L}\cdot\hat{\vec S}\tag{5.7.1}$$

The coupling must take the form “field made by the orbit” dotted into “moment carried by the spin” — hence the name spin–orbit coupling. The coefficient $\xi(r)$ depends on the electron’s distance from the nucleus; its precise form only becomes useful next chapter, together with the hydrogen wavefunctions. This section answers the two more pressing questions first: how big is this term, and what does it change?

How big is the effect

Order-of-magnitude estimate: where α² comes frombasic~8 minexpand

Use Bohr-model numbers as parts (we want the magnitude, not the decimal places).

Step 1: the field the electron sees. Estimate the field at the centre of a current loop from the Biot–Savart law: a nuclear charge $+e$ circling at speed $v$ on radius $r$ gives

$$B\sim\frac{\mu_0}{4\pi}\,\frac{ev}{r^2}\tag{5.7.2}$$

Insert hydrogen ground-state numbers: $v=\alpha c\approx2.2\times10^6\ \text{m/s}$ ($\alpha\approx\tfrac1{137}$ is the fine-structure constant; in the Bohr model the first-orbit speed is exactly $\alpha c$), and $r=a_0\approx5.3\times10^{-11}\ \text{m}$:

$$B\sim10^{-7}\times\frac{1.6\times10^{-19}\times2.2\times10^6} {(5.3\times10^{-11})^2}\ \text{T}\approx13\ \text{T}\tag{5.7.3}$$

A dozen tesla — several times stronger than a hospital MRI’s 3 T. The inside of an atom is a rather ferocious magnetic environment.

Step 2: the moment’s energy in that field. The electron’s moment is about one Bohr magneton, $\mu_B\approx5.8\times10^{-5}\ \text{eV/T}$:

$$\Delta E\sim\mu_B B\approx5.8\times10^{-5}\times13\ \text{eV} \approx7\times10^{-4}\ \text{eV}\tag{5.7.4}$$

Step 3: say it dimensionlessly. Compare with the level energy itself (the Rydberg energy, $13.6\ \text{eV}$):

$$\frac{\Delta E}{E_1}\sim\frac{7\times10^{-4}}{13.6}\approx5\times10^{-5} \approx\alpha^2\tag{5.7.5}$$

Not a coincidence but an identity: substitute $B\sim\frac{\mu_0}{4\pi}\frac{e\,\alpha c}{a_0^2}$ and $\mu_B=\frac{e\hbar}{2m_e}$ term by term and simplify, and you get exactly $\mu_B B=\alpha^2\times13.6\ \text{eV}$ (each source contributes one $\alpha$: the speed is $\alpha c$ — the scale of relativistic corrections; and the magnetic field is itself a relativistic effect of moving charge).

Conclusion:

$$\Delta E_{\text{SO}}\sim\alpha^2\,E_{\text{Ry}}\approx7\times10^{-4}\ \text{eV} \quad(\text{hydrogen ground-state scale; shrinking as}\ 1/n^3\ \text{for larger}\ n)\tag{5.7.6}$$

This is exactly how the fine-structure constant $\alpha$ got its name: its square is the scale ratio of the fine structure to the main levels.

What it changes: a changeover of good quantum numbers

With the magnitude settled, turn to the structural consequences — the real payload of this section. Without $\hat H_{\text{SO}}$, both $\hat L_z$ and $\hat S_z$ commute with the Hamiltonian, and $m_l,m_s$ are good quantum numbers (conserved labels, safe for naming levels). What happens once $\hat{\vec L}\cdot\hat{\vec S}$ is added?

Who still conserves: a commutator check-upadvanced~7 minexpand

Expand the dot product: $\hat{\vec L}\cdot\hat{\vec S} =\hat L_x\hat S_x+\hat L_y\hat S_y+\hat L_z\hat S_z$.

Check $\hat L_z$. Orbital and spin operators commute with each other (two separate ledgers), so all that matters is $\hat L_z$ against the various $\hat L_i$:

$$[\hat L_z,\hat{\vec L}\cdot\hat{\vec S}] =[\hat L_z,\hat L_x]\hat S_x+[\hat L_z,\hat L_y]\hat S_y =\ii\hbar\big(\hat L_y\hat S_x-\hat L_x\hat S_y\big)\neq0\tag{5.7.7}$$

$\hat L_z$ is no longer conserved. The physical picture: the effective field generated by the spin points along $\vec S$, so $\vec L$ precesses about it (last section’s Larmor picture run in reverse); likewise $\vec S$ precesses about $\vec L$. The two vectors drag each other around, and each one’s z component wobbles. $\hat S_z$ fails the check-up too:

$$[\hat S_z,\hat{\vec L}\cdot\hat{\vec S}] =\ii\hbar\big(\hat L_x\hat S_y-\hat L_y\hat S_x\big)\neq0\tag{5.7.8}$$

Check $\hat J_z=\hat L_z+\hat S_z$. Add the two lines — and they cancel term by term:

$$[\hat J_z,\hat{\vec L}\cdot\hat{\vec S}] =\ii\hbar\big(\hat L_y\hat S_x-\hat L_x\hat S_y\big) +\ii\hbar\big(\hat L_x\hat S_y-\hat L_y\hat S_x\big)=0\tag{5.7.9}$$

Neither component is conserved on its own, but their sum is — internal drag leaves the total untouched, like two people shoving each other on a boat while the boat’s total momentum stays fixed.

Check $\hat J^2$. The fastest route is to rewrite the dot product as a difference of squares: $\hat J^2=(\hat{\vec L}+\hat{\vec S})^2=\hat L^2+\hat S^2+2\hat{\vec L}\cdot\hat{\vec S}$, hence

$$\hat{\vec L}\cdot\hat{\vec S}=\frac{\hat J^2-\hat L^2-\hat S^2}{2}\tag{5.7.10}$$

The three terms on the right commute with one another ($\hat L^2$ commutes with every $\hat L_i,\hat S_i$, likewise $\hat S^2$, and $\hat J^2$ is built from them and the dot product). So $\hat J^2,\hat L^2,\hat S^2$ all commute with $\hat H_{\text{SO}}$.

Election result: the good quantum numbers change from $(l,m_l,s,m_s)$ to $(l,s,j,m_j)$ — exactly the labels of last section’s total angular momentum basis. The CG coefficients are no longer a paper exercise: $\ket{j,m_j}$ is now a genuine energy eigenstate, and the product states are not.

The rewrite above also hands us the splitting formula for free. In the state $\ket{l,s,j,m_j}$:

$$\langle\hat{\vec L}\cdot\hat{\vec S}\rangle =\frac{\hbar^2}{2}\big[j(j+1)-l(l+1)-s(s+1)\big]\tag{5.7.11}$$

∎Key formulas

• Coupling term

  $$\hat H_{\text{SO}}=\xi(r)\,\hat{\vec L}\cdot\hat{\vec S}$$

  In the electron frame the circling nucleus is a current → B ∝ L, dotted into the spin moment; Thomas precession fixes a 1/2

• Magnitude

  $$\Delta E_{\text{SO}}\sim\alpha^2 E_{\text{Ry}}\approx 7\times10^{-4}\ \text{eV}$$

  The origin of α ≈ 1/137's name: α² is the scale of the fine structure

• Rewriting the dot product

  $$\hat{\vec L}\cdot\hat{\vec S}=\tfrac12\big(\hat J^2-\hat L^2-\hat S^2\big)$$

  Read off at a glance: J², L², S², J_z conserved; L_z, S_z not

• Level-shift formula

  $$\langle\hat{\vec L}\cdot\hat{\vec S}\rangle=\tfrac{\hbar^2}{2}\big[j(j+1)-l(l+1)-s(s+1)\big]$$

  p states: j=3/2 gets +ħ²/2, j=1/2 gets −ħ² — the doublet splits open here

?Self-check4 questions

1. 1.

   Where does the magnetic field responsible for spin–orbit coupling come from?

   • In the electron's frame, the circling nucleus forms a current loop, producing at the electron a field proportional to L
   • An external field applied by the apparatus
   • The spin magnetic moment of the nucleus
   • The field generated by the electron's own spin

   Submit
2. 2.

   After adding the L·S term, which statements about the conserved quantities (good quantum numbers) are correct? (Select all that apply.)

   Select all that apply

   • L_z and S_z no longer commute with the Hamiltonian; m_l and m_s stop being good quantum numbers
   • J_z = L_z + S_z is still conserved: the two commutators cancel exactly, term by term
   • L² and S² also stop being conserved
   • The energy eigenstates should now be labelled |l, s, j, m_j⟩

   Submit
3. 3.

   The sodium yellow D line is a doublet — but at which level does the splitting responsible for it occur?

   • The upper level 3p: it splits into j = 3/2 and 1/2; the lower level 3s does not split, since l = 0
   • The lower level 3s splits in two
   • Both 3p and 3s split in two
   • The two lines come from two different sodium isotopes

   Submit
4. 4.

   Hydrogen's 2p level splits into 2p_{3/2} and 2p_{1/2} by spin–orbit coupling, with a gap of about 4.5×10⁻⁵ eV. Converting to frequency (divide by h = 4.136×10⁻¹⁵ eV·s), roughly how many GHz? (One decimal place is enough.)

   GHz50% relative tolerance

   Submit

What comes next

This chapter has assembled the full angular momentum kit for a three-dimensional world: the orbital $(l,m)$ with its spherical harmonics, the two-dimensional algebra of spin $\tfrac12$, the CG coefficients that bolt them together, and the commutator check-up that decides which set of labels is the “good” one.

Time to meet quantum mechanics’ flagship problem — the hydrogen atom. The next chapter puts the Coulomb potential into the three-dimensional Schrödinger equation: the angular part is exactly this chapter’s spherical harmonics, not one line needs recomputing; the remaining radial equation will deliver the $E_n=-13.6\ \text{eV}/n^2$ that Bohr once hit by guesswork — this time, derived. And this section’s fine structure will return for the finishing touch: etching onto the broad strokes of $-13.6/n^2$ the delicate hairlines a spectrometer actually sees.