University Quantum Mechanics II · The Fine Structure of Hydrogen and the Lamb Shift · 8.7
Expanding the exact Dirac energy to order α⁴
This is where the fine-structure calculation gets audited. The Dirac equation hands us an exact hydrogen energy that we quote rather than derive. Expanding it in (Zα)² checks every factor of the perturbative sum, measures what that sum leaves out, and shows why no order of the Dirac–Coulomb energy will ever split 2S₁/₂ from 2P₁/₂.
Build the model
Connect the measurement to the mechanism.
The Dirac equation for one electron bound to a fixed point charge Ze can be solved exactly, and its energies, Eₙⱼ = mc²[1 + Z²α²/(n − j − ½ + √((j + ½)² − Z²α²))²]⁻¹⁄², are quoted here from relativistic quantum mechanics. The formula is not a rival to the week's perturbation theory but its referee. It contains Zα only through (Zα)², so it expands as a power series in (Zα)²: the zeroth term is the rest energy, the first is the Bohr energy of the Schrödinger equation, and the second is exactly the fine-structure term that the relativistic kinetic, spin-orbit and Darwin corrections summed to.
That match certifies the three-term sum, quoted Thomas factor and Darwin coefficient included, through order α⁴mc². The next term, of order α⁶mc², is 1.2 MHz for 1S and 0.38 MHz for 2S: under a thousandth of the Lamb shift and comparable to the proton-size shift, so for hydrogen the truncated series is as good as the model itself. The model stops in three places.
It holds a point nucleus fixed, so recoil, nuclear size and the proton's magnetic moment are missing. It depends on n and j but never on l, so 2S₁/₂ and 2P₁/₂ stay degenerate at every order, and the measured 1057.8 MHz between them must come from outside the model, almost all of it from the quantised field. And as Zα grows the series converges ever more slowly, until at Zα = 1 the root for j = ½ reaches zero, beyond which the formula has no real answer.
- Simple definition
- The Dirac–Coulomb energy is the exact bound-state energy, rest energy included, of a Dirac electron in the field of a fixed point charge Ze; expanded in powers of (Zα)² it gives the rest energy, the Bohr energy and the fine-structure shift in turn.
- Example
- For hydrogen's ground state it reduces to E = mc²√(1 − α²), so mc² − E = 13.6057 eV + 0.1811 meV + 4.82 neV + … = 13.6059 eV: the Bohr binding, the fine-structure term, and the first term beyond it.
Exact for one electron and a fixed point nucleus. It contains n, j and (Zα)², and never l: every later card is read off from this one.
m electron mass (kg), c (m s⁻¹), E in J; α = e²/(4πε₀ħc) and Zα are pure numbers; n = 1, 2, …; j = ½, …, n − ½
Rewrites the exact energy as mc²[1 + (Zα/n*)²]⁻¹⁄² with n* = n − δⱼ: Bohr's form with n lowered slightly, the whole j dependence carried by δⱼ.
Dimensionless. Hydrogen: δ = 2.66 × 10⁻⁵ for j = ½ and 1.33 × 10⁻⁵ for j = 3/2
Here the expansion is one binomial series, mc² − E = mc²(x²/2 + x⁴/8 + x⁶/16 + …) with x = Zα: the quickest test of any general coefficient.
n = 1, j = ½. Hydrogen: mc² − E = 13.6059 eV, against 13.6057 eV from the Bohr formula
The third term equals [Eₙ²/(2mc²)][3 − 4n/(j + ½)], the perturbative sum of the previous topic, so that sum is right through order α⁴mc².
Rest energy, Bohr energy Eₙ, fine structure, each in J. Hydrogen 1S: 511.0 keV, −13.6 eV, −0.181 meV
Under a thousandth of the Lamb shift and comparable to the proton-size shift, so in hydrogen it matters only alongside corrections the Dirac model leaves out.
Hydrogen, as a frequency E/h: 1S −1.17 MHz; 2S₁/₂ and 2P₁/₂ −0.383 MHz; 2P₃/₂ −0.018 MHz
Every coefficient of the series is a function of n and j alone, so any splitting of equal-j levels must come from outside the Dirac–Coulomb problem.
Measured in hydrogen: 2S₁/₂ lies 1057.8 MHz above 2P₁/₂ (the Lamb shift)
What the quoted formula contains
The Dirac equation for one electron in the static Coulomb field of a point charge Ze has an exact bound-state spectrum. Solving it takes four-component spinors, so this course quotes the result from relativistic quantum mechanics rather than deriving it: Eₙⱼ = mc²[1 + Z²α²/(n − j − ½ + √((j + ½)² − Z²α²))²]⁻¹⁄², with n = 1, 2, … and j = ½, …, n − ½. Three assumptions are built in. The energy includes the rest energy, so the binding is mc² − E. The nucleus is a fixed point with no size, spin or recoil. The field is a classical Coulomb field, not a quantised one. Test the formula before trusting it. Zα is a pure number, so E has the units of mc². As Zα → 0 the bracket tends to 1 and E → mc², a free electron at rest. For n = 1, j = ½ the denominator shrinks to √(1 − Z²α²), the bracket becomes 1/(1 − Z²α²), and E = mc²√(1 − Z²α²): for hydrogen mc² − E = 13.6059 eV, a hair above Bohr's 13.6057 eV.
Expanding in (Zα)², one factor at a time
Write x = Zα and k = j + ½, and expand from the inside out. First the root: √(k² − x²) = k − x²/(2k) − x⁴/(8k³) − …, so the denominator is n* = n − δⱼ with quantum defect δⱼ ≈ x²/(2j + 1). For hydrogen δⱼ = 2.66 × 10⁻⁵ at j = ½: the relativistic atom is a Bohr atom with n lowered by a few parts in 10⁵, lowered more for small j. Next the ratio: x²/n*² = (x²/n²)[1 + x²/(nk) + …]. Last the outer power: (1 + u)⁻¹⁄² = 1 − u/2 + 3u²/8 − … . Collecting, E = mc² − mc²x²/(2n²) − [mc²x⁴/(2n⁴)](n/k − ¾) + O(x⁶). Every step is a Taylor series in x², so only even powers of Zα can appear, at any order. The x² term is the Bohr energy Eₙ, the Schrödinger result reappearing as the low-speed limit of a relativistic equation. The x⁴ term for hydrogen's 1S is −mc²α⁴/8 = −0.181 meV, or 43.8 GHz.
Matching the α⁴ term to the perturbative sum
The previous topic added three first-order shifts, the relativistic kinetic correction and spin-orbit coupling derived in this unit and the Darwin term quoted, and got Eₙⱼ⁽¹⁾ = [Eₙ²/(2mc²)][3 − 4n/(j + ½)]. Put in Eₙ = −mc²α²/(2n²): the prefactor is mc²α⁴/(8n⁴), and multiplying out gives −[mc²α⁴/(2n⁴)][n/(j + ½) − ¾], the α⁴ term of the Dirac expansion symbol for symbol. The match certifies two things. The three-term sum is right through order α⁴mc². And the quoted pieces carry the right factors: drop the Thomas factor ½ from spin-orbit, or double the Darwin term, and the sum starts to depend on l as well as j, which the exact formula forbids. It does not make the sum exact. A perturbation of order α²Eₙ treated at first order has nothing to say about order α⁶, and agreement between two theories is not yet agreement with a measurement.
The first term the perturbation theory misses
Carry each of the three series one step further and the (Zα)⁶ term appears: E⁽⁶⁾ = −mc²(Zα)⁶[1/(8k³n³) + 3/(8k²n⁴) − 3/(4kn⁵) + 5/(16n⁶)]. Test it on 1S, where the exact form is one binomial series: n = k = 1 gives 1/8 + 3/8 − 3/4 + 5/16 = 1/16, the x⁶ coefficient of 1 − √(1 − x²). With mc²α⁶ = 77.2 neV, hydrogen's 1S moves by −4.82 neV, or −1.17 MHz; 2S₁/₂ and 2P₁/₂ by −0.383 MHz; 2P₃/₂ by −0.018 MHz. Set that beside what the model leaves out. The 1S Lamb shift, 8.17 GHz, is 7000 times larger: its order, α⁵ln(1/α)mc², beats α⁶mc² by ln(1/α)/α ≈ 670, and the coefficients supply the other factor of ten. The proton's 0.84 fm radius lifts 1S by about 1.1 MHz, a first-order estimate quoted here. Recoil beyond the reduced mass is of order (m/M)α⁴mc², ten times α⁶mc². For hydrogen the α⁴ truncation is already as accurate as the Dirac model itself.
Degenerate at every order
Read off what the exact formula depends on: n, j and (Zα)², never l. The 2S₁/₂ (l = 0) and 2P₁/₂ (l = 1) states share n = 2 and j = ½, so the Dirac equation gives them identical energies, not only at order α⁴ but exactly. Every coefficient of the series, including the −21/1024 that multiplies mc²α⁶ for both, is a function of n and j alone. Relativistic quantum mechanics traces this to an extra conserved operator of the Dirac–Coulomb problem, a cousin of the Runge–Lenz vector; that is quoted, not shown here. In 1947 Lamb and Retherford found 2S₁/₂ about 1 GHz above 2P₁/₂, and the modern value is 1057.8 MHz. Two features of that splitting rule out a Dirac origin: it depends on l, and it scales as α⁵ln(1/α)mc², an odd power with a logarithm that no series in (Zα)² can produce. It belongs to quantum electrodynamics, whose value the next topic quotes alongside Welton's estimate.
When Zα is no longer small
For a hydrogen-like ion the expansion parameter is Zα, not α, and the (Zα)⁴ truncation degrades as Z grows. In U⁹¹⁺, with Z = 92 and Zα = 0.671, the exact point-nucleus 1S binding mc²[1 − √(1 − Z²α²)] is 132.28 keV. The Bohr term gives 115.16 keV, adding (Zα)⁴ reaches 128.13 keV, and adding (Zα)⁶ reaches 131.06 keV. The (Zα)⁶ term, 2.92 keV, is only 4.4 times smaller than the (Zα)⁴ term before it, where in hydrogen the ratio is 38 000. The 4.1 keV that the (Zα)⁴ series misses is about nine times the measured 1S Lamb shift of 0.46 keV, which for uranium includes the nuclear-size shift, so at high Z the exact formula is mandatory. At Zα = 1, Z = 137, the root √(1 − Z²α²) reaches zero, and beyond it the point-nucleus formula for j = ½ has no real value at all; a nucleus of finite size removes that breakdown. The figure plots the exact and the truncated binding against Zα.
Change one variable at a time
Make the relationship visible.
At Z = 1 the curves coincide and everything past (Zα)⁴ is 0.0027% of the 1S fine structure. Drag Z to 92 with n = 1 and j = ½: the (Zα)⁴ series now misses 4.1 keV. Then raise j + ½ and watch every correction shrink.
COUPLING Zα0.0073
EXACT BINDING13.61 eV
FINE STRUCTURE ÷ BOHR0.0013 %
BEYOND (Zα)⁴ ÷ FINE STRUCTURE0.0027 %
Live interpretationCOUPLING Zα: 0.0073. EXACT BINDING: 13.61 eV. FINE STRUCTURE ÷ BOHR: 0.0013 %. BEYOND (Zα)⁴ ÷ FINE STRUCTURE: 0.0027 %
Catch the common trap
Explain before calculating.
The exact 1S energy of a hydrogen-like ion is mc²√(1 − Z²α²). If its expansion is stopped after the (Zα)⁴ term, how much binding energy does the truncation miss for Ar¹⁷⁺, with Z = 18 and mc² = 511.0 keV?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyThe exact Dirac energy of hydrogen can be read as a Bohr formula with n replaced by n* = n − δⱼ. Use the quantum defects of the n = 2 levels with j = ½ and j = 3/2 to predict the 2P₃/₂–2P₁/₂ interval as a frequency, and compare it with the measured 10.969 GHz.
- The defect is δ ≈ α²/(2j + 1): δ(½) = α²/2 = 2.6626 × 10⁻⁵ and δ(3/2) = α²/4 = 1.3313 × 10⁻⁵, so n* = 1.999 973 for j = ½ and 1.999 987 for j = 3/2. The j = ½ level has the smaller n*, so it is more tightly bound.
- To lowest order E = mc² − mc²α²/(2n*²), whose slope is dE/dn* = mc²α²/n*³ = 27.211 eV ÷ 8 = 3.4014 eV per unit of n* at n = 2.
- The two levels differ in n* by δ(½) − δ(3/2) = α²/4 = 1.3313 × 10⁻⁵, so ΔE = 3.4014 eV × 1.3313 × 10⁻⁵ = 45.28 μeV, which is mc²α⁴/32, the same interval the α⁴ term of the expansion gives.
- As a frequency, ν = ΔE/h = 45.28 × 10⁻⁶ eV ÷ 4.1357 × 10⁻¹⁵ eV s = 10.949 GHz. Scaling by the reduced-mass factor 0.999 456, the leading recoil correction, which is quoted, gives 10.943 GHz.
- The measured 10.969 GHz is 0.23% higher. Keeping every order of the exact formula adds only 0.36 MHz, so the remaining 25.3 MHz, or 0.23%, is not a Dirac effect. It is quantum electrodynamics, quoted: the electron's anomalous moment, g − 2 ≈ α/π = 0.23%, enlarges the spin-orbit doublet by that fraction, to 10.969 GHz.
Answer2P₃/₂ lies 45.3 μeV, or 10.95 GHz, above 2P₁/₂; with the reduced mass, 10.943 GHz, 0.23% below the measured 10.969 GHz, a gap the quoted QED anomalous moment closes.
MediumCarry the Dirac expansion for the n = 2, j = ½ level of hydrogen to order α⁶. Find the coefficient, express the shift as a frequency, and compare it with the measured 1057.8 MHz between 2S₁/₂ and 2P₁/₂.
- With k = j + ½ = 1, √(1 − α²) = 1 − α²/2 − α⁴/8 − …, so n* = 2 − 1 + √(1 − α²) = 2 − α²/2 − α⁴/8.
- Write n* = 2(1 − ε) with ε = α²/4 + α⁴/16. Then (1 − ε)⁻² = 1 + 2ε + 3ε² + … = 1 + α²/2 + 5α⁴/16, and u = α²/n*² = α²/4 + α⁴/8 + 5α⁶/64.
- Expand (1 + u)⁻¹⁄² = 1 − u/2 + 3u²/8 − 5u³/16. With u² = α⁴/16 + α⁶/16 + … and u³ = α⁶/64 + …, the α⁶ pieces are −5/128 from −u/2, +3/128 from 3u²/8 and −5/1024 from −5u³/16: in total −40/1024 + 24/1024 − 5/1024 = −21/1024.
- Check the order below: the α⁴ pieces are −1/16 + 3/128 = −5/128, which is −(1/32)(2 − ¾), the fine-structure term for n = 2, j = ½. So E⁽⁶⁾ = −(21/1024)mc²α⁶ = −(21/1024)(77.16 neV) = −1.582 neV, and ΔE/h = −0.383 MHz.
- Nothing in the calculation used l, so 2P₁/₂ moves by the same −0.383 MHz and the pair stays degenerate. The shift is 3.6 × 10⁻⁴ of the 1057.8 MHz Lamb shift and 2.8 × 10⁻⁵ of this level's 13.69 GHz fine-structure term.
AnswerE⁽⁶⁾ = −(21/1024)mc²α⁶ = −1.58 neV, a shift of −0.383 MHz shared by 2S₁/₂ and 2P₁/₂; it is 3.6 × 10⁻⁴ of the measured Lamb shift and splits nothing.
HardA student redoing the n = 2 fine structure forgets the Thomas factor, so the spin-orbit term comes out twice its correct size, while the relativistic kinetic and Darwin terms are right. Find the student's shifts of 2S₁/₂, 2P₁/₂ and 2P₃/₂ in μeV, show how the α⁴ term of the Dirac expansion exposes the error without any experiment, and name a measurement that also exposes it.
- Work in units of u = E₂²/(2mc²) = mc²α⁴/128 = 11.321 μeV. From the neighbouring topics: Eᵣₑₗ = −u[4n/(l + ½) − 3]; Eₛₒ = 2un[j(j + 1) − l(l + 1) − ¾]/[l(l + ½)(l + 1)] for l ≥ 1; ED = 4nu for l = 0 only.
- Correct values at n = 2: 2S₁/₂ = −13u + 8u = −5u; 2P₁/₂ = −7u/3 − 8u/3 = −5u; 2P₃/₂ = −7u/3 + 4u/3 = −u. These match the Dirac α⁴ term −(mc²α⁴/32)(2/k − ¾) = −4u(2/k − ¾), which is −5u for k = 1 and −u for k = 2.
- Doubling spin-orbit: 2P₁/₂ becomes −7u/3 − 16u/3 = −23u/3 = −86.79 μeV and 2P₃/₂ becomes −7u/3 + 8u/3 = +u/3 = +3.77 μeV, while 2S₁/₂ keeps −5u = −56.60 μeV.
- The Dirac expansion requires 2S₁/₂ and 2P₁/₂ to coincide at −56.60 μeV and 2P₃/₂ to sit at −11.32 μeV. The student's 2P₁/₂ lies 30.19 μeV, or 7.30 GHz, below 2S₁/₂: an energy that depends on l at fixed n and j, which no order of the exact formula allows. Pen and paper catch the error.
- In the laboratory the P doublet decides: the student predicts 8u = 90.57 μeV, or 21.90 GHz, against the Dirac 45.28 μeV, or 10.95 GHz, and the measured 10.969 GHz.
AnswerStudent: 2S₁/₂ −56.60 μeV, 2P₁/₂ −86.79 μeV, 2P₃/₂ +3.77 μeV. The Dirac α⁴ term demands −56.60, −56.60 and −11.32 μeV, and the student's 21.90 GHz doublet is twice the measured 10.969 GHz.