
Advanced Quantum Mechanics course sequence · Interactive lesson
The coupled radial equations for G(r) and F(r)
This is the step that turns the Dirac atom into a calculation. Two real first-order equations for G(r) and F(r) replace a four-component wave equation, and the rest of the unit, exact energies included, is read off them. Derive them once, carefully, and test them on the free particle and on Quantum Mechanics II.
UAQM-07.04 · Lesson 4 of 9 in this unit. Position is not a completion score.Build the model
Four components in three dimensions reduce to two real functions of r.
The key idea
The coupled radial equations are the two real first-order equations, one from each block row of the Dirac equation, obeyed by the radial functions G(r) and F(r) of an electron's upper and lower components in any central potential.
Example. For s(1/2) states, κ = −1, they read dG/dr = G/r + (E + mc² − V)F/ħc and dF/dr = −F/r − (E − mc² − V)G/ħc; a free electron of kinetic energy 1.00 MeV solves them with G = sin kr and F = −0.703 kr j₁(kr).
The complete physical model and assumptions
Once J², Jz, parity and Dirac's K are diagonal, nothing about an electron in a central field is left open except two functions of r. The angles are fixed by spinor harmonics: the upper pair of components carries χκm with orbital l, the lower pair χ(−κ)m with orbital l′ = 2j − l and the opposite parity, and the radial amplitudes are written G/r and iF/r. The one derivative in the Dirac Hamiltonian, cσ⋅p, splits as σ⋅p = (σ⋅r̂)(r̂⋅p + iσ⋅L/r) into −iħ∂/∂r and an angular operator the harmonics already diagonalise, while σ⋅r̂ swaps χκm and χ(−κ)m. Each block row of Hψ = Eψ then collapses to one real first-order equation, the lower row giving dG/dr and the upper dF/dr, with κ entering the two with opposite signs. Nothing is approximated: the pair holds for any central V. Its checks are its limits. With V = 0 it decouples exactly into Quantum Mechanics II's free radial equations of orders l and l′. When E − mc² and V are small beside 2mc² = 1.022 MeV, F is slaved to G, F ≈ (ħ/2mc)(G′ + κG/r), and G obeys Quantum Mechanics II's radial equation with l(l + 1) = κ(κ + 1) for both signs of κ, which is why 2P(1/2) and 2P(3/2) share a radial function at that order. That reduction fails where |V| nears 2mc², within about Z × 1.41 fm of a point nucleus, and there the equations must be solved as they stand.
1Two radial functions are all that is left
SI units throughout, with E the total energy including mc²; where four-vectors appear the metric is (+, −, −, −), though nothing here needs one. This unit's earlier topics supplied the labels, quoted here: in a central V(r), J², Jz, Dirac parity and K = β(Σ⋅L + ħ) commute with H, and K's eigenvalue −κħ fixes j = |κ| − ½ and the orbital l of the upper pair. So the upper two components are a radial function times the spinor harmonic χκm. The lower two share j and m but, because Dirac parity is β times spatial inversion, have the opposite orbital parity, so they carry χ(−κ)m with l′ = 2j − l. Write ψ = ((G/r)χκm, i(F/r)χ(−κ)m)ᵀ. The 1/r does the job of QM-II's u = rR, and the i is a choice that will make both equations real. The harmonics are orthonormal on the sphere, so ∫ψ†ψ d³r = ∫₀^∞ (G² + F²) dr: G and F carry m(−1/2), and one integral normalises the pair. Four complex functions of three coordinates have become two real functions of one.
2Splitting σ⋅p into a radial derivative and an angular label
The only derivative in H = cα⋅p + βmc² + V sits in cσ⋅p, which mixes radius and angle. Insert (σ⋅r̂)² = 1 on the left and use (σ⋅a)(σ⋅b) = a⋅b + iσ⋅(a × b), valid for operators if their order is kept: σ⋅p = (σ⋅r̂)[r̂⋅p + iσ⋅(r̂ × p)]. With r̂ standing to the left of p, r̂⋅p = −iħ∂/∂r and r̂ × p = L/r, so σ⋅p = (σ⋅r̂)(−iħ∂/∂r + iσ⋅L/r), derived. The angular part is already diagonal: since (σ⋅L + ħ)χκm = −κħχκm, σ⋅L is −(1 + κ)ħ on χκm and (κ − 1)ħ on χ(−κ)m. And σ⋅r̂, which commutes with J but is odd under inversion, swaps the two harmonics, σ⋅r̂ χκm = −χ(−κ)m in the previous topic's phase convention. On (f/r)χκm the bracket gives −iħ[f′/r − f/r² + (1 + κ)f/r²] = −(iħ/r)(f′ + κf/r), and σ⋅r̂ then supplies −χ(−κ)m. In one line: σ⋅p[(f/r)χ(±κ)m] = (iħ/r)(f′ ± κf/r)χ(∓κ)m.
3Each block row becomes one real first-order equation
Call the upper pair φ = (G/r)χκm and the lower pair χ = i(F/r)χ(−κ)m. The lower row of Hψ = Eψ is cσ⋅pφ = (E + mc² − V)χ: its left side is (iħc/r)(G′ + κG/r)χ(−κ)m and its right side i(E + mc² − V)(F/r)χ(−κ)m, so the i, the 1/r and the harmonic cancel and dG/dr = −(κ/r)G + (E + mc² − V)F/ħc. The upper row, cσ⋅pχ = (E − mc² − V)φ, gives i × (iħc/r)(F′ − κF/r) = −(ħc/r)(F′ − κF/r) on χκm, so dF/dr = (κ/r)F − (E − mc² − V)G/ħc. Both are derived, for any central V, with nothing dropped. Dimensions: κ/r and an energy over ħc are both m⁻¹. Every coefficient is real, so G and F can be taken real; the radial current cψ†(α⋅r̂)ψ ∝ iGF(|χ(−κ)m|² − |χκm|²) then vanishes, since σ⋅r̂ is unitary and the two angular densities are equal. The system's matrix is traceless, so G₁F₂ − F₁G₂ is constant for two solutions at one E: two solutions regular at r = 0 are proportional, one per κ and E, as in QM-II.
4Free-field check: Bessel functions of orders l and l′
With V = 0 the pair decouples exactly. Differentiate the first equation, substitute the second, and replace (E + mc²)F/ħc by G′ + κG/r: the κG′/r terms cancel and G″ + [k² − κ(κ + 1)/r²]G = 0, with k² = (E² − m²c⁴)/(ħc)². That is QM-II's free radial equation for u = rR, and κ(κ + 1) is l(l + 1) for both signs of κ. The same steps give F″ + [k² − κ(κ − 1)/r²]F = 0, and κ(κ − 1) = l′(l′ + 1): F carries the other orbital. The regular solutions are G = kr jₗ(kr) and F = B kr j(l′)(kr), and the first equation fixes B = sign(κ) cp/(E + mc²), derived. As p → 0, B → p/2mc, Unit 6's v/2c; as p → ∞, B → 1, and the small component of an ultrarelativistic electron is as large as the other. At a kinetic energy T = 1.00 MeV, B = 0.703 and F holds T/2E = 33% of the probability. Since E² = p²c² + m²c⁴ is exact, this case tests every sign in the derivation with no approximation to hide behind.
5The non-relativistic limit returns QM-II's radial equation
Solve the first equation for F, still exactly: F = ħc(G′ + κG/r)/(2mc² + ε − V), with ε = E − mc². For a slow electron ε and V are electron-volts against 2mc² = 1.022 MeV, so F ≈ (ħ/2mc)(G′ + κG/r): F has no freedom of its own. Put this into the second equation. The κG′/r terms cancel as in the free case, and multiplying by −ħc leaves −(ħ²/2m)G″ + [V + ħ²κ(κ + 1)/2mr²]G = εG, derived. That is QM-II's radial equation for u = rR, and κ(κ + 1) is l(l + 1) whether κ = −(l + 1) or κ = l, so G is QM-II's uₙₗ for either j. For hydrogen it returns ε = −13.6 eV/n², whose 2p → 1s line at 10.204 eV matches the measured Lyman-α, 10.199 eV at 121.567 nm, to 5 parts in 10⁴, the reduced-mass correction. Integrating F² by parts with the same cancellation gives ∫F² dr = ⟨T⟩/2mc² for every κ: 3.33 × 10⁻⁶ for any n = 2 level of hydrogen, Unit 6's result recovered from the radial equations.
6What the leading order leaves out, and where it fails
κ(κ + 1) = 2 for both κ = −2 and κ = +1, so at leading order 2P(3/2) and 2P(1/2) share a G and an energy. Hydrogen's measured splitting is 10.969 GHz, or 4.536 × 10⁻⁵ eV, which is 1.33 × 10⁻⁵ of the 3.40 eV binding, α²/4: the relative size, (v/c)², of the (ε − V)/2mc² dropped from the denominator. Keep that term to first order and the radial equation gains two things: a factor 1 − (ε − V)/2mc² on its kinetic and centrifugal terms, and a new term −(ħ²V′/4m²c²)(G′ + κG/r). The new term's piece −κħ²V′G/4m²c²r is the only one that tells the two signs of κ apart, and writing −κ = 1 + σ⋅L/ħ shows it is Unit 6's spin–orbit coupling, Thomas factor included, now emerging from the radial equations; the Hard example evaluates it. The expansion itself fails where |V| is not small beside 2mc²: for Coulomb, inside r = Zαħ/2mc = Z × 1.41 fm, which is 1.41 fm in hydrogen but 130 fm in uranium, far outside its nucleus. There F is no longer slaved to G, and the exact equations must be solved as they stand, the next topic's work at the origin and at infinity.
Key equations
Read the units, assumptions and limits before substituting.
Separation in a central field
ψ = ( (G(r)/r) χκm , i (F(r)/r) χ(−κ)m )ᵀ
J², Jz, K and parity fix every angle. The 1/r is QM-II's u = rR, the i makes the equations real, and ∫ψ†ψ d³r becomes ∫(G² + F²) dr.
G, F real, in m(−1/2); χκm and χ(−κ)m orthonormal spinor harmonics of one j and m, orbitals l and l′ = 2j − l.σ⋅p on a radial–angular product
σ⋅p [ (f/r) χ(±κ)m ] = (iħ/r) (df/dr ± κf/r) χ(∓κ)m
From σ⋅p = (σ⋅r̂)(r̂⋅p + iσ⋅L/r): a radial derivative, a κ term from σ⋅L, and σ⋅r̂ swapping the two harmonics.
f(r) any radial function; σ⋅L = −(1 + κ)ħ on χκm. The phase σ⋅r̂ χκm = −χ(−κ)m is quoted from the previous topic; the formula is derived here.From the lower row: dG/dr
dG/dr = −(κ/r) G + (E + mc² − V) F / ħc
The coefficient of F is about 2mc/ħ = 5.18 × 10¹² m⁻¹, so a small F still shapes G: this term carries the kinetic energy.
E total energy and V(r) in J; ħc = 3.16 × 10⁻²⁶ J m; κ = ∓(j + ½), a nonzero integer. Each term is in m(−3/2).From the upper row: dF/dr
dF/dr = (κ/r) F − (E − mc² − V) G / ħc
Its small coefficient of G, against the large coefficient of F above, is the asymmetry that keeps F of order v/c times G.
Same symbols, with κ entering at the opposite sign. E − mc² − V is the local kinetic energy, electron-volts in an atom.Small component of a slow electron
F ≈ (ħ/2mc) (dG/dr + κG/r)
F is slaved to G. Hydrogen 2P(3/2), κ = −2: F = −(α/4)G at every radius; 2P(1/2), κ = +1, has the same G but an s-wave F.
Valid where |E − mc² − V| ≪ 2mc² = 1.022 MeV; ħ/2mc = 1.93 × 10⁻¹³ m. Exact: F = ħc(dG/dr + κG/r)/(E + mc² − V).Non-relativistic limit for G
−(ħ²/2m) d²G/dr² + [ V + ħ²κ(κ + 1)/2mr² ] G = (E − mc²) G
Derived by eliminating F. G is QM-II's uₙₗ, so 2P(1/2) and 2P(3/2) coincide at this order; their measured 10.97 GHz splitting is the α² left out.
κ(κ + 1) = l(l + 1) for κ = −(l + 1) and for κ = l: QM-II's radial equation for u = rR, energies in J.Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. At κ = −1 slide p down to 0.05: the dashed p wave shrinks to a ripple of amplitude p/2mc. Push p to 4 and its amplitude, 0.781, nears the solid curve's without reaching it. Then step to κ = +1: G becomes the p wave and F the s wave, leaving r = 0 with a finite slope.
Live results
- κ
- -1
- F/G AMPLITUDE cp/(E + mc²)
- 0.236
- PROBABILITY SHARE IN F
- 0.053
- KINETIC ENERGY
- 60.3 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. Momentum p: 0.50 mc; State: κ = −2, −1, +1, +2: 2
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
Because F is about v/2c of G, it is tempting to drop it and keep dG/dr = −(κ/r)G. That equation gives G ∝ r(−κ): a pure power with no energy in it, growing without limit for κ < 0 and blowing up at the origin for κ > 0, so no bound state at all. The error is judging a term by the size of the function instead of the size of the product. F is small, but it is multiplied by (E + mc² − V)/ħc ≈ 2mc/ħ = 5.18 × 10¹² m⁻¹, and that product, (2mc/ħ)F ≈ G′ + κG/r, is exactly as large as the terms beside it. The small component is negligible in the normalisation, where it enters as F², a few parts in 10⁶ for hydrogen's n = 2 levels; it is not negligible in the equations, where it carries the whole kinetic energy into the equation for G. Eliminating F, not deleting it, is what returns Schrödinger's equation.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
An electron with 50.0 keV of kinetic energy moves through a field-free region in the p(1/2) partial wave, κ = +1. From the coupled radial equations, say which Riccati–Bessel functions G and F are, find the ratio of their amplitudes and the wavenumber k, and compare the ratio with the slow-electron estimate p/2mc.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- With V = 0 the pair decouples into G″ + [k² − κ(κ + 1)/r²]G = 0 and F″ + [k² − κ(κ − 1)/r²]F = 0. For κ = +1, κ(κ + 1) = 2 gives l = 1 and κ(κ − 1) = 0 gives l′ = 0, so G = kr j₁(kr) and F = B kr j₀(kr) = B sin kr.
- Fix B from the first equation, G′ + G/r = (E + mc²)F/ħc. With x = kr, d(x j₁)/dx + j₁ = x j₀, so the left side is k × kr j₀(kr) and B = ħck/(E + mc²) = cp/(E + mc²), positive because κ > 0.
- Numbers: E + mc² = 50.0 + 1022.0 = 1072.0 keV and pc = √(T(T + 2mc²)) = √(50.0 × 1072.0) keV = 231.5 keV, so B = 231.5/1072.0 = 0.216. Then k = pc/ħc = 231.5 keV / (197.33 keV pm) = 1.173 pm⁻¹, a wavelength 2π/k = 5.36 pm.
- Slow-electron estimate: p/2mc = 231.5/1022.0 = 0.227, 4.9% high, because it replaces E + mc² = 2mc² + T by 2mc²; the factor 1 + T/2mc² = 1.049 is the whole difference. Checks: B is dimensionless, B → 0 as T → 0, and F = B sin kr leaves r = 0 with a finite slope, so F/r is finite at the origin, as an s wave must be.
Answer. G = kr j₁(kr), a p wave; F = 0.216 sin kr, an s wave; k = 1.17 × 10¹² m⁻¹. The estimate p/2mc = 0.227 is 4.9% high at 50 keV.
MediumExample 2
At leading order hydrogen's 2P(3/2) (κ = −2) and 2P(1/2) (κ = +1) share QM-II's G(r) = N r² e(−r/2a₀), N = (24a₀⁵)(−1/2). Use F ≈ (ħ/2mc)(dG/dr + κG/r) to find F for each, compare their weights ∫F² dr, and find the 2P(1/2) density at the nucleus as a fraction of the 2s density 1/(8πa₀³).
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.
- Both levels have κ(κ + 1) = 2, hence one G. From G = N r² e(−r/2a₀), dG/dr = (2/r − 1/2a₀)G, and ħ/2mc = αa₀/2 because a₀ = ħ/αmc.
- 2P(3/2), κ = −2: dG/dr − 2G/r = −G/2a₀, so F = (αa₀/2)(−1/2a₀)G = −(α/4)G = −1.82 × 10⁻³ G at every radius. Its weight is (α/4)² = α²/16 = 3.33 × 10⁻⁶.
- 2P(1/2), κ = +1: dG/dr + G/r = (3/r − 1/2a₀)G, so F = (α/2)(3a₀/r − 1/2)G. It changes sign at r = 6a₀, and near r = 0, F ≈ (3α/2)a₀N r: F/r stays finite, the mark of the s-wave harmonic χ(−1)m in the lower pair.
- Weight: ∫F² dr = (α²/4)[9a₀²⟨r⁻²⟩ − 3a₀⟨r⁻¹⟩ + 1/4], and QM-II's 2p values ⟨r⁻²⟩ = 1/12a₀² and ⟨r⁻¹⟩ = 1/4a₀ give (α²/4)(3/4 − 3/4 + 1/4) = α²/16. The same weight, ⟨T⟩/2mc² = 3.40 eV/1.022 MeV = 3.33 × 10⁻⁶, in a different shape.
- Density at r = 0: (F/r)²|χ(−1)m|² with |χ(−1)m|² = 1/4π and N² = 1/24a₀⁵ is (9α²/4)a₀²/(24a₀⁵ × 4π) = 3α²/128πa₀³. Divided by 1/8πa₀³ this is 3α²/16 = 9.98 × 10⁻⁶. The 2P(3/2) lower function goes as r², so its density there is zero.
Answer. F = −(α/4)G for 2P(3/2) and F = (α/2)(3a₀/r − 1/2)G for 2P(1/2). Both weigh α²/16 = 3.33 × 10⁻⁶, but only 2P(1/2) reaches the nucleus, with 9.98 × 10⁻⁶ of the 2s density.
HardExample 3
Keep the first correction in the exact elimination F = ħc(dG/dr + κG/r)/(2mc² + ε − V), ε = E − mc², and find the term in the radial equation for G that depends on the sign of κ. Use it to compute hydrogen's 2P(3/2) − 2P(1/2) splitting and compare it with the measured 10.969 GHz.
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.
- Write F = (ħ/2mc)f(G′ + κG/r) with f = [1 + (ε − V)/2mc²]⁻¹ ≈ 1 − (ε − V)/2mc², so df/dr ≈ V′/2mc². Substitute into dF/dr − κF/r = −(ε − V)G/ħc; the κG′/r terms cancel as before.
- Multiplying by −ħc: −(ħ²/2m)f[G″ − κ(κ + 1)G/r²] − (ħ²V′/4m²c²)(G′ + κG/r) + VG = εG. Every piece depends on κ only through κ(κ + 1) = 2, shared by both 2p levels, except W = −(ħ²κ/4m²c²)(V′/r)G. With −κ = 1 + σ⋅L/ħ, W is (1/2m²c²)(V′/r)L⋅S plus a κ-free part: Unit 6's spin–orbit term.
- First order: both levels start from the same G, so ΔE = −(ħ²/4m²c²)(κ(3/2) − κ(1/2))⟨V′/r⟩ = (3ħ²/4m²c²)⟨V′/r⟩, with κ(3/2) = −2 and κ(1/2) = +1. For V = −αħc/r, V′/r = αħc/r³, and QM-II quotes ⟨r⁻³⟩ = 1/24a₀³ for 2p.
- So ΔE = 3αħ³c/(96m²c²a₀³) = αħ³/(32m²c a₀³), and a₀ = ħ/αmc turns this into ΔE = α⁴mc²/32 = 4.528 × 10⁻⁵ eV, or ΔE/h = 10.95 GHz, with 2P(3/2) on top. Dimensions: α⁴mc² is an energy. Limit: ΔE → 0 as c → ∞, as a relativistic effect must.
- Measured: 10.969 GHz, 0.18% higher. The gap is physics a Dirac electron around a fixed point nucleus omits: the anomalous moment raises the spin–orbit coupling by 2aₑ = 0.23%, and the reduced mass lowers the splitting by 0.05%, together giving 10.969 GHz.
Answer. ΔE = α⁴mc²/32 = 4.53 × 10⁻⁵ eV, or 10.95 GHz, from the κ-odd term alone; the measured 10.969 GHz is 0.18% higher, the anomalous moment less the recoil correction.
Quick check
Select an answer and explain it before checking.
Review & continue
The coupled radial equations are the two real first-order equations, one from each block row of the Dirac equation, obeyed by the radial functions G(r) and F(r) of an electron's upper and lower components in any central potential.
- 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.