University Quantum Mechanics II · The Fine Structure of Hydrogen and the Lamb Shift · 8.2
The relativistic kinetic-energy correction
This is the fine-structure term you can compute from start to finish without quoting anything. Take the kinetic energy one order past Newton, trade p⁴ for the potential, and two radial averages you already own give every level's shift: all negative, of order α² times the Bohr energy, and largest for s states.
Build the model
Connect the measurement to the mechanism.
Quantum Mechanics I built hydrogen on the Newtonian kinetic energy p²/2m. Relativity replaces it with √(p²c² + m²c⁴) − mc², and because the electron's momentum in level n is only about α/n of mc, the difference can be expanded: the first term past p²/2m is H′ᵣ = −p⁴/(8m³c²), smaller than the kinetic energy by roughly α²/4n², so of order α⁴mc². It is a scalar with no spin in it, so it commutes with every component of L and S.
Inside the 2n²-fold degenerate level it is therefore already diagonal, in the uncoupled and the coupled basis alike, and first-order degenerate perturbation theory collapses to one expectation value per l. That value needs no fourth derivatives: p² acting on an eigenstate is 2m(Eₙ − V), so ⟨p⁴⟩ becomes 4m²⟨(Eₙ − V)²⟩, and the known averages ⟨1/r⟩ and ⟨1/r²⟩ turn it into Eᵣ⁽¹⁾ = −(Eₙ²/2mc²)[4n/(l + ½) − 3]. Every level moves down and the s levels most, so this term alone would break the Coulomb l-degeneracy.
That is exactly where it stops being the whole story: hydrogen's measured 2S₁/₂–2P₁/₂ gap is one gigahertz, not the 29 GHz this term puts between 2s and 2p, so spin-orbit coupling and the Darwin term must cancel its l-dependence, exactly between states of equal j as later topics show, leaving that gigahertz to the Lamb shift of quantum electrodynamics. The expansion itself fails for heavy hydrogen-like ions, where Zα is not small, and one order further in s states, where ⟨p⁶⟩ diverges.
- Simple definition
- The relativistic kinetic-energy correction is the first-order shift produced by H′ᵣ = −p⁴/(8m³c²), the leading amount by which the true kinetic energy √(p²c² + m²c⁴) − mc² falls below p²/2m.
- Example
- For hydrogen's 1s state the bracket 4n/(l + ½) − 3 equals 5, so Eᵣ = −5α⁴mc²/8 = −0.906 meV: 6.7 × 10⁻⁵ of the 13.6 eV binding energy, which is (5/4)α².
Derived from the binomial series for mc²√(1 + p²/m²c²). The second term is the perturbation, smaller than the first by p²/4m²c².
p in kg m s⁻¹, m in kg, c in m s⁻¹; every term in J. The series converges for p below mc.
Derived with QM I's virial theorem: 7.30 × 10⁻³ in the ground state. The correction is about α²/4n² of the kinetic energy, so of order α⁴mc².
Dimensionless; α = 1/137.036. From the virial theorem ⟨p²⟩/2m = −Eₙ = α²mc²/2n².
Commuting with L², Lz and Sz makes it diagonal in |n l mₗ mₛ⟩; commuting with L², J² and Jz makes it diagonal in |n l j mⱼ⟩. No secular matrix either way.
H′ᵣ in J. A rotational scalar with no spin operator in it, so it commutes with L², every Lᵢ and every Sᵢ.
Derived. It swaps a fourth derivative for two radial averages, and the virial theorem fixes the first: ⟨V⟩ = 2Eₙ for every l.
Eₙ and V in J. Uses p²|ψ⟩ = 2m(Eₙ − V)|ψ⟩ and ⟨p⁴⟩ = ⟨p²ψ|p²ψ⟩, since p² is Hermitian.
The only l-dependent input. Low l means large ⟨1/r²⟩: an s electron reaches the nucleus, where |V| and the local kinetic energy Eₙ − V are largest.
In J². From ⟨1/r²⟩ = 1/[(l + ½)n³a²] (Unit 6) and e²/4πε₀a = 2n²|Eₙ|, with a = 5.29 × 10⁻¹¹ m.
Derived here. Negative for every state and largest for s: −906 μeV for 1s, −147.2 μeV for 2s, −26.4 μeV for 2p, whatever mₗ, mₛ or j.
In eV; Eₙ²/2mc² = α⁴mc²/8n⁴ = 181.1 μeV/n⁴. The bracket is ⟨p⁴⟩/⟨p²⟩², never below 1.
One term past p²/2m, and how small it is
Subtract the rest energy from E² = p²c² + m²c⁴: T = mc²[√(1 + p²/m²c²) − 1]. The binomial series gives T = p²/2m − p⁴/(8m³c²) + p⁶/(16m⁵c⁴) − …, derived here and convergent for p below mc. Keep p²/2m in H₀ and take the next term as the perturbation, H′ᵣ = −p⁴/(8m³c²). Its size follows from QM I's virial theorem: ⟨p²⟩/2m = −Eₙ = α²mc²/2n², so √⟨p²⟩/mc = α/n, which is 7.30 × 10⁻³ in the ground state and 3.65 × 10⁻³ at n = 2. H′ᵣ is smaller than p²/2m by p²/4m²c² ≈ α²/4n², so the shift is of order α²|Eₙ| ~ α⁴mc² = 1.449 meV, the fine-structure rung of the previous topic's ladder, against 13.6 eV for the Bohr energy. The sign is known before any integral: the true kinetic energy grows more slowly with p than p²/2m does, because dT/dp is the speed v = pc²/E, always less than p/m. Every level must move down.
Why no secular matrix is needed
Level n holds 2n² states degenerate under H₀, so first-order theory in principle means diagonalising H′ᵣ in that subspace. Here it arrives diagonal. p² = p⋅p is a rotational scalar, so [Lᵢ, p²] = 0 for every component, and H′ᵣ contains no spin, so [Sᵢ, H′ᵣ] = 0. Commuting with L² means H′ᵣ never connects different l; commuting with Lz and Sz, never different mₗ or mₛ. Commuting with L₊ and L₋ forbids any mₗ dependence: the diagonal element for mₗ + 1 is ⟨mₗ|L₋H′ᵣL₊|mₗ⟩ divided by ⟨mₗ|L₋L₊|mₗ⟩, and moving H′ᵣ through L₋ turns it into the element for mₗ. So on each (n, l) block of 2(2l + 1) states H′ᵣ is a multiple of the identity, and every basis of the block is good: |n l mₗ mₛ⟩ and |n l j mⱼ⟩ alike, since coupling only recombines states of one l. What H′ᵣ fails to commute with is H₀ itself, because [p⁴, V] ≠ 0. That matters only at second order, which mixes different n and is of order α⁶mc².
Trading p⁴ for the potential
Fourth derivatives of Rₙₗ are avoidable. The unperturbed state obeys (p²/2m + V)|ψ⟩ = Eₙ|ψ⟩, so p²|ψ⟩ = 2m(Eₙ − V)|ψ⟩, and because p² is Hermitian, ⟨p⁴⟩ = ⟨p²ψ|p²ψ⟩ = 4m²⟨(Eₙ − V)²⟩. Hence ⟨H′ᵣ⟩ = −(Eₙ² − 2Eₙ⟨V⟩ + ⟨V²⟩)/(2mc²). The virial theorem, or ⟨1/r⟩ = 1/n²a from Unit 2, gives ⟨V⟩ = 2Eₙ. Unit 6's ⟨1/r²⟩ = 1/[(l + ½)n³a²], with e²/4πε₀a = 2n²|Eₙ|, gives ⟨V²⟩ = 4nEₙ²/(l + ½). So Eᵣ⁽¹⁾ = −(Eₙ²/2mc²)[1 − 4 + 4n/(l + ½)] = −(Eₙ²/2mc²)[4n/(l + ½) − 3], derived. Dimensions: Eₙ²/mc² is an energy. Limit: it vanishes as c → ∞. Order: Eₙ²/2mc² = α⁴mc²/8n⁴. The norm ⟨p²ψ|p²ψ⟩ is used on purpose: applying p⁴ directly to an s state throws off a delta function at the origin that is easy to lose, while the norm needs only (Eₙ − V)ψ, which grows as 1/r at the origin but is square-integrable.
The bracket measures the spread of p²
Since ⟨p²⟩ = 2m|Eₙ| for every l, the result reads Eᵣ⁽¹⁾ = −(⟨p²⟩²/8m³c²) × ⟨p⁴⟩/⟨p²⟩², and the bracket 4n/(l + ½) − 3 is exactly ⟨p⁴⟩/⟨p²⟩² = 1 + Var(p²)/⟨p²⟩². A limiting case checks it. For the circular orbit l = n − 1 the bracket is (n + 3/2)/(n − ½), which tends to 1 at large n: a classical circular orbit has constant speed, no spread in p², and shift −Eₙ²/2mc². For s states the bracket is 8n − 3, so 5 for 1s, where p² fluctuates with a standard deviation twice its mean. The reason is where the electron goes. With no centrifugal barrier an s electron passes through the region near the nucleus where the local kinetic energy Eₙ − V is huge, and the correction weights the square of that local kinetic energy. Hence the order s lowest, then p, d, f. The same algebra gives the bracket as 4⟨1/r²⟩/⟨1/r⟩² − 3, and that ratio is 2n for s states: relative to its size, a higher-n s electron spends more of its time close in.
What the measured spectrum says about this term alone
At n = 2, E₂²/2mc² = 11.32 μeV, so 2s moves by −147.2 μeV and all six 2p states by −26.4 μeV, a splitting of 120.8 μeV or 29.2 GHz. The measured 2S₁/₂–2P₁/₂ separation is 1057.8 MHz, 28 times smaller, and the measured 10.97 GHz between 2P₃/₂ and 2P₁/₂ gets nothing from this term, which cannot tell j = ½ from j = 3/2. So the l-dependence derived here must be cancelled by spin-orbit coupling (the next two topics) and the Darwin term (quoted from relativistic theory later in this unit): between states of equal j the cancellation turns out to be exact, and the 1057.8 MHz that survives is the Lamb shift, quoted from quantum electrodynamics. The ground state shows it in one number. Hydrogen's measured ionisation energy, 13.598 434 6 eV in the NIST tables, exceeds the reduced-mass Bohr value 13.598 287 3 eV by 147.3 μeV. The kinetic term alone gives 905.65 μeV of extra binding: right sign, right order α⁴mc², six times too large. Spin-orbit coupling vanishes for l = 0. Add the quoted Darwin shift, +724.52 μeV, and the 1s Lamb shift quoted from quantum electrodynamics, +33.80 μeV, and the total is −147.33 μeV, the measured 147.3 μeV of extra binding.
Where the expansion stops
Three limits bound the result. It is one of three α⁴ terms and never appears alone in a spectrum. Its expansion parameter is really Zα/n: in a hydrogen-like ion every length shrinks by Z, the formula keeps its form with Eₙ = −Z²α²mc²/2n², and the fractional shift becomes (Zα)²[4n/(l + ½) − 3]/4n². For 1s that is 1.25(Zα)²: 1% at Z = 12, but 56% for hydrogen-like uranium, Z = 92, where first-order theory is meaningless and only the Dirac energy, quoted later in this unit, will do. And the series cannot be taken one order further in s states. The wavefunction has a cusp at r = 0, so the momentum distribution of every s state falls only as p⁻⁸; ⟨p⁴⟩ converges, but ⟨p⁶⟩, weighted by p⁶ × p² dp, diverges. Hydrogen's α⁶ correction is finite, and it has to be taken from the Dirac equation rather than from the next term of this series.
Change one variable at a time
Make the relationship visible.
Put l at 3 and step n from 1 to 4: the marked highest-l level climbs from −5 to −1.57 units toward the faint ⟨p²⟩² line as the orbit turns circular, while the s level sinks to −29 units, even though the unit itself shrinks from 181 to 0.71 μeV.
SHIFT Eᵣ⁽¹⁾-46.96 μeV
AS A FREQUENCY Eᵣ⁽¹⁾/h-11.355 GHz
⟨p⁴⟩/⟨p²⟩²21.000
RMS p/mc = α/n2.432 × 10⁻³
Live interpretationSHIFT Eᵣ⁽¹⁾: −46.96 μeV. AS A FREQUENCY Eᵣ⁽¹⁾/h: −11.355 GHz. ⟨p⁴⟩/⟨p²⟩²: 21.000. RMS p/mc = α/n: 2.432 × 10⁻³
Catch the common trap
Explain before calculating.
The n = 3 level of hydrogen holds 18 states, degenerate under H₀. Which statement about the first-order effect of H′ᵣ = −p⁴/(8m³c²) on them is correct?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyHydrogen's n = 4 level contains 4s, 4p, 4d and 4f. Find the rms electron momentum as a fraction of mc, then the first-order relativistic kinetic shifts of 4s and 4f in μeV and GHz, and say which of the two lies closer to the value a classical circular orbit would give.
- Virial theorem: ⟨p²⟩/2m = |E₄| = α²mc²/32, so √⟨p²⟩/mc = α/4 = 1.82 × 10⁻³. The expansion in p/mc is safe, and the shift is about α²/64 = 8.3 × 10⁻⁷ of the kinetic energy, times the bracket.
- The unit for this shell: E₄²/2mc² = α⁴mc²/(8 × 4⁴) = 1.449 × 10⁻³ eV ÷ 2048 = 0.708 μeV.
- Brackets 4n/(l + ½) − 3: for 4s, 16/0.5 − 3 = 29; for 4f, 16/3.5 − 3 = 11/7 = 1.571.
- Shifts: E(4s) = −29 × 0.7075 μeV = −20.5 μeV and E(4f) = −1.571 × 0.7075 μeV = −1.11 μeV. Dividing by h = 4.1357 × 10⁻¹⁵ eV s gives −4.96 GHz and −0.269 GHz.
- A classical circular orbit has a sharp p², so ⟨p⁴⟩ = ⟨p²⟩², the bracket is 1 and the shift is −0.708 μeV. The 4f state, bracket 1.57, is the near-circular one; 4s, bracket 29, is dominated by its passes close to the nucleus.
Answer√⟨p²⟩/mc = 1.82 × 10⁻³; E(4s) = −20.5 μeV (−4.96 GHz) and E(4f) = −1.11 μeV (−0.269 GHz). 4f lies within a factor 1.6 of the circular-orbit value −0.708 μeV.
MediumHydrogen-like He⁺ (Z = 2) has its n = 2 level at −13.61 eV, the same Bohr energy as hydrogen's 1s. Redo ⟨V⟩ and ⟨V²⟩ with Z, find the relativistic kinetic shifts of He⁺ 2s and 2p, the 2s–2p splitting this term alone would give in GHz, and explain why He⁺ 2s and H 1s shift by different amounts.
- With V = −Ze²/4πε₀r every length scales as a/Z: ⟨1/r⟩ = Z/n²a, ⟨1/r²⟩ = Z²/[(l + ½)n³a²] and Eₙ = −Z²α²mc²/2n². The virial theorem still gives ⟨V⟩ = 2Eₙ, and Z²e²/4πε₀a = 2n²|Eₙ| gives ⟨V²⟩ = 4nEₙ²/(l + ½). The formula −(Eₙ²/2mc²)[4n/(l + ½) − 3] survives unchanged, with Z hidden in Eₙ.
- Shell unit: E₂²/2mc² = (13.606 eV)²/(2 × 510 999 eV) = 181.1 μeV, which is Z⁴ = 16 times hydrogen's n = 2 unit of 11.32 μeV.
- Brackets 13 for 2s and 7/3 for 2p: E(2s) = −13 × 181.13 μeV = −2354.7 μeV = −2.355 meV and E(2p) = −(7/3) × 181.13 μeV = −422.6 μeV.
- Splitting (32/3) × 181.13 μeV = 1932.1 μeV; divided by h = 4.1357 × 10⁻¹⁵ eV s that is 4.672 × 10¹¹ Hz = 467 GHz, sixteen times hydrogen's 29.2 GHz.
- H 1s has the same Eₙ, hence the same unit 181.1 μeV and the same ⟨p²⟩, but its bracket is 5, giving −905.7 μeV. He⁺ 2s has the wider spread of p²: the bracket is 4⟨1/r²⟩/⟨1/r⟩² − 3, and ⟨1/r²⟩/⟨1/r⟩² = 2n is 4 for it against 2 for 1s. Same ⟨p²⟩, different ⟨p⁴⟩, ratio 13/5 = 2.6.
- Order check: 2.355 meV/13.606 eV = 1.73 × 10⁻⁴ = (13/4)(Zα/n)² with Zα/n = 7.30 × 10⁻³, as small as in hydrogen's 1s, so first order is just as reliable.
AnswerE(2s) = −2.355 meV and E(2p) = −0.423 meV; the splitting is 1.932 meV, or 467 GHz, Z⁴ = 16 times hydrogen's. He⁺ 2s shifts 2.6 times as much as H 1s despite equal Bohr energies, because its ⟨p⁴⟩/⟨p²⟩² is 13, not 5.
HardCheck the 1s result without the (Eₙ − V) trick. Apply ∇² to ψ₁₀₀ = e(−r/a)/√(πa³) to find ⟨p⁴⟩ = ⟨p²ψ|p²ψ⟩, compare it with ⟨p²⟩², evaluate Eᵣ in meV, and then test whether the next term of the series, p⁶/(16m⁵c⁴), has a finite expectation value in this state.
- For a function of r alone ∇²f = f″ + (2/r)f′. With f = e(−r/a), f′ = −f/a and f″ = f/a², so ∇²ψ = (1/a² − 2/ar)ψ and p²ψ = −ħ²(1/a² − 2/ar)ψ. No delta function appears: e(−r/a) ≈ 1 − r/a near the origin, and ∇² of r is 2/r, an integrable singularity with no δ function in it.
- ⟨p⁴⟩ = ħ⁴⟨(1/a² − 2/ar)²⟩ = ħ⁴[1/a⁴ − (4/a³)⟨1/r⟩ + (4/a²)⟨1/r²⟩]. For 1s, ⟨1/r⟩ = 1/a and ⟨1/r²⟩ = 2/a², so ⟨p⁴⟩ = (ħ⁴/a⁴)(1 − 4 + 8) = 5ħ⁴/a⁴.
- ⟨p²⟩ = −ħ²⟨1/a² − 2/ar⟩ = −ħ²(1/a² − 2/a²) = ħ²/a², so ⟨p⁴⟩/⟨p²⟩² = 5, the bracket 4n/(l + ½) − 3 at n = 1, l = 0. The standard deviation of p² is √(5 − 1) = 2 times its mean.
- With a = ħ/αmc, ħ/a = αmc and Eᵣ = −5ħ⁴/(8m³c²a⁴) = −(5/8)α⁴mc² = −(5/8)(1.449 meV) = −0.906 meV, the general formula's value. Dimensions: (ħ/a)⁴ is (kg m s⁻¹)⁴, and dividing by m³c² leaves J.
- Next order: ⟨p⁶⟩ = ⟨p²ψ|p²|p²ψ⟩ = ħ⁶∫|∇(∇²ψ)|² d³r. Near r = 0, ∇²ψ ≈ −2ψ(0)/ar, whose gradient has magnitude 2ψ(0)/ar², so the integrand behaves as 4ψ(0)²/(a²r⁴) and ∫ r⁻⁴ 4πr² dr diverges at the origin.
- So the series in p/mc gives a finite α⁴ term but an infinite α⁶ term for every s state. Hydrogen's α⁶ correction is finite, so it must be quoted from the Dirac energy, not built from the next term of this expansion.
Answer⟨p⁴⟩ = 5ħ⁴/a⁴ = 5⟨p²⟩², so Eᵣ(1s) = −(5/8)α⁴mc² = −0.906 meV, agreeing with the formula; ⟨p⁶⟩ diverges, so for s states the expansion is usable only up to the p⁴ term.