University Quantum Mechanics II · The Fine Structure of Hydrogen and the Lamb Shift · 8.8
The Lamb shift: the 1947 measurement and Welton's estimate
The Dirac equation insists that hydrogen's 2S₁/₂ and 2P₁/₂ share one energy; a 1947 microwave experiment put them a gigahertz apart. Learn how a 4.4 μeV gap was read off a beam of metastable atoms, and how Welton's zero-point jiggle gets its sign, its s-only reach and its 1/n³ before quantum electrodynamics supplies the last digits.
Build the model
Connect the measurement to the mechanism.
Take the Dirac energy at its word and hydrogen's 2S₁/₂ and 2P₁/₂ are exactly degenerate, because the fine structure depends on n and j alone. In 1947 Lamb and Retherford broke the tie with a beam of metastable 2s atoms, a microwave field and a tungsten plate that counted the survivors: 2S₁/₂ lies about 1000 MHz above 2P₁/₂, now measured as 1057.8 MHz, or 4.4 μeV. Welton's model names what the Dirac equation leaves out. The radiation field is itself quantised, each of its modes keeps a zero-point energy ħω/2, and that fluctuating field shakes the electron.
A free electron driven at ω moves as eE/(mω²), and against a field spectrum rising as ω³ every octave contributes equally, so ⟨(δr)²⟩ = (2α/π)(ħ/mc)² ln(ωₘₐₓ/ωₘᵢₙ), an rms jiggle of 70 to 80 fm. Averaging the Coulomb potential over it adds (1/6)⟨(δr)²⟩∇²V, a delta function at the nucleus, so only s states move. They move up, by (4α⁵mc²/(3πn³)) times the logarithm: one power of α below fine structure, falling as 1/n³ like the density at the nucleus, and near a gigahertz for n = 2. The model stops at its inputs.
The zero-point energy is assumed, the free-electron response fails below the atom's own frequencies and above mc²/ħ, and a logarithm between guessed cutoffs fixes the number only to tens of percent. Bethe's 1040 MHz and the quantum-electrodynamic value that matches experiment are quoted, not derived.
- Simple definition
- The Lamb shift is the displacement of a hydrogen level from its Dirac energy by the electron's coupling to the quantised electromagnetic field, seen most directly as a splitting of levels that share n and j but differ in l.
- Example
- 2S₁/₂ lies 1057.8 MHz above 2P₁/₂: hν = 1.0578 × 10⁹ s⁻¹ × 4.1357 × 10⁻¹⁵ eV s = 4.375 × 10⁻⁶ eV, about a tenth of the 10.97 GHz fine-structure gap from 2P₁/₂ up to 2P₃/₂.
Assumed, not derived: it needs the field quantised, each mode a QM I oscillator. Half the energy is magnetic, so ε₀⟨E²⟩, not half of it, is the whole energy density.
Per unit ω, in J s m⁻³: ħω/2 per mode times ω²/(π²c³) modes per unit volume and ω, two polarisations counted.
From mδr̈ = −eE at frequency ω. Squared, its ω⁻⁴ meets the field's ω³ and leaves dω/ω, so every octave shakes the electron equally.
δr in m, E in V m⁻¹, m in kg. Valid where ω far exceeds the atom's own frequencies and ħω stays below mc².
Derived from the two cards above with e² = 4πε₀αħc. A logarithm near 8 gives an rms jiggle of about 74 fm, roughly 1/700 of a Bohr radius.
δr in m; ħ/mc = 3.862 × 10⁻¹³ m and 2α/π = 4.646 × 10⁻³. Only the ratio of the cutoffs enters.
For V = −αħc/r, ∇²V = 4παħc δ³(r): a contact term, felt only by states with ψ(0) ≠ 0, which is why levels with l ≥ 1 do not move.
V in J, ∇²V in J m⁻². Uses ⟨δr⟩ = 0 and ⟨δrᵢδrⱼ⟩ = ⟨(δr)²⟩δᵢⱼ/3 for a jiggle with no preferred direction.
Uses |ψₙ₀₀(0)|² = 1/(πn³a³) and a = ħ/(αmc). Positive, falling as 1/n³, zero for l ≥ 1, and one power of α below the fine structure.
ΔE in J, quoted in μeV. 4α⁵mc²/(3π) = 4.488 μeV; for n = 2 that is 0.561 μeV, or 135.6 MHz, per unit of the logarithm.
Quoted, not derived here: a level whose population decays as e(−t/τ) gives a Lorentzian line of this full width. The 2p decay that empties the beam also broadens the dip, to a tenth of the shift; optical Balmer-α from a 300 K gas is Doppler-broadened to 5.6 GHz.
Δν in Hz, τ₂ₚ in s. The quoted τ₂ₚ = 1.60 ns gives a full width at half maximum of about 100 MHz.
A degeneracy that survives the Dirac equation
The exact Dirac energy of the previous topic depends on n and j only, so it puts 2S₁/₂ and 2P₁/₂ at one energy although their l differ, and the α⁴ fine structure derived term by term earlier in this unit agrees. Any splitting must come from physics outside one electron in a classical Coulomb field. It is small. The 1057.8 MHz measured today is hν = 4.375 μeV, a tenth of the 10.97 GHz interval from 2P₁/₂ up to 2P₃/₂ and 2.3 × 10⁻⁶ of the 1.89 eV Balmer-α photon. Optical work in the 1930s hinted at a shift of this size but could not settle it, because Balmer-α from hydrogen at 300 K is Doppler-broadened to a full width of 5.6 GHz, five times the splitting sought. Driving a transition directly between the n = 2 levels removes that obstacle: Doppler width is proportional to the transition frequency, so at 1 GHz and 300 K it shrinks to about 12 kHz.
Lamb and Retherford count metastable atoms
Their beam began as hydrogen molecules dissociated in a hot tungsten oven. Electron bombardment lifted a small fraction of the atoms to 2s, which is metastable: no single electric-dipole photon connects it to 1s, since both have l = 0, and the two-photon decay left takes 0.12 s (quoted). The atoms crossed a microwave region and struck a tungsten plate. Each 2s atom carries 10.2 eV of excitation, more than tungsten's 4.5 eV work function, and can eject an electron; a 1s atom ejects none, so the plate current counts 2s atoms. When the microwaves match a 2s–2p interval they drive atoms into 2p, which decays to 1s in 1.6 ns (quoted), about 13 μm of flight at 8 km s⁻¹, and the current falls. That same lifetime makes the dip about 100 MHz wide, a tenth of the effect. In practice the frequency was held fixed while a magnetic field Zeeman-shifted the levels through resonance, and the resonances were extrapolated to zero field. The result: 2S₁/₂ lies about 1000 MHz above 2P₁/₂, higher and not lower.
The zero-point field shakes a free electron
Welton's input is assumed rather than derived. Quantise each mode of the radiation field as a QM I oscillator and even its ground state keeps ħω/2. Counting ω²/(π²c³) modes per unit volume and per unit ω, both polarisations included, gives an energy density ħω³/(2π²c³) per unit ω, and since half of it is magnetic this is also ε₀⟨E²⟩ω. Far above the atom's own frequencies the binding force is negligible, so the electron answers as if free: mδr̈ = −eE gives δr = eE/(mω²) at each frequency. Modes with random phases add in mean square, so ⟨(δr)²⟩ = ∫ (e/(mω²))² ħω³/(2π²ε₀c³) dω. The ω³ of the field against the ω⁻⁴ of the response leaves ∫ dω/ω, and with e² = 4πε₀αħc the result is (2α/π)(ħ/mc)² ln(ωₘₐₓ/ωₘᵢₙ). Dimensions check: α and the logarithm are pure numbers and (ħ/mc)² is a length squared. Every octave contributes equally, which is why the cutoffs will survive only inside a logarithm.
Averaging the Coulomb well over the jiggle
The jiggle is fast compared with the orbital motion, so the bound electron effectively samples V at r + δr. Expand: V(r + δr) = V + δr⋅∇V + ½(δr⋅∇)²V + …. Averaged over a jiggle with no preferred direction, ⟨δr⟩ = 0 removes the first-order term and ⟨δrᵢδrⱼ⟩ = ⟨(δr)²⟩δᵢⱼ/3 turns the second into (1/6)⟨(δr)²⟩∇²V. For V = −αħc/r, Poisson's equation gives ∇²V = 4παħc δ³(r), a contact term. Its first-order shift ⟨nlm|ΔV|nlm⟩ needs only |ψ(0)|², which is 1/(πn³a³) for ns states (Unit 2) and zero for every l ≥ 1, since those wavefunctions vanish as rˡ. So ΔE = (1/6)(2α/π)(ħ/mc)² ln(ωₘₐₓ/ωₘᵢₙ) × 4αħc/(n³a³), and a = ħ/(αmc) collapses it to (4α⁵mc²/(3πn³)) ln(ωₘₐₓ/ωₘᵢₙ). The sign is positive: the Coulomb well is deepest at the nucleus, and averaging it over a small sphere there can only make it shallower, so s electrons are bound less and 2S₁/₂ rises.
Only the logarithm knows the cutoffs
The integral diverges at both ends, and physics outside the model cuts it off. Above ħω ≈ mc² the non-relativistic response fails and quantum electrodynamics takes over, so ωₘₐₓ ≈ mc²/ħ. Below the atom's transition frequencies the electron no longer answers as if free, so ωₘᵢₙ is an atomic frequency, but which one is a judgment. The atomic unit ħωₘᵢₙ = α²mc² = 27.2 eV gives ln(1/α²) = 9.84 and a 2S₁/₂ shift of 1335 MHz, 26% above experiment. Bethe's 1947 calculation of the relevant mean excitation energy for 2s, 17.8 Ry (quoted), gives a logarithm of 7.65 and 1038 MHz; Bethe himself reported 1040 MHz. The forgiveness is the point: each decade of ω is worth 135.6 MHz × ln 10 = 312 MHz, so halving either cutoff moves the n = 2 answer by only 94 MHz. The picture is self-consistent too: the rms jiggle is about 73 fm, 1/700 of a Bohr radius, so ψ is effectively constant across one jiggle, which is all the contact form needs.
Derived, assumed, quoted, and checked
Derived here: the jiggle once ħω/2 is granted, the contact form, the restriction to s states, the 1/n³ and the α⁵ scaling. Assumed: the zero-point energy, which needs the field quantised, and both cutoffs. Quoted from quantum electrodynamics at leading order: the 2S₁/₂ self-energy, Welton's form with the logarithm made exact (the mean excitation energy computed accurately as 16.6 Ry) plus a constant, about 1066 MHz; vacuum polarisation, −27 MHz, again on 2S₁/₂ only; and a 13 MHz lowering of 2P₁/₂, mostly from the electron's anomalous magnetic moment. That totals 1052 MHz, and higher-order terms supply the rest of the measured 1057.8 MHz. The checks: α⁵mc² is an energy; 1057.8/10 969 = 0.096, one power of α times a logarithm below fine structure; and 1/n³ turns the 2S₁/₂ shift, 1057.8 − 12.9 ≈ 1045 MHz, into 8.36 GHz for 1S, against a measured 8.17 GHz. The 2% excess is the mean excitation energy, 19.8 Ry for 1s rather than 16.6 Ry.
Change one variable at a time
Make the relationship visible.
Start at ħωₘᵢₙ = 2 Ry, the atomic unit α²mc², and 2S₁/₂ lands 26% above the measured line. Slide ħωₘᵢₙ to Bethe's 17.8 Ry and it reaches 1038 MHz, within 2%. Halving ħωₘₐₓ costs only 94 MHz: a cutoff enters through its logarithm.
ln(ωₘₐₓ/ωₘᵢₙ)9.84
RMS JIGGLE83 fm
2S₁/₂ SHIFT, WELTON1335 MHz
WELTON ÷ MEASURED1.26
Live interpretationln(ωₘₐₓ/ωₘᵢₙ): 9.84. RMS JIGGLE: 83 fm. 2S₁/₂ SHIFT, WELTON: 1335 MHz. WELTON ÷ MEASURED: 1.26
Catch the common trap
Explain before calculating.
Welton's argument puts the 2S₁/₂ shift near 1.04 GHz. What does the same argument predict for the 3S₁/₂ and 3D₅/₂ levels of hydrogen?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyIn a beam apparatus of the Lamb–Retherford type, hydrogen atoms move at 8.0 km s⁻¹ and fly 0.10 m from the excitation region to a tungsten detector. Using the quoted lifetimes τ₂ₛ = 0.12 s and τ₂ₚ = 1.6 ns and tungsten's tabulated 4.5 eV work function, show why the detector current counts 2s atoms and why a microwave resonance appears as a dip.
- Flight time: t = L/v = 0.10 m ÷ 8.0 × 10³ m s⁻¹ = 1.25 × 10⁻⁵ s = 12.5 μs.
- An undisturbed 2s atom can decay only by two-photon emission. The fraction lost in flight is 1 − e(−t/τ₂ₛ) ≈ t/τ₂ₛ = 12.5 × 10⁻⁶ s ÷ 0.12 s = 1.0 × 10⁻⁴, so 99.99% of the metastables arrive. Limiting case: with no microwaves and no stray electric field to mix in 2p, the 2s flux is conserved.
- At the plate each 2s atom can release E₂ − E₁ = 13.6 eV × (1 − 1/4) = 10.2 eV, more than the 4.5 eV work function, so it can eject an electron; a 1s atom has nothing to give. The current measures the 2s flux alone.
- An atom driven to 2p decays in τ₂ₚ = 1.6 ns, covering vτ₂ₚ = 8.0 × 10³ m s⁻¹ × 1.6 × 10⁻⁹ s = 1.3 × 10⁻⁵ m, about 13 μm. Its chance of surviving the whole flight is e(−12.5 μs/1.6 ns) = e(−7800): zero for every practical purpose.
- On resonance the microwaves turn detectable 2s atoms into undetectable 1s atoms, so the current dips. The lever is enormous: a 4.375 μeV photon switches off a 10.2 eV signal, a ratio of 10.2 eV ÷ (4.375 × 10⁻⁶ eV) = 2.3 × 10⁶.
AnswerFlight time 12.5 μs; about 99.99% of undisturbed 2s atoms reach the plate, while an atom driven to 2p decays within about 13 μm. The current counts 2s atoms, so a resonance shows as a dip.
MediumA student applies Welton's formula to hydrogen 2s with ħωₘₐₓ = mc² = 511.0 keV and the lower cutoff set at the 2s ionisation energy, ħωₘᵢₙ = 3.40 eV. Find the logarithm, the rms jiggle and its ratio to a = 52 918 fm, and the shift in μeV and MHz. Then redo the shift with Bethe's quoted mean excitation energy of 17.8 Ry and compare both with the measured 1057.8 MHz.
- Logarithm: ln(ωₘₐₓ/ωₘᵢₙ) = ln(510 999 eV ÷ 3.40 eV) = ln(1.503 × 10⁵) = 11.92.
- Jiggle: ⟨(δr)²⟩ = (2α/π)(ħ/mc)² × 11.92 = 4.646 × 10⁻³ × 11.92 × (386.16 fm)² = 0.05538 × (386.16 fm)², so √⟨(δr)²⟩ = 0.2353 × 386.16 fm = 90.9 fm. That is 1.7 × 10⁻³ a, so expanding V to second order in δr is safe.
- Shift: 4α⁵mc²/(3π) = 4.488 μeV, and n³ = 8 leaves 0.5610 μeV per unit of the logarithm. So ΔE = 0.5610 μeV × 11.92 = 6.69 μeV, and ΔE/h = 6.69 × 10⁻⁶ eV ÷ 4.1357 × 10⁻¹⁵ eV s = 1.62 × 10⁹ Hz, or 1617 MHz.
- Checks: α⁵mc² is an energy and the logarithm a pure number; the shift is order α⁵ ln(1/α), some 7 times below the 45.3 μeV fine-structure interval at n = 2; and it is positive. Against the measured 1057.8 MHz, though, it is 53% high.
- Bethe's cutoff: 17.8 × 13.606 eV = 242.2 eV, so ln(510 999/242.2) = 7.654 and ΔE = 0.5610 μeV × 7.654 = 4.29 μeV, or 1038 MHz, within 2% of experiment. A cutoff 71 times higher lowered the logarithm by only ln 71 = 4.27, which is 579 MHz.
Answerln = 11.92, rms jiggle 91 fm (1.7 × 10⁻³ a), ΔE ≈ 6.7 μeV ≈ 1.62 GHz, 53% above experiment. Bethe's 17.8 Ry gives 4.29 μeV ≈ 1038 MHz: a factor of 71 in the cutoff moved the answer by 36%.
HardHe⁺ is hydrogen-like with Z = 2, and its 2S₁/₂–2P₁/₂ interval is measured at about 14.04 GHz. Redo Welton's estimate for nuclear charge Z, scaling Bethe's hydrogen cutoff of 17.8 Ry the way atomic frequencies scale. Compare the result with experiment and with a naive Z⁴ scaling of hydrogen's 1057.8 MHz, then test the α hierarchy by comparing the Lamb-to-fine-structure ratio in He⁺ and in H.
- Track each factor. The jiggle's prefactor (2α/π)(ħ/mc)² belongs to the vacuum field and a free electron, so Z cannot touch it; only its lower cutoff moves. V = −Zαħc/r gives ∇²V = 4πZαħc δ³(r), one power of Z, and |ψₙ₀₀(0)|² = Z³/(πn³a³) adds three. Atomic frequencies scale as Z², so ħωₘᵢₙ → Z² × 17.8 Ry while ħωₘₐₓ stays mc². Hence ΔE = (4Z⁴α⁵mc²/(3πn³)) ln(mc²/(Z² × 17.8 Ry)).
- Numbers: Z² × 17.8 Ry = 71.2 × 13.606 eV = 968.7 eV, so the logarithm is ln(510 999/968.7) = ln 527.5 = 6.268, down by ln 4 = 1.386 from hydrogen's 7.654. Limiting case: Z = 1 returns hydrogen's 1038 MHz.
- Shift: Z⁴ × 135.64 MHz × 6.268 = 16 × 850.2 MHz = 1.36 × 10⁴ MHz, or 13.6 GHz (56.3 μeV). That is 3% below the measured 14.04 GHz, a miss of the same kind as hydrogen's 2%.
- The naive 16 × 1057.8 MHz = 16.9 GHz is 21% high: it forgets that the lower cutoff rises as Z², shrinking the logarithm. The premise still holds: with the smaller logarithm the rms jiggle is 386.16 fm × √(4.646 × 10⁻³ × 6.268) = 65.9 fm, only 2.5 × 10⁻³ of the He⁺ Bohr radius a/Z = 26 459 fm.
- Hierarchy: fine structure scales as Z⁴α⁴mc², so the He⁺ n = 2 interval is 16 × 10.95 GHz = 175.2 GHz at lowest order. The Lamb-to-fine ratio is 14.04/175.2 = 0.080 in He⁺ against 1.058/10.95 = 0.097 in H, a fall by a factor 0.83; Welton predicts the ratio of the logarithms, 6.268/7.654 = 0.82. Both ratios are order α times a logarithm.
AnswerWelton gives about 13.6 GHz (56 μeV) for He⁺ 2s, 3% below the measured 14.04 GHz, while naive Z⁴ scaling gives 16.9 GHz. The Lamb-to-fine-structure ratio falls from 0.097 to 0.080, as the Z-dependent logarithm predicts.