
Advanced Quantum Mechanics course sequence · Interactive lesson
The ground state: weak singularity and small component
The 1s spinor is the simplest Dirac–Coulomb solution, one term you can write down whole, and it turns γ = √(1 − (Zα)²) into numbers you can weigh: how much of the electron lives in the small component, how steeply its density climbs at the nucleus, and why uranium's electron feels its nucleus's size while hydrogen's barely notices.
UAQM-07.08 · Lesson 8 of 9 in this unit. Position is not a completion score.Build the model
One power of r and one exponential carry the whole relativistic ground state.
The key idea
The Dirac ground state of a hydrogen-like ion is a single power-times-exponential spinor, G ∝ rγ e(−Zr/a₀) with F = −(1 − γ)G/(Zα), in which γ = √(1 − (Zα)²) fixes the energy, the weight of the small component and how sharply the density rises at the nucleus.
Example. Hydrogen-like uranium, Z = 92: Zα = 0.6714 and γ = 0.7411, so E = 0.7411 × 511.0 keV = 378.7 keV (a binding of 132.3 keV), F/G = −0.386 at every radius, and the small component holds (1 − 0.7411)/2 = 12.9% of the probability.
The complete physical model and assumptions
For n = 1 the termination condition leaves nothing to terminate: nᵣ = 0 with κ = −1, and the Coulomb radial pair is solved by one power times one exponential, G = A rγ e(−λr) and F = B rγ e(−λr). The 1/r terms of the two equations each fix B/A, and the two values agree only if γ = √(1 − (Zα)²); the constant terms then fix λ = Z/a₀, exactly Bohr's, and E = γmc². So everything relativistic about the ground state sits in one number, γ, which falls below 1 by about (Zα)²/2. Three consequences follow, all derived here. The lower component is a fixed multiple of the upper at every radius, F/G = −(1 − γ)/(Zα), and holds a share (1 − γ)/2 of the probability: 1.3 × 10⁻⁵ in hydrogen, 13% in uranium. The density near the nucleus goes as r(2γ−2), divergent but integrable, a weak singularity whose log–log slope is −5.3 × 10⁻⁵ in hydrogen and −0.52 in uranium. And because both theories share the exponential, the ratio of the Dirac to the Schrödinger density is 2(2Zr/a₀)(2γ−2)/Γ(2γ + 1): 1.0006 at the proton's radius, about 10 at the surface of a uranium nucleus. That is where the model stops. A real nucleus has a size, inside it the potential is not −Zαħc/r, and the r(2γ−2) rise is cut off; in uranium the cut-off moves the 1s level by about 200 eV, a correction the next topic takes up.
1One term of the series solves the ground state
Work in SI units, quoting energies in keV and lengths in fm; where four-vectors appear the metric is (+, −, −, −). The nucleus is a point charge of infinite mass, V = −Zαħc/r, and topic 7.4's spinor, G(r)/r times the spinor harmonic of κ above and iF(r)/r times that of −κ below, obeys ħc(G′ + κG/r) = (E + mc² − V)F and ħc(F′ − κF/r) = −(E − mc² − V)G. For n = 1, topic 7.6 leaves nᵣ = 0 and κ = −1, so the series has a single term: try G = A rγ e(−λr) and F = B rγ e(−λr). Collect the 1/r terms. The first equation gives (γ − 1)A = ZαB, the second (γ + 1)B = −ZαA. These are two values of B/A, (γ − 1)/(Zα) and −Zα/(γ + 1), and they agree only if γ² = 1 − (Zα)². Topic 7.5's exponent at the origin returns, now as the condition that one term solve both equations. Dimensions: γ and Zα are pure numbers, and ħc/r, mc² and V are energies, as each equation requires.
2The constant terms fix the decay length and the energy
The terms without 1/r give −ħcλA = (E + mc²)B and −ħcλB = −(E − mc²)A. Their product gives (ħcλ)² = m²c⁴ − E², and eliminating B/A with the 1/r result gives E = γmc² and ħcλ = Zαmc², so λ = Z/a₀ with a₀ = ħ/(mcα) = 52 918 fm. The decay length is Bohr's; everything relativistic is in γ. For hydrogen the binding mc²(1 − γ) is 13.6059 eV against Bohr's 13.6057 eV, and γmc² = mc² − (Zα)²mc²/2 − (Zα)⁴mc²/8 − … reproduces Quantum Mechanics II's 1s fine-structure shift, −(Zα)⁴mc²/8 = −0.18 meV. Scaled by the reduced mass the Dirac value is 13.59847 eV, 34 μeV above the measured ionisation energy, 13.59843 eV; that gap is the quoted 1S Lamb shift, 8.17 GHz. At Z = 92 the binding is 132.28 keV, against 115.16 keV from Bohr and 128.13 keV through order (Zα)⁴.
3Normalise with a gamma function, then weigh the small part
Because F/G is one constant at every radius, G² + F² = G²[1 + (1 − γ)/(1 + γ)] = 2G²/(1 + γ), and normalisation needs a single integral, ∫₀^∞ r(2γ) e(−2Zr/a₀) dr = Γ(2γ + 1)/(2Z/a₀)(2γ+1). Hence A² = (1 + γ)(2Z/a₀)(2γ+1)/[2Γ(2γ + 1)], in m(−2γ−1) so that G² dr is a probability. At γ = 1, Γ(3) = 2 and A = 2(Z/a₀)³⁄², Schrödinger's 1s, as it must be. The share in the lower component is a ratio of identical integrals, Pₛₘₐₗₗ = [(1 − γ)/(1 + γ)] ÷ [2/(1 + γ)] = (1 − γ)/2. Hydrogen: 1.33 × 10⁻⁵. Iron, Z = 26: 0.91%. Uranium: 12.9%. Unit 6's order-v/c estimate ⟨T⟩/2mc² = (Zα)²/4 agrees with hydrogen's value to a part in 75 000, but gives 0.113 for uranium, 13% short, because (1 − γ)/2 = (Zα)²/4 + (Zα)⁴/16 + … and the higher terms are no longer small.
4A second route: the Coulomb virial theorem
Quantum Mechanics II found the virial theorem from ⟨[H, r⋅p]⟩ = 0 in a stationary state; repeat it for H = cα⋅p + βmc² + V. Since [pᵢ, rⱼ] = −iħδᵢⱼ, [cα⋅p, r⋅p] = −iħ cα⋅p, while [V, r⋅p] = iħ r⋅∇V and β commutes with r⋅p. For V ∝ 1/r, r⋅∇V = −V, so ⟨cα⋅p⟩ = −⟨V⟩ and E = mc²⟨β⟩ for every bound state of the point Coulomb field. In the Dirac representation β is +1 on the upper pair and −1 on the lower, so ⟨β⟩ = 1 − 2Pₛₘₐₗₗ. With E = γmc², Pₛₘₐₗₗ = (1 − γ)/2 follows without an integral, and the two routes check each other. The same wavefunction gives ⟨1/r⟩ = Z/(γa₀), so ⟨V⟩ = −(Zα)²mc²/γ: in uranium −310.8 keV against Schrödinger's −230.3 keV, the electron sitting 1/γ = 1.35 times deeper in the potential.
5A weak singularity at the nucleus
The upper radial function is g = G/r ∝ r(γ−1), so for γ < 1 the Dirac 1s wavefunction diverges at the nucleus where Schrödinger's is finite. The density ρ = (G² + F²)/(4πr²) goes as r(2γ−2), yet stays integrable because 4πr²ρ ∝ r(2γ). Both theories share e(−2Zr/a₀), so their ratio is exact and simple: ρD/ρS = 2(2Zr/a₀)(2γ−2)/Γ(2γ + 1), which at γ = 1 is 2/Γ(3) = 1 at every r, as it must be. On a log–log plot the Dirac density near the origin is a line of slope 2γ − 2: −5.3 × 10⁻⁵ for hydrogen, −0.518 for uranium. At the proton's charge radius, 0.841 fm, the hydrogen ratio is 1.0006, and doubling it would take r ≈ 10(−5653) a₀. In uranium the ratio is 10.0 at the nuclear surface, 7.56 fm, and falls below 1 only beyond 1.13 a₀/Z ≈ 650 fm, so the extra density near the nucleus is drawn from farther out: ⟨r⟩ = (2γ + 1)a₀/(2Z) is 714 fm against Schrödinger's 863 fm.
6Where the size of the nucleus takes over
Quantum Mechanics II estimated the nuclear-size shift of an s level from the Schrödinger contact density: for 1s, ΔE = (2/3)(Zα)⁴mc²(rᵣₘₛ/ƛC)², with ƛC = ħ/mc = 386.16 fm. For hydrogen, with rᵣₘₛ = 0.841 fm, that is 4.6 neV, or 1.1 MHz, and relativity alters it by the 6 × 10⁻⁴ just found, about 0.7 kHz: small, though large enough that precision hydrogen theory includes it. For uranium, rᵣₘₛ = 5.857 fm gives 15.9 eV, but the density that estimate assumes is about ten times too small across the nucleus, and the full Dirac calculation for a finite nucleus, quoted here, gives about 199 eV, 12.5 times more. That shift is part of the 1s Lamb shift of U⁹¹⁺ measured as 460 ± 5 eV (Gumberidze and co-workers, 2005); the rest is almost all radiative. Inside a real nucleus the potential is finite, g tends to a constant and the r(2γ−2) rise stops. The point-nucleus spinor is a good description outside the nucleus and the wrong one inside it; topic 7.9 takes it from there.
Key equations
Read the units, assumptions and limits before substituting.
Ground-state large component (derived)
G(r) = A rγ e(−Zr/a₀), with γ = √(1 − (Zα)²)
The whole series for nᵣ = 0, κ = −1. The decay constant Z/a₀ is Bohr's, unchanged; only the power of r near the origin is relativistic.
r in m; a₀ = ħ/(mcα) = 5.29 × 10⁻¹¹ m; A in m(−γ−1/2) so that G² dr is a probability; Zα and γ are pure numbers.Small-to-large ratio, the same at every r (derived)
F(r)/G(r) = −(1 − γ)/(Zα) = −Zα/(1 + γ)
Each 1/r balance gives one form, and they agree only because γ² = 1 − (Zα)². A ratio constant in r is special to states with nᵣ = 0.
Dimensionless. Hydrogen −3.65 × 10⁻³; uranium (Z = 92) −0.386; tends to −Zα/2 as Zα → 0.Ground-state energy (derived)
E₁ = γmc² = mc² √(1 − (Zα)²)
Expands to mc² − (Zα)²mc²/2 − (Zα)⁴mc²/8 − …: Bohr's level, then Quantum Mechanics II's 1s fine-structure shift, then what order α⁴ misses.
mc² = 511.0 keV. Binding mc²(1 − γ): 13.6059 eV at Z = 1 and 132.28 keV at Z = 92, point nucleus of infinite mass.Normalisation (derived)
A² = (1 + γ)(2Z/a₀)(2γ+1) / [2 Γ(2γ + 1)]
At γ = 1, Γ(3) = 2 and A = 2(Z/a₀)³⁄²: Schrödinger's 1s, so the non-relativistic limit is built in.
From ∫₀^∞ r(2γ) e(−2Zr/a₀) dr = Γ(2γ + 1)/(2Z/a₀)(2γ+1) and G² + F² = 2G²/(1 + γ); A² in m(−2γ−1).Probability in the small component (derived)
Pₛₘₐₗₗ = ∫₀^∞ F² dr = (1 − γ)/2
Also follows with no integral from the Coulomb virial result E = mc²⟨β⟩, since ⟨β⟩ = 1 − 2Pₛₘₐₗₗ in the Dirac representation.
Dimensionless. Hydrogen 1.33 × 10⁻⁵; iron (Z = 26) 0.91%; uranium 12.9%. Lowest order (Zα)²/4.Dirac over Schrödinger density (derived)
ρD(r)/ρS(r) = 2 (2Zr/a₀)(2γ−2) / Γ(2γ + 1)
Its log–log slope 2γ − 2 is the weak singularity. Where the ratio is large, the size of the nucleus, not the point charge, decides the density.
Both densities in m⁻³; the shared e(−2Zr/a₀) cancels. Hydrogen at 0.841 fm: 1.0006. Uranium at 7.56 fm: 10.0.Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Start at Z = 1 with r = 0.8 fm: the curves coincide and the enhancement reads 1.0006. Raise Z to 92 with r = 7.6 fm, the edge of a uranium nucleus: the solid line tilts to slope −0.52 and the density there is ten times Schrödinger's. Then shrink r and watch the gap keep growing.
Live results
- F/G AT EVERY r
- -0.3856
- SMALL-COMPONENT SHARE
- 12.9433 %
- ρD/ρS AT PROBE r
- 9.9944
- 1s BINDING mc²(1 − γ)
- 132.2799 keV
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; Probe radius r: 7.6 fm
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
Seeing g ∝ r(γ−1) diverge at r = 0, students conclude that the Dirac ground state is unphysical, or that relativity must matter strongly at every nucleus because the contact density behind the hyperfine and nuclear-size shifts has become infinite. Neither follows. The divergence is integrable: 4πr²ρ ∝ r(2γ) vanishes at the origin and the norm is finite. And it is weak. In hydrogen the density's exponent 2γ − 2 is −5.3 × 10⁻⁵, so the density at the proton's radius exceeds Schrödinger's by 6 parts in 10⁴, and it would double only at r ≈ 10(−5653) a₀, far inside any nucleus. No measurement samples r = 0 itself: the nuclear-size shift samples the density across the nucleus, where the enhancement is a finite number set by γ: 1.0006 for hydrogen, about 10 for uranium. The infinity belongs to the point-charge idealisation, and the size of the nucleus removes it before anything can see it.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
Hydrogen-like iron, Fe²⁵⁺ (Z = 26), with a point nucleus of infinite mass. Find γ, the 1s binding energy, the ratio F/G and the share of probability in the small component, and compare each with its lowest-order value.
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.
- Zα = 26/137.036 = 0.18973, so (Zα)² = 0.035998 and γ = √(1 − 0.035998) = √0.964002 = 0.98184. The log–log slope of the density at the origin, 2γ − 2 = −0.0363, is already about 680 times hydrogen's.
- Binding: mc²(1 − γ) = 510.999 keV × 0.018164 = 9.282 keV. Bohr's (Zα)²mc²/2 = 9.197 keV; adding the order-(Zα)⁴ term (Zα)⁴mc²/8 = 0.083 keV gives 9.280 keV, and the last 1.5 eV is the next term, (Zα)⁶mc²/16.
- F/G = −Zα/(1 + γ) = −0.18973/1.98184 = −0.0957, against the lowest-order −Zα/2 = −0.0949.
- Pₛₘₐₗₗ = (1 − γ)/2 = 0.018164/2 = 0.00908, or 0.908%. The order-v/c estimate (Zα)²/4 = 0.00900 is 0.9% low, close to the relative correction (Zα)²/4 itself.
- Checks: γ, F/G and Pₛₘₐₗₗ are pure numbers and the binding is an energy. Setting Z = 1 in the same formulas returns hydrogen's 13.6059 eV and 1.33 × 10⁻⁵.
Answer. γ = 0.98184; binding 9.282 keV (Bohr 9.197 keV, through order α⁴ 9.280 keV); F/G = −0.0957; the small component holds 0.908% of the probability, against 0.900% at lowest order.
MediumExample 2
Hydrogen-like lead, Pb⁸¹⁺ (Z = 82), with a point nucleus. Use the Coulomb virial relation E = mc²⟨β⟩ and the 1s wavefunction to find ⟨β⟩, ⟨V⟩ and ⟨cα⋅p⟩, check that they add up to E, and compare ⟨V⟩ and ⟨r⟩ with Schrödinger's values.
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.
- Zα = 82/137.036 = 0.59838, (Zα)² = 0.35806 and γ = √0.64194 = 0.80121. Then E = γmc² = 0.80121 × 510.999 keV = 409.42 keV, a binding of 101.58 keV.
- E = mc²⟨β⟩ gives ⟨β⟩ = γ = 0.80121, so Pₛₘₐₗₗ = (1 − ⟨β⟩)/2 = 0.0994: about a tenth of the probability is in the lower pair.
- ⟨1/r⟩ = ∫r(2γ−1)e(−2Zr/a₀)dr ÷ ∫r(2γ)e(−2Zr/a₀)dr = (2Z/a₀)Γ(2γ)/Γ(2γ + 1) = Z/(γa₀). With ħc/a₀ = αmc², ⟨V⟩ = −Zαħc⟨1/r⟩ = −(Zα)²mc²/γ = −0.35806 × 510.999 keV/0.80121 = −228.37 keV, and the virial relation gives ⟨cα⋅p⟩ = +228.37 keV.
- Check: ⟨cα⋅p⟩ + mc²⟨β⟩ + ⟨V⟩ = 228.37 + 409.42 − 228.37 = 409.42 keV = E. Every term is an energy, and as γ → 1, ⟨V⟩ → −(Zα)²mc², Schrödinger's value and twice the binding.
- Schrödinger's ⟨V⟩ = −(Zα)²mc² = −182.97 keV, so the Dirac electron sits 1/γ = 1.248 times deeper. ⟨r⟩ = (2γ + 1)a₀/(2Z) = 2.6024 × 52 918 fm/164 = 839.7 fm, against 3a₀/(2Z) = 968.0 fm: a contraction of 13.3%.
Answer. ⟨β⟩ = 0.8012 (Pₛₘₐₗₗ = 9.9%), ⟨V⟩ = −228.4 keV and ⟨cα⋅p⟩ = +228.4 keV, which with mc²⟨β⟩ = 409.4 keV sum to E; ⟨V⟩ is 1.25 times Schrödinger's −183.0 keV, and ⟨r⟩ = 840 fm is 13% below 968 fm.
HardExample 3
Treat uranium-238 as a sphere of radius R = 7.56 fm, the uniform-charge radius that matches its measured rms charge radius of 5.857 fm, but keep the point-nucleus 1s wavefunctions of U⁹¹⁺. Find the probability that the electron lies inside r < R in the Dirac and in the Schrödinger theory, their ratio, and how that ratio relates to the density enhancement at r = R.
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.
- Both radial densities are normalised forms of r(2γ)e(−2Zr/a₀), so with t = 2Zr/a₀ the probability inside R is P = ∫₀x t(2γ)e(−t)dt / Γ(2γ + 1), where x = 2ZR/a₀ = 184 × 7.56 fm/52 918 fm = 0.02629. Schrödinger is the case γ = 1.
- For x ≪ 1 expand e(−t) ≈ 1 − t, giving P ≈ [xs/Γ(s + 1)][1 − sx/(s + 1)] with s = 2γ + 1; the next correction is of order x², below 10⁻³.
- Schrödinger, s = 3: PS = (0.02629³/6)(1 − 0.75 × 0.02629) = 3.03 × 10⁻⁶ × 0.9803 = 2.97 × 10⁻⁶.
- Dirac, γ = 0.74113 and s = 2.48227: xs = exp(2.48227 × ln 0.02629) = exp(−9.032) = 1.195 × 10⁻⁴; Γ(s + 1) = 2.48227 × Γ(2.48227) = 2.48227 × 1.3130 = 3.2592; the bracket is 1 − 0.7128 × 0.02629 = 0.9813. So PD = (1.195 × 10⁻⁴/3.2592) × 0.9813 = 3.60 × 10⁻⁵.
- Ratio PD/PS = 12.1: the electron spends 1 part in 27 800 of its time inside the nucleus, not 1 in 337 000. To leading order the ratio is the surface enhancement 2x(2γ−2)/Γ(2γ + 1) = 10.0 times 3/(2γ + 1) = 1.209, because the enhancement grows toward the centre and its average over the sphere exceeds its surface value.
- Checks: P is dimensionless, and as Z → 1, γ → 1 and the ratio → 1. The same factor shows in the energy: the full finite-nucleus Dirac calculation of the 1s size shift, about 199 eV (quoted), is 12.5 times Schrödinger's 15.9 eV, and that shift is part of the measured 1s Lamb shift of U⁹¹⁺, 460 ± 5 eV.
Answer. PD ≈ 3.60 × 10⁻⁵ against PS ≈ 2.97 × 10⁻⁶ inside 7.56 fm, a ratio of 12.1, which is the surface enhancement 10.0 times 3/(2γ + 1); it is close to the factor of about 12.5 by which the full finite-nucleus shift exceeds Schrödinger's estimate.
Quick check
Select an answer and explain it before checking.
Review & continue
The Dirac ground state of a hydrogen-like ion is a single power-times-exponential spinor, G ∝ rγ e(−Zr/a₀) with F = −(1 − γ)G/(Zα), in which γ = √(1 − (Zα)²) fixes the energy, the weight of the small component and how sharply the density rises at the nucleus.
- 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.