University Quantum Mechanics II · The Fine Structure of Hydrogen and the Lamb Shift · 8.5

The l = 0 problem and the Darwin term

The spin-orbit formula has a hole exactly where the electron comes closest to the nucleus. This lesson shows why the rest-frame argument cannot fill it, brings in the term the Dirac equation supplies instead, and evaluates it: a contact interaction that lifts s states, and only s states, by the amount the fine structure needs.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Spin-orbit coupling, derived from the field an electron sees in its rest frame, breaks down at l = 0. There ⟨L⋅S⟩ = 0 while ⟨1/r³⟩ diverges, because an s electron has finite density at the nucleus, so the shift is 0 × ∞; the classical picture, a radial orbit plunging through a point charge, has nothing to offer. The Dirac equation fills the gap.

Reduced to order (v/c)², a step this course quotes rather than performs, it gives the kinetic and spin-orbit corrections already met plus a third, the Darwin term (ħ²/8m²c²)∇²V, which carries no spin and no momentum. By Poisson's equation ∇²V follows the charge density that makes V, so for a point nucleus the term collapses onto the origin as a delta function and moves a level in proportion to |ψ(0)|². Only s wavefunctions reach the origin, so only s levels move, upward by mc²(Zα)⁴/(2n³): 0.7245 meV for hydrogen's 1S and one eighth of that for 2S.

The size is plausible if the electron's position is smeared over about its Compton wavelength ħ/mc = 386 fm, and it equals what the j = l + ½ spin-orbit formula gives at l = 0 once l cancels, which is what lets the fine structure depend on n and j alone. The smearing picture fixes only the form and the order; the coefficient is the Dirac equation's. And the model stops at order α⁴mc² with a point nucleus: the 1057.8 MHz that separates 2S₁/₂ from 2P₁/₂ lies beyond it.

Simple definition: The Darwin term is the correction (ħ²/8m²c²)∇²V that the Dirac equation adds to the hydrogen Hamiltonian at order (v/c)²; for a Coulomb potential it is a contact interaction at the nucleus, so it shifts only states whose wavefunction is nonzero there.
Example: In hydrogen's 1S state |ψ(0)|² = 1/(πa³), so the Darwin term raises the level by mc²α⁴/2 = 0.7245 meV, or 175 GHz; 2S rises by 90.6 μeV, one eighth as much, and every 2P state by exactly zero.

Spin-orbit for j = l + ½, before cancelling lEₛₒ = (Eₙ²/mc²) n l / [l(l + ½)(l + 1)]

At l = 0 it reads 0/0: L⋅S annihilates an s state while ⟨1/r³⟩ diverges as ∫dr/r, because Rₙ₀(0) ≠ 0. Cancelling l first gives 2nEₙ²/mc², a value the argument cannot justify.

Eₙ = −mc²(Zα)²/(2n²) in J. The upper l is from ⟨L⋅S⟩ = lħ²/2, the lower from ⟨1/r³⟩ = Z³/[l(l + ½)(l + 1)n³a³]

Darwin term, quoted from the Dirac equationH_D = (ħ²/8m²c²) ∇²V

From the order-(v/c)² reduction of the Dirac equation, quoted not derived. No spin and no momentum: it acts wherever ∇²V ≠ 0, which by Poisson's equation is where charge sits.

ħ in J s, m in kg, c in m s⁻¹, V the potential energy in J; ħ/mc = 386.16 fm is a length, so H_D is in J

Coulomb contact form, derived hereH_D = (πħ²/2m²c²)(Ze²/4πε₀) δ³(r)

The whole correction sits at the nucleus, so a state feels it only through |ψ(0)|². Near the origin ψₙₗₘ ∝ rˡ, so every l ≥ 1 state gets exactly zero.

From ∇²(1/r) = −4πδ³(r). Ze²/4πε₀ in J m, δ³(r) in m⁻³

Darwin shift of an ns level, derived hereE_D = (πħ²/2m²c²)(Ze²/4πε₀)|ψₙ₀₀(0)|² = mc²(Zα)⁴/(2n³)

Upward, falling as 1/n³ with the density at the nucleus and growing as Z⁴. It equals 2nEₙ²/mc², the value the j = l + ½ spin-orbit formula takes at l = 0.

|ψₙ₀₀(0)|² = Z³/(πn³a³) from Unit 2. Hydrogen: 1S +724.5 μeV (175.2 GHz), 2S +90.57 μeV, 3S +26.83 μeV

Size in the α hierarchyE_D/|Eₙ| = (Zα)²/n, for l = 0

An order-α² correction to the Bohr energy, the same order as the kinetic and spin-orbit terms, so no consistent calculation can keep those two and drop this one.

Dimensionless. Hydrogen: 5.33 × 10⁻⁵ for 1S, 2.66 × 10⁻⁵ for 2S

Smeared-electron reading, a heuristic⟨V(r + δr)⟩ = V(r) + (⟨δr²⟩/6) ∇²V + …

Reproduces the form and order of H_D, but the ¾ is read off the quoted Dirac result, not predicted. Outside all charge ∇²V = 0, so to this order smearing changes nothing there.

δr an isotropic random displacement, ⟨δr²⟩ in m². Matching H_D needs ⟨δr²⟩ = ¾(ħ/mc)², an rms of 334 fm

Why the rest-frame argument has no answer at l = 0

Spin-orbit coupling came from the magnetic field B = −v × E/c² that the electron meets in its own rest frame. In the Coulomb field it is proportional to L/r³, so the shift factors into an angular part and a radial part, Eₛₒ ∝ ⟨L⋅S⟩⟨1/r³⟩. For j = l + ½ the angular part is lħ²/2, the radial part is Z³/[l(l + ½)(l + 1)n³a³], and the two factors of l cancel. At l = 0 they cannot. L⋅S annihilates an s state, so the first factor is exactly zero, while the radial integral ∫|Rₙ₀|² r⁻¹ dr diverges logarithmically at the origin, because Rₙ₀(0) is finite. The shift is 0 × ∞. The classical picture offers no rescue: an orbit with L = 0 is a straight line through the nucleus, where the field the argument relies on is infinite. Setting the spin-orbit shift of an s state to zero is defensible for the operator, but then something else has to supply the s-state shift that experiment demands.

What the Dirac equation adds, quoted

The Dirac equation treats the electron relativistically from the start. Expanding its Hamiltonian in powers of v/c, by the Foldy–Wouthuysen sequence of unitary transformations on four-component spinors, gives to order (v/c)²: H = mc² + p²/2m + V − p⁴/8m³c² + (1/2m²c²)(1/r)(dV/dr) S⋅L + (ħ²/8m²c²)∇²V. This course quotes that result rather than deriving it. The p⁴ term is the kinetic correction, the next is spin-orbit coupling with the Thomas factor already inside it, and the last is new: the Darwin term. Check it before using it. ħ/mc is a length, the reduced Compton wavelength 386.16 fm, and ∇²V is an energy per length squared, so H_D is an energy. In an atom ∇²V is of size V/a², so H_D/V is of size (ħ/mca)² = α²: the term enters at order α⁴mc², like the other two, and cannot be dropped while they are kept. It contains neither S nor p, so it treats both spin states alike.

For Coulomb, a contact term at the nucleus

Put in the Coulomb potential energy V = −Ze²/(4πε₀r). The Laplacian of 1/r vanishes everywhere except the origin, and Gauss's theorem fixes its weight there: by spherical symmetry the flux of ∇(1/r) = −r̂/r² through a sphere of any radius centred on the origin is −4π, so ∇²(1/r) = −4πδ³(r). So ∇²V = (Ze²/ε₀)δ³(r), which is Poisson's equation saying that ∇²V follows the charge that produces V, here all at one point. The Darwin term becomes H_D = (πħ²/2m²c²)(Ze²/4πε₀)δ³(r). Choosing a basis costs nothing: H_D depends on r alone and carries no spin, so it commutes with L², L_z, S² and J, and it is already diagonal in the coupled basis used for spin-orbit. Its first-order shift is (πħ²/2m²c²)(Ze²/4πε₀)|ψ(0)|². Near the origin the wavefunction goes as rˡ, so |ψ(0)|² = 0 for every l ≥ 1 state and those levels do not move at all. Only l = 0 survives, and Unit 2 supplies |ψₙ₀₀(0)|² = Z³/(πn³a³).

Evaluating the s-state shift, and checking it

Substitute, using e²/4πε₀ = αħc and a = ħ/(mcα): E_D = (ħ²/2m²c²)(Zαħc)(Z³/n³)(mcα/ħ)³ = mc²(Zα)⁴/(2n³). For hydrogen that is +724.5 μeV (175.2 GHz) for 1S, +90.57 μeV (21.90 GHz) for 2S and +26.83 μeV (6.49 GHz) for 3S. Four checks. Dimensions: mc² times a pure number. Order: E_D/|Eₙ| = (Zα)²/n, 5.3 × 10⁻⁵ for 1S, an α⁴mc² effect. Scaling: 1/n³ is the density at the nucleus, not the 1/n² of the energy, and Z⁴ is three powers of Z from |ψ(0)|² and one from the charge. The ground state: 1S has no spin-orbit shift, and the kinetic correction −5mc²α⁴/8 = −905.7 μeV plus the Darwin +724.5 μeV gives −181.1 μeV, the α⁴ term of the quoted Dirac ground-state energy mc²√(1 − α²) = mc² − mc²α²/2 − mc²α⁴/8 − …. Leave the Darwin term out and that check fails by a factor of five.

The same number as spin-orbit at l = 0

Return to the j = l + ½ formula and cancel l before setting l = 0: Eₛₒ = (Eₙ²/mc²) n/[(l + ½)(l + 1)] becomes 2nEₙ²/mc², and with Eₙ = −mc²α²/(2n²) that is mc²α⁴/(2n³), the Darwin shift exactly. The equality is not a derivation. The spin-orbit operator gives zero on an s state, and the Darwin term has no spin in it. It is a property of the Dirac equation, whose energies depend on n and j only, and it is why the s levels land on the same formula as the l ≥ 1 levels of equal j, the result of the next topic. Measurement decides between the two readings of 0/0. With no s-state spin-orbit shift and no Darwin term, hydrogen's 2S₁/₂ would sit 4E₂²/mc² = 90.57 μeV, or 21.90 GHz, below 2P₁/₂; with the Darwin term the two coincide. Lamb and Retherford found 2S₁/₂ about 1 GHz above 2P₁/₂, and the modern value is 1.058 GHz. The Darwin term closes 21.90 GHz of a 22.96 GHz gap, and the 1.06 GHz left over is the Lamb shift, which belongs to quantum electrodynamics.

Reading it as a smeared electron

Why should an electron respond to ∇²V? A heuristic, not a derivation: a Dirac electron cannot be localised more finely than about its Compton wavelength ħ/mc without mixing in negative-energy states, so its effective position jitters over that scale. Average V over a small random displacement δr with no preferred direction. The first-order term averages to zero and the second gives ⟨V(r + δr)⟩ = V(r) + (⟨δr²⟩/6)∇²V, the Darwin form. It matches the quoted coefficient if ⟨δr²⟩ = ¾(ħ/mc)², an rms of 334 fm, which is (√3/2)α times the Bohr radius. Two features follow at once. Outside all charge V obeys Laplace's equation, and the average of such a function over a sphere lying wholly in that region equals its value at the centre, so smearing changes V only where the averaging sphere reaches the nucleus, within δr of it; to the order kept, that is a contact term at the nucleus itself. There it blunts the −1/r spike, which raises the energy. The picture gives the form, the sign and the order; the ¾ comes from the Dirac equation.

Change one variable at a time

Make the relationship visible.

Interactive model

Smearing radius δ200 fm

Principal quantum number n1

Orbital quantum number l (at most n − 1)0

Drag δ to 334 fm, where the shell's shift equals the quoted Darwin term. Then change n: both scale as 1/n³, so the match holds for every n. With n = 2 or 3, set l = 1 and the integrand vanishes, because a p state has no density at the nucleus.

SHELL-MODEL SHIFT259.1 μeV

DARWIN TERM (QUOTED)724.5 μeV

SHELL ÷ DARWIN FOR l = 00.358

DARWIN SHIFT AS FREQUENCY175.19 GHz

Live interpretationSHELL-MODEL SHIFT: 259.1 μeV. DARWIN TERM (QUOTED): 724.5 μeV. SHELL ÷ DARWIN FOR l = 0: 0.358. DARWIN SHIFT AS FREQUENCY: 175.19 GHz

Catch the common trap

Explain before calculating.

Misconception check

Faced with 0/0 at l = 0, it is tempting to settle it by fiat, in one of two ways. Setting the s-state spin-orbit shift to zero because L = 0, and stopping there, leaves only the kinetic correction: it makes hydrogen's 1S fine-structure shift five times too large, −905.7 μeV against the Dirac value of −181.1 μeV, and puts 2S₁/₂ 21.9 GHz below 2P₁/₂, where measurement finds it 1.06 GHz above. Taking the l → 0 limit of the j = l + ½ formula and calling the result spin-orbit coupling for s states gets the right number for the wrong reason. The term that shifts s states contains no spin operator and no angular momentum. It comes from ∇²V, the charge density at the nucleus, and it would shift any state with density there. The numbers agree because the Dirac equation makes the energy a function of n and j alone, not because an s electron has spin-orbit coupling.

Within the n = 3 level of hydrogen, which states does the Darwin term shift, and by how much? Take mc²α⁴ = 1.449 meV.

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyEvaluate the Darwin shift of hydrogen's 2S level directly from the contact term, using |ψ₂₀₀(0)|² = 1/(8πa³), ħ/mc = 3.8616 × 10⁻⁴ nm, e²/4πε₀ = 1.4400 eV nm and a = 0.052918 nm. Give it in μeV and in GHz, and compare it with |E₂| = 3.4014 eV.

The contact term gives E_D = (πħ²/2m²c²)(e²/4πε₀)|ψ₂₀₀(0)|² = (ħ/mc)²(e²/4πε₀)/(16a³): the π cancels and 2 × 8 makes the 16.
Numerator: (3.8616 × 10⁻⁴ nm)² = 1.4912 × 10⁻⁷ nm², times 1.4400 eV nm, is 2.1473 × 10⁻⁷ eV nm³. Denominator: 16 × (0.052918 nm)³ = 2.3710 × 10⁻³ nm³.
E_D = 2.1473 × 10⁻⁷ ÷ 2.3710 × 10⁻³ eV = 9.057 × 10⁻⁵ eV = 90.57 μeV. Check against the closed form: mc²α⁴/(2 × 2³) = 1.4490 meV ÷ 16 = 90.56 μeV.
As a frequency, ν = E_D/h = 9.057 × 10⁻⁵ eV ÷ 4.1357 × 10⁻¹⁵ eV s = 21.90 GHz. Relative to the binding, 90.57 μeV ÷ 3.4014 eV = 2.663 × 10⁻⁵ = α²/2, the order-α² size expected.
Every 2P state has ψ(0) = 0, so the Darwin term leaves 2P₁/₂ and 2P₃/₂ exactly where the kinetic and spin-orbit corrections put them.

AnswerE_D(2S) = +90.6 μeV, or 21.9 GHz, which is 2.66 × 10⁻⁵ of |E₂|, that is α²/2; the 2P states are not shifted at all.

MediumFor He⁺ (Z = 2, nucleus treated as infinitely heavy), evaluate at n = 2 the Darwin shift of 2S and the value the j = l + ½ spin-orbit formula takes at l = 0, and show they agree. Then use the kinetic and spin-orbit results of the earlier topics to find how far 2S₁/₂ would sit from 2P₁/₂ without the Darwin term, and compare with the measured separation, 2S₁/₂ above 2P₁/₂ by 14.04 GHz.

E₂ = −Z² × 13.6057 eV/2² = −13.6057 eV, so E₂²/mc² = 185.115 eV² ÷ 510 999 eV = 362.26 μeV.
Spin-orbit formula at l = 0 once l cancels: 2nE₂²/mc² = 4 × 362.26 μeV = 1449.0 μeV. Darwin term: mc²(Zα)⁴/(2n³) = 16 × 1.4490 meV ÷ 16 = 1449.0 μeV. They agree, as the Dirac equation requires.
Kinetic correction, quoted from the earlier topic as −(Eₙ²/2mc²)[4n/(l + ½) − 3]: 2S gets −181.13 μeV × 13 = −2354.7 μeV, and 2P gets −181.13 μeV × (16/3 − 3) = −422.6 μeV.
Spin-orbit for 2P₁/₂, where j(j + 1) − l(l + 1) − ¾ = −2: 362.26 μeV × 2 × (−2)/(1 × 1.5 × 2) = −483.0 μeV, so 2P₁/₂ totals −905.7 μeV. Without the Darwin term 2S₁/₂ sits at −2354.7 μeV, 1449.0 μeV below it; adding +1449.0 μeV brings it to −905.7 μeV exactly.
As a frequency 1449.0 μeV is 350.4 GHz. The measured 14.04 GHz, with 2S₁/₂ on top, is 4.0% of that: the Darwin term closes 350 GHz of the gap, and the Lamb shift, from quantum electrodynamics, accounts for the rest.

AnswerBoth equal 1.449 meV, or 350.4 GHz. Without the Darwin term 2S₁/₂ would lie 350 GHz below 2P₁/₂; with it the two coincide, and the measured 14.04 GHz is the Lamb shift.

HardModel the Darwin term by spreading the electron's charge uniformly over a spherical shell of radius δ about its nominal position. (a) Find the resulting shift of hydrogen's 1S level. (b) Choose δ to reproduce the quoted 1S Darwin shift, 0.7245 meV. (c) The proton's own charge is spread with rms radius rₚ = 0.8414 fm; apply the same averaging to it, estimate the finite-size shift of 1S, and compare the two lengths.

Write ke² = e²/4πε₀. By the shell theorem the averaged potential energy is −ke²/r wherever the shell does not enclose the proton and −ke²/δ wherever it encloses it, so smearing changes V only within δ of the nucleus, by ΔV = ke²(1/r − 1/δ), which is positive.
δ is a few hundred fm and a = 5.2918 × 10⁴ fm, so |ψ₁₀₀|² stays at 1/(πa³) to within 2% across the region. Then ΔE = (1/πa³)∫₀^δ ke²(1/r − 1/δ) 4πr² dr = (4ke²/a³)(δ²/2 − δ²/3) = (2/3)ke²δ²/a³.
Set this equal to 0.7245 meV, with a³ = 1.4818 × 10¹⁴ fm³ and ke² = 1.4400 × 10⁶ eV fm: δ² = 1.5 × 0.7245 × 10⁻³ eV × a³/ke² = 1.118 × 10⁵ fm², so δ = 334 fm. That is 0.866 ħ/mc, or ⟨δr²⟩ = ¾(ħ/mc)², the value the heuristic needs.
A spherical charge distribution is a stack of shells and the shift is linear in δ², so for the proton δ² becomes its mean square radius: ΔE_size = (2/3)ke²rₚ²/a³ = 0.7245 meV × (0.8414/334.4)² = 0.7245 meV × 6.33 × 10⁻⁶ = 4.59 neV, or 1.11 MHz.
The electron's effective smearing is 397 times the proton's radius, so the Darwin shift is 1.58 × 10⁵ times the finite-size shift. The rₚ used is a measured value; the averaging that gives both shifts is derived here to leading order in δ/a.

Answer(a) ΔE = (2/3)ke²δ²/a³. (b) δ = 334 fm, 0.866 ħ/mc. (c) The proton's size lifts 1S by 4.59 neV (1.11 MHz), 1.6 × 10⁵ times less than the Darwin shift.