
Advanced Quantum Mechanics course sequence · Interactive lesson
Where the point-nucleus Dirac atom ends, and what it omits
Every exact solution has edges, and this one has two. Past Z = 137 a point nucleus no longer defines an atom, and only the nucleus's real size can. Long before that, real hydrogen departs from the Dirac atom by effects some forty times Dirac's own correction. This topic locates both edges and puts a size on each.
UAQM-07.09 · Lesson 9 of 9 in this unit. Position is not a completion score.Build the model
Z = 137 is where a point nucleus stops defining an atom, not where the electron reaches c.
The key idea
The point-nucleus Dirac atom ends where the exponent γ = √(κ² − (Zα)²) of its radial functions at r = 0 stops being real, at Zα = 1 for the j = ½ levels; beyond that the equation cannot fix a spectrum without the nucleus's size.
Example. At Z = 137, Zα = 0.99974 and γ = 0.0229, so the 1s energy mc²γ is 11.7 keV and its binding 499.3 keV; at Z = 138, γ² = 1 − (1.00703)² = −0.0141, and a point nucleus has no real 1s energy at all.
The complete physical model and assumptions
Topic 7.06's exact energies rest on the exponent γ = √(κ² − (Zα)²) with which G and F leave a point nucleus, and they hold only while γ is real. For the j = ½ levels, |κ| = 1, γ reaches zero at Zα = 1, Z = 137.04: the 1s energy mc²γ falls to zero with infinite slope, 2s₁/₂ and 2p₁/₂ end at mc²/√2, and 2p₃/₂, with |κ| = 2, carries on. Beyond, γ is imaginary, both solutions near r = 0 oscillate as cos(σ ln r + δ) with nothing to fix δ, and the Hamiltonian no longer defines a spectrum: Unit 3's fall to the centre, moved by the electron's spin from l + ½ to j + ½. The repair is the fact the model idealised away. A nucleus of radius R keeps the potential finite inside, and matching at R weights the irregular root by R(2γ), which vanishes as R → 0 and so justifies topic 7.05's choice; for imaginary γ that weight only oscillates, and no point limit exists. With a real nucleus the 1s level falls on through Z = 137 and reaches −mc² near Z = 172, quoted, where a one-particle reading fails and Unit 8 takes over. Hydrogen leaves the model for gentler reasons, ranked: nuclear motion moves 1s by 1.79 THz, the Lamb shift by 8.17 GHz, hyperfine coupling splits it by 1.42 GHz and the proton's size moves it by 1.1 MHz, so Dirac's own 43.8 GHz correction to 1s means something only once the reduced mass is restored.
1Where γ vanishes, and why only the j = ½ levels end there
Units are SI, with x⁰ = ct and metric diag(1, −1, −1, −1), though nothing here needs four-vectors. Topic 7.05 found G and F leaving a point nucleus as rγ with γ = √(κ² − (Zα)²), and topic 7.06's energies, E = mc²[1 + (Zα)²/(nᵣ + γ)²](−1/2), are built on it. γ depends on |κ| = j + ½ alone, so it reaches zero at Zα = j + ½: for every j = ½ level at once, at Z = 1/α = 137.04, derived. The 1s energy, mc²γ, falls to zero there with slope −mc²α Zα/γ, which is −3.4 keV per unit Z at uranium and −163 keV at Z = 137: a vertical tangent. The nᵣ = 1 pair, 2s₁/₂ and 2p₁/₂, ends at mc²/√2 = 361 keV. Nothing happens to 2p₃/₂, whose γ = √(4 − (Zα)²) is still 1.73. As Zα → 0, the 1s exponent γ → 1 returns Quantum Mechanics II's r(l+1) for an s state, the limit check. Unit 3's spinless 1s ended at Zα = l + ½ = ½, Z = 68.5; the spin has moved the edge to j + ½.
2Beyond Zα = 1, nothing at the origin fixes the solution
Above the edge γ² is negative, so write γ = iσ with σ = √((Zα)² − 1). Both near-origin solutions, r(±iσ) = cos(σ ln r) ± i sin(σ ln r), have modulus 1: both are bounded, both square-integrable, and the real combinations are G ∝ cos(σ ln(r/r₀)) with r₀ any length, derived. Even the small component stops being small, since topic 7.05's indicial pair gives |F/G| = |iσ − 1|/(Zα) = 1. Near the origin the angular term κ/r and the Coulomb term Zα/r scale alike, so the equation holds no length that could fix r₀; each choice yields a different set of levels, and the energy formula itself turns complex, E(1s) = imc²σ. This is Unit 3's fall to the centre, nodes piling up without end as r → 0, though slowly: at Z = 140, σ = 0.209 and successive nodes lie a factor e(π/σ) = 3.3 × 10⁶ apart in r. The point-nucleus Hamiltonian no longer defines the atom, so the theory ends near Z = 137, derived. What it lacks is a statement about short distances, and the nucleus is the only one physics supplies.
3A finite nucleus chooses the root, and removes the edge
Spread the charge within a radius R. Inside, the potential stays finite and the solution is regular; for R ≪ ħ/mc = 386 fm its ratio ρ = F/G at r = R is a pure number set by Zα and the charge profile. Outside, G = A rγ + B r(−γ), with F/G equal to a₊ = (γ − 1)/(Zα) on the first root and a₋ = −(1 + γ)/(Zα) on the second. Matching F/G at R gives B/A = R(2γ)(a₊ − ρ)/(ρ − a₋), derived. For real γ this vanishes as R → 0, so the point limit picks rγ for every Zα below 1, including the band from 0.866 to 1 where normalisability could not: topic 7.05's choice, settled. The approach slows near the edge, since the irregular share at radius r goes as (R/r)(2γ), and 2γ is 1.48 for uranium but 0.63 at Z = 130. For imaginary γ, R(2iσ) = cos(2σ ln R) + i sin(2σ ln R) has modulus 1 and no limit, so the levels depend on R however small it is made. With R finite, nothing singular happens at Z = 137 at all.
4Diving to −mc², where a one-particle theory stops
With a finite nucleus the 1s level keeps falling through Z = 137. Calculations with R ≈ 1.2 A¹⁄³ fm, quoted, bring it to −mc², the top of the negative-energy continuum, near Z = 172, and bring 2p₁/₂ there near Z = 185. Below −mc², topic 7.05's decay constant ħcλ = √(m²c⁴ − E²) turns imaginary: the state no longer decays at infinity and dissolves into a resonance in the negative-energy continuum, which a one-particle equation cannot interpret. Unit 8's hole theory reads it as a vacuum that, offered an empty 1s, keeps an electron and emits a positron: a charged vacuum, quoted. No such nucleus exists, the heaviest yet made having Z = 118, and uranium-uranium collisions, which assemble Z = 184 for about 10⁻²¹ s, have not shown the effect cleanly, quoted. The point nucleus could not even approach this: its 1s ends at E = 0, half-way down the gap, so the dive is a finite-size phenomenon from start to finish.
5What the Dirac atom omits for hydrogen's 1s, ranked
Dirac's own correction to Bohr's 1s is −α⁴mc²/8, 43.8 GHz in frequency, and four omissions sit around it. Nuclear motion is the largest: the reduced mass lowers the binding by 13.6059 eV × mₑ/(mₑ + mₚ) = 7.41 meV, 1.79 THz, derived, 41 times Dirac's correction; what the reduced mass misses, of order (mₑ/mₚ)α⁴mc², 0.19 GHz in scale, is 24 MHz for 1s, quoted. The Lamb shift, from the electron's coupling to the quantised field, raises 1s by 8.17 GHz, quoted here; Unit 13 estimates it as Bethe did, without the full QED corrections. Coupling to the proton's magnetic moment splits 1s by 1.42 GHz, F = 1 above F = 0, without moving the centroid, quoted. The proton's size, at first order with |ψ(0)|² = 1/(πa³), raises 1s by (2/3)α⁴mc²(mcrₚ/ħ)² = 1.1 MHz for rₚ = 0.841 fm, derived; mcrₚ/ħ is a pure number, so the dimensions check, and the shift vanishes for a point proton. As fractions of the binding the list runs from 5 × 10⁻⁴ down to 3 × 10⁻¹⁰.
6Which omission splits 2S₁/₂ from 2P₁/₂
Dirac's degeneracy of 2S₁/₂ and 2P₁/₂ survives most of the list. The reduced mass rescales every level by mᵣ/m, and the leading recoil term beyond it depends, like the Dirac energy, on n and j alone, so neither separates the pair. Hyperfine coupling splits each into F = 0 and F = 1 without moving either centroid. The proton's size acts on s states only, but at n = 2 it is 1.1 MHz/8 = 0.14 MHz. That leaves the radiative shift, measured at 1057.8 MHz with 2S₁/₂ above 2P₁/₂: the one omission that breaks the degeneracy, which Unit 13 estimates as Bethe did, at about 1040 MHz. The ranking is not universal, because the omissions grow differently with Z: the Lamb and size shifts roughly as Z⁴, hyperfine as Z³, the reduced-mass shift as Z²mₑ/M. In hydrogen-like uranium nuclear motion is worth 0.3 eV of the 132.3 keV binding, ²³⁸U has spin zero and so no hyperfine structure, and nuclear size and the radiative shift, about 0.20 keV and 0.27 keV, quoted, lead the list.
Key equations
Read the units, assumptions and limits before substituting.
Exponent at the origin, and where it vanishes
γ = √(κ² − (Zα)²), real only for Zα ≤ |κ|
Derived in topic 7.05 and used here as the gauge: every j = ½ level loses its real γ at once, while j = 3/2 states keep theirs up to Zα = 2.
κ = ∓(j + ½) and Zα are pure numbers; α = 1/137.036, so for j = ½ the edge Zα = 1 is Z = 137.04.The ground state at the edge
E(1s) = mc²√(1 − (Zα)²) → 0 as Zα → 1
Derived in topic 7.08. Its slope, −mc²α Zα/√(1 − (Zα)²), diverges at the edge: −3.4 keV per unit Z at uranium, −163 keV at Z = 137.
mc² = 0.511 MeV; E includes the rest energy, so the binding mc² − E reaches mc², half the 2mc² gap.Past the edge: a fall to the centre
G ∝ cos(σ ln(r/r₀)), with σ = √((Zα)² − κ²)
Derived: both roots r(±iσ) are bounded and nothing at r = 0 fixes r₀, so no spectrum is defined; Unit 3's collapse at l + ½, here at j + ½.
r₀ is any length, in m; G in m(−1/2); the nodes pile up at r = 0, a factor e(π/σ) apart in r.A finite nucleus chooses the root
B/A = R(2γ)(a₊ − ρ)/(ρ − a₋), for G = A rγ + B r(−γ) beyond R
Derived: R(2γ) → 0 as R → 0 selects rγ, settling topic 7.05's choice; for imaginary γ, |R(2iσ)| = 1 and the point limit never settles.
R is the nuclear radius in m; a± = (±γ − 1)/(Zα) for κ = −1; ρ = F/G at R from the interior, a pure number.Nuclear motion, the largest omission
ΔE(1s) = B_∞ mₑ/(mₑ + mₚ), with B_∞ = 13.6059 eV
Derived from the two-body separation: 41 times Dirac's own α⁴mc²/8 = 43.8 GHz. Recoil the reduced mass misses, of order (mₑ/mₚ)α⁴mc², is 24 MHz for 1s, quoted.
B_∞ is the infinite-mass Dirac binding; mₚ/mₑ = 1836.15; ΔE = 7.41 meV, h × 1.79 THz, raising 1s.Finite nuclear size, the smallest
ΔE(ns) = (2/(3n³)) (Zα)⁴ mc² ⟨r²⟩ (mc/ħ)²
Derived at first order with Quantum Mechanics II's |ψ(0)|² = Z³/(πa³n³). It acts on s states only, lifting 2S₁/₂ above 2P₁/₂ by 0.14 MHz of the measured 1057.8 MHz.
⟨r²⟩¹⁄² = 0.841 fm for the proton, against ħ/mc = 386.16 fm; ΔE = h × 1.1 MHz for hydrogen 1s.Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. From Z = 92 drag right: at 119 the open root crosses the dashed line and normalisability no longer rejects it; at 137 the roots meet at s = 0 as Dirac 1s reaches E = 0 with a vertical tangent; past it they split along Im s, and only 2p₃/₂ survives.
Live results
- COUPLING Zα
- 0.671
- REAL PART γ OF s
- 0.741
- IMAGINARY PART σ OF s
- 0.000
- 2p₃/₂ ENERGY / mc²
- 0.942
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Nuclear charge Z: 92
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The tempting explanation of Z = 137 is Bohr's: the 1s electron moves at Zαc, so at Zα = 1 it would reach c. It is wrong on every count. Bohr's speed for n = 2 is Zαc/2, yet 2s₁/₂ ends at Zα = 1 exactly as 1s does, while 2p₃/₂, with the same n and the same Bohr speed, carries on to Zα = 2: the edge follows j, not n or speed. A spinless particle fails at Zα = ½, where Bohr's speed is only c/2. At Z = 137 itself the Dirac 1s is an ordinary normalisable state with E = 11.7 keV. What fails is the singular −Zαħc/r overpowering the angular barrier at r = 0, a property of a point nucleus: give the nucleus a radius and nothing happens at Z = 137 at all.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
For hydrogen-like ions with a point nucleus, take mc² = 511.0 keV and α = 1/137.036. Find γ and the 1s energy at Z = 118 and at Z = 137, and decide whether normalisability alone still rejects r(−γ) at Z = 118. Then say whether Z = 138 has a 1s level and whether it has a 2p₃/₂ level.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Z = 118: Zα = 0.86109, so γ = √(1 − 0.74147) = 0.5085 and E(1s) = mc²γ = 259.8 keV, bound by 251.2 keV. Since 2γ = 1.017 exceeds 1, the integral of r(−2γ) diverges at r = 0: normalisability still rejects r(−γ), barely.
- Z = 137: Zα = 0.99974, (Zα)² = 0.999475 and γ = √0.000525 = 0.0229, so E(1s) = 11.7 keV, bound by 499.3 keV. The slope −mc²α Zα/γ is −163 keV per unit Z: the curve is almost vertical.
- Z = 138: Zα = 1.00703 and γ² = 1 − 1.01412 = −0.0141, so γ = 0.119i and mc²γ is imaginary. There is no real 1s energy, and 2s₁/₂ and 2p₁/₂, which share |κ| = 1, are gone too.
- 2p₃/₂ has κ = −2 and nᵣ = 0: γ₂ = √(4 − 1.01412) = 1.7280 and E = mc²γ₂/|κ| = 0.8640 mc² = 441.5 keV, bound by 69.5 keV, an ordinary level.
- Checks: Zα and γ are pure numbers, so mc²γ is an energy; as Zα → 0, mc²γ → mc²(1 − (Zα)²/2), which is Bohr's binding.
Answer. Z = 118: γ = 0.508, E(1s) = 260 keV, r(−γ) still rejected; Z = 137: γ = 0.0229, E(1s) = 11.7 keV; Z = 138: no 1s level, but 2p₃/₂ at 441.5 keV.
MediumExample 2
Hydrogen's measured ionisation energy from 1s is 109 678.771 74 cm⁻¹. Take R_∞c = 3.289 841 960 3 × 10¹⁵ Hz, α = 1/137.035 999 and mₚ/mₑ = 1836.152 67. Find the point-nucleus Dirac binding frequency of 1s, correct it for nuclear motion, adding the quoted recoil term of 23.8 MHz, and identify what the measurement still disagrees with.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Infinite mass: B_∞ = mc²(1 − √(1 − α²)) = R_∞hc × 2(1 − √(1 − α²))/α², and that ratio is 1 + α²/4 + α⁴/8 = 1.000 013 313. So B_∞/h = 3.289 885 759 × 10¹⁵ Hz, 43.80 GHz above Bohr's R_∞c.
- Nuclear motion: divide by 1 + mₑ/mₚ = 1.000 544 617 to get 3.288 095 006 × 10¹⁵ Hz, a drop of 1.790 753 THz, then add the 23.8 MHz recoil: 3.288 095 030 × 10¹⁵ Hz.
- Measured: 109 678.771 74 cm⁻¹ × 2.997 924 58 × 10¹⁰ cm s⁻¹ = 3.288 086 857 × 10¹⁵ Hz, which is 13.598 435 eV.
- Theory minus measurement is 8.173 GHz, or 33.8 μeV, 2.5 parts in 10⁶, with the real atom less bound. That is the quoted 1s Lamb shift, 8.173 GHz. Hyperfine structure does not enter, because the measured value refers to the centroid.
- Limits: without relativity the reduced-mass Bohr value, 3.288 051 232 × 10¹⁵ Hz, is 35.6 GHz too small; without the reduced mass the Dirac value is 1.80 THz too large. Only with both does the residual shrink to the Lamb shift.
Answer. Dirac with reduced mass and recoil gives 3.288 095 030 × 10¹⁵ Hz against the measured 3.288 086 857 × 10¹⁵ Hz; the residual, 8.173 GHz (33.8 μeV), is the 1s Lamb shift.
HardExample 3
Model a Z = 130 nucleus as a thin shell of charge of radius R = 8.0 fm, inside which the potential is the constant −Zαħc/R, and take R ≪ ħ/mc so the mass terms drop out near the nucleus. For κ = −1, show that normalisability cannot reject r(−γ), find B/A for G = A rγ + B r(−γ) outside the shell, and the irregular term's size relative to the regular one at r = 50 fm. By what factor does halving R reduce that ratio?
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- Zα = 130/137.036 = 0.94866, so γ = √(1 − 0.89995) = 0.31631 and 2γ = 0.6326. With 2γ below 1, the integral of r(−2γ) converges at r = 0: both roots are normalisable, the band where topic 7.05 needed a finite ⟨V⟩ to choose.
- Inside, with k = Zα/R, the radial equations are dG/dr = G/r + kF and dF/dr = −F/r − kG. Writing G = rg and F = rf gives g″ + (2/r)g′ + k²g = 0, so the regular solution is g = j₀(kr), f = −j₁(kr), and at r = R, where kR = Zα, ρ = F/G = −j₁(Zα)/j₀(Zα).
- At x = 0.94866, sin x = 0.81263 and cos x = 0.58278, so j₀ = sin x/x = 0.85662, j₁ = sin x/x² − cos x/x = 0.28866 and ρ = −0.33698. For small Zα, ρ → −Zα/3.
- Outside, a₊ = (γ − 1)/(Zα) = −0.72069 and a₋ = −(1 + γ)/(Zα) = −1.38755. Then B/A = R(2γ)(a₊ − ρ)/(ρ − a₋) = R(2γ)(−0.38372/1.05058) = −0.365 R(0.633), which is −1.36 fm(0.633) for R = 8.0 fm.
- The ratio of irregular to regular term at radius r is (B/A)r(−2γ) = −0.365(R/r)(0.633). At r = 50 fm, (8.0/50)(0.633) = 0.314, so the irregular term is −0.115 times the regular one, at a radius still inside ħ/mc = 386 fm, where the power laws roughly hold. Halving R multiplies it by 2(−0.633) = 0.645.
- Checks: for small Zα the coefficient tends to −(Zα)²/12, and a numerical integration of the same equations reproduces B/A. For uranium the halving factor is 2(−1.48) = 0.36; at Z = 137 it is 0.97, and at γ = 0 nothing is suppressed.
Answer. B/A = −0.365 R(0.633), or −1.36 fm(0.633) at R = 8.0 fm; the irregular term is about −11% of the regular one at 50 fm, and halving R multiplies it by 0.645, so it vanishes as R → 0, but slowly.
Quick check
Select an answer and explain it before checking.
Review & continue
The point-nucleus Dirac atom ends where the exponent γ = √(κ² − (Zα)²) of its radial functions at r = 0 stops being real, at Zα = 1 for the j = ½ levels; beyond that the equation cannot fix a spectrum without the nucleus's size.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.