
Advanced Quantum Mechanics course sequence · Interactive lesson
Behaviour at infinity and at the origin: γ = √(κ² − (Zα)²)
Before solving the Dirac hydrogen atom you find what its radial functions do at the two ends, and this is where relativity first changes the rule for an acceptable wavefunction rather than just shifting an energy. Two 2 × 2 determinants do the work; the choice between two roots is the hard part.
UAQM-07.05 · Lesson 5 of 9 in this unit. Position is not a completion score.Build the model
The tail is set by the mass gap, the core by a power that is no longer an integer.
The key idea
Near a point Coulomb charge the Dirac radial functions go as rγ with γ = √(κ² − (Zα)²), far away they fall as e(−λr) with ħcλ = √(m²c⁴ − E²), and the acceptable solution is the one with a finite norm and a finite mean potential energy.
Example. For the 1s state of hydrogen-like uranium (Z = 92, κ = −1), Zα = 0.6714 and γ = 0.7411, so G/r and F/r grow as r(−0.259) toward the nucleus: infinite at r = 0, yet ∫(G² + F²) dr ∝ ∫ r1.48 dr converges there.
The complete physical model and assumptions
Topic 7.04 left two coupled first-order equations for G and F. Before solving them for a point nucleus, V = −Zαħc/r, find what they do at the two ends, as Quantum Mechanics II did for the Schrödinger radial equation. Far out, V and the κ/r terms die away, the pair collapses to G″ = λ²G, and the decay constant obeys ħcλ = √(m²c⁴ − E²): a bound state must sit in the gap between −mc² and mc², and when its binding is small beside mc² this is QM-II's √(2mB)/ħ. Near the nucleus the Coulomb term grows exactly as fast as κ/r, so neither can be dropped; a trial power rs turns the equations into a 2 × 2 determinant whose roots are s = ±γ with γ = √(κ² − (Zα)²). That one change from QM-II's integer exponents carries the physics. γ is no longer an integer, and for |κ| = 1 it is below 1, so s₁/₂ and p₁/₂ spinors diverge at the origin as r(γ−1), mildly but measurably, and the two roots are separated by 2γ rather than by 2l + 1. While γ is at least ½, normalisability throws out r(−γ), as it threw out QM-II's r(−l) for l ≥ 1; for |κ| = 1 that stops at Zα = √3/2, Z ≈ 119, and from there the root is chosen by demanding a finite mean potential energy. The analysis ends where γ does: at Zα = |κ| the roots merge and beyond it they turn complex, the fall to the centre that topic 7.09 takes up with the finite nucleus that settles the choice physically.
1Conventions, and why the ends come first
Units are SI, with ħc = 197.327 MeV fm and α = e²/(4πε₀ħc) = 1/137.036; the metric signature is (+, −, −, −), and it enters here only through E² − p²c² = m²c⁴. E includes the rest energy, and κ follows topic 7.02: −1 for s₁/₂, +1 for p₁/₂, −2 for p₃/₂. Topic 7.04 derived, for any central V, dG/dr = −κG/r + (E + mc² − V)F/(ħc) and dF/dr = κF/r − (E − mc² − V)G/(ħc), normalised by ∫(G² + F²) dr = 1. Here V = −Zαħc/r, a point nucleus. Quantum Mechanics II solved its radial equation by finding the behaviour as r → ∞ and r → 0, factoring both off, and terminating a series in between; topic 7.06 does the last step. What changes is the technique at each end. The equations are first order and coupled, so a trial form turns them into a 2 × 2 linear system for the two amplitudes, and the exponent comes from requiring a non-zero solution: a vanishing determinant.
2Far out: the mass gap fixes the decay
As r → ∞, κ/r and V/(ħc) fall away, leaving dG/dr = (E + mc²)F/(ħc) and dF/dr = (mc² − E)G/(ħc). Try G = a e(−λr) and F = b e(−λr): −ħcλ a = (E + mc²)b and −ħcλ b = (mc² − E)a, and a non-zero (a, b) needs (ħcλ)² = (mc² + E)(mc² − E) = m²c⁴ − E², derived. A normalisable solution needs real, positive λ, so a bound state has |E| below mc²; outside that window λ is imaginary and the solutions oscillate, the positive- and negative-energy continua. Dimensions: ħcλ is an energy. Limit: with E = mc² − B, (ħcλ)² = B(2mc² − B) → 2mc²B as B/mc² → 0, so λ → √(2mB)/ħ, QM-II's value; for hydrogen's 13.6 eV that is 1/a₀ = 1.89 × 10¹⁰ m⁻¹, and the relativistic factor √(1 − B/2mc²) differs from 1 by 6.7 × 10⁻⁶. The first equation fixes the ratio: F/G = b/a → −ħcλ/(E + mc²) = −√((mc² − E)/(mc² + E)). The 1/r terms dress e(−λr) with a power of r, which topic 7.06's series carries.
3Near the nucleus: a determinant gives γ
As r → 0, Zαħc/r outgrows E ± mc² while κ/r grows at the same rate, so both 1/r terms stay: dG/dr = (−κG + ZαF)/r and dF/dr = (κF − ZαG)/r. Try G = a rs and F = b rs. Then (s + κ)a − Zα b = 0 and Zα a + (s − κ)b = 0, and a non-zero (a, b) requires (s + κ)(s − κ) + (Zα)² = 0: s² = κ² − (Zα)², so s = ±γ with γ = √(κ² − (Zα)²), derived. Dimensions: every term is a pure number. Limit Zα → 0: γ → |κ|, the smaller orbital label plus one, l + 1 for negative κ, where the upper component leads, and l′ + 1 for positive κ, where the lower one does; so the dominant component starts as QM-II's r(l+1), with its own orbital label. What is new is that γ is not an integer once Zα ≠ 0: for |κ| = 1 it is 1 − 2.66 × 10⁻⁵ in hydrogen and 0.7411 in uranium, and the two roots ±γ differ by 2γ rather than by the integer 2l + 1.
4Which component carries the singularity
The eigenvector of the same determinant fixes the ratio at the origin: for the rγ root, F/G → (γ + κ)/(Zα), derived. For s₁/₂, κ = −1, this is −(1 − γ)/(Zα) ≈ −Zα/2: −0.0036 in hydrogen and −0.386 in uranium, so the upper component leads, as non-relativistic intuition expects. For p₁/₂, κ = +1, it is (1 + γ)/(Zα): 274 in hydrogen and 2.59 in uranium. The small component of a p₁/₂ state is the large one at the nucleus. Topic 7.04's estimate F ≈ (ħ/2mc)(dG/dr + κG/r), with G ∝ r², puts the crossover at r ≈ 1.5ħ/mc ≈ 580 fm in hydrogen. F carries l′ = 0, and with it a density |ψ|² ∝ r(2γ−2) that does not vanish at the origin, which is why p₁/₂ levels, unlike p₃/₂, feel the nuclear size and parity-violating contact interactions. For |κ| = 2, γ is close to 2 and |ψ|² vanishes as r(2γ−2) ≈ r², as a p₃/₂ density should; a d₃/₂ density does the same, through its l′ = 1 lower component, not as QM-II's r⁴.
5Two roots, and the condition that chooses
Near r = 0 a solution can mix rγ and r(−γ), and one must go. Since G² + F² ∝ r(2s), the norm ∫ r(2s) dr converges at 0 only if 2s exceeds −1. The rγ root passes at every Zα below |κ|, where γ is real; r(−γ) fails, and is excluded, only while γ ≥ ½. For |κ| ≥ 2 that always holds, but for |κ| = 1 it needs Zα ≤ √3/2, that is Z ≤ 118.7, just beyond oganesson, Z = 118, the heaviest named element. From Z = 119 to 137 both roots are square-integrable. The condition that still decides is a finite mean potential energy: ⟨V⟩ = −Zαħc ∫ (G² + F²) r⁻¹ dr has integrand r(2γ−1) for the rγ root, finite for any positive γ, and r(−2γ−1) for r(−γ), infinite for every γ, derived. QM-II met the same situation in the s wave, where ψ ∝ 1/r is square-integrable and was excluded because ∇²(1/r) = −4πδ³(r). Operator theory, quoted, agrees: up to Zα = √3/2 the Hamiltonian has a single self-adjoint form, and beyond it finite ⟨V⟩ picks the distinguished one, as topic 7.09's finite nucleus confirms.
6A measured trace of the power law
The near-origin power is not hidden: an observable that weights small r exposes it. The magnetic hyperfine energy of an s₁/₂ electron involves ∫ G F r⁻² dr, quoted, and with the exponents above its integrand near the origin goes as r(2γ−2): it converges only for γ above ½, the same line at which r(−γ) turns normalisable, derived. For hydrogen-like bismuth, ²⁰⁹Bi⁸²⁺, Zα = 0.6057 and γ = 0.7957. The point-nucleus Dirac 1s spinor multiplies the non-relativistic Fermi value, 2.748 eV, by 1/(γ(2γ − 1)) = 2.125, quoted, giving 5.839 eV; the measured splitting is 5.084 eV (Klaft and co-workers, 1994), 1.85 times the Fermi value. Relativity accounts for the bulk of that enhancement. The point theory overshoots by 15%, mostly because r(γ−1) is steepest inside the nucleus, where the real charge and magnetisation are spread over about 5.5 fm, the finite-nucleus physics that topic 7.09 introduces; radiative corrections make up a small remainder. The far end is checked the same way: the measured 1s binding of U⁹¹⁺ fixes λ, as the second worked example shows.
Key equations
Read the units, assumptions and limits before substituting.
Decay constant at large r
ħcλ = √(m²c⁴ − E²)
Derived. λ is real only for E between −mc² and mc², the bound-state window; with E = mc² − B and B ≪ mc² it becomes √(2mB)/ħ, Quantum Mechanics II's decay constant.
λ in m⁻¹; E (rest energy included) and mc² in J, with ħc = 197.327 MeV fmRatio of the components far out
F/G → −√((mc² − E)/(mc² + E)) as r → ∞
Derived. About −√(B/2mc²) for a weakly bound level, −1 at E = 0, and unbounded as E → −mc², where F takes over: the components swap roles across the gap.
dimensionless; G and F in m(−1/2), normalised so that ∫(G² + F²) dr = 1Exponent at the origin
γ = √(κ² − (Zα)²), with G and F ∝ r(±γ) as r → 0
Derived from a 2 × 2 determinant. Not an integer: it tends to |κ|, the smaller orbital label plus one, as Zα → 0, and is real only while Zα is below |κ|.
κ the Dirac label (−1 for s₁/₂, +1 for p₁/₂, −2 for p₃/₂); Z the nuclear charge number; α = 1/137.036Which component leads at the origin
F/G → (γ + κ)/(Zα) as r → 0, for the rγ root
Derived. Negative κ: the upper component leads. Positive κ: the lower one does, by 274 in hydrogen's p₁/₂, which gives that state an s-like density at the nucleus.
dimensionless; −(1 − γ)/(Zα) ≈ −Zα/2 for s₁/₂, and (1 + γ)/(Zα) ≈ 2/(Zα) for p₁/₂When the second root is normalisable
∫₀ r(−2γ) dr is finite only when γ is below ½
Derived. Up to Z = 118 the probability integral alone discards r(−γ); from Z = 119 to 137 it discards nothing, and a second condition has to.
for |κ| = 1 that is Zα above √3/2 = 0.866, Z above 118.7; for |κ| ≥ 2, γ ≥ √3 while Zα ≤ 1, so neverFinite mean potential energy
⟨V⟩ = −Zαħc ∫ (G² + F²) r⁻¹ dr
Derived. Finite for rγ at every positive γ, infinite for r(−γ) at every Zα: this condition, not normalisability, selects the root all the way to Zα = 1.
⟨V⟩ in J; near r = 0 the integrand goes as r(2γ−1) for the rγ root and r(−2γ−1) for r(−γ)Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. With |κ| = 1, drag Z from 92 to 119: the open dot in the norm row crosses the dashed line, so r(−γ) turns normalisable, yet in the ⟨V⟩ row it stays left at every Z. At Z = 137 the roots almost merge. On the left, push E past ±mc²: no real λ remains, and the continua begin.
Live results
- FAR-FIELD DECAY CONSTANT
- 0.673 mc/ħ
- EXPONENT AT THE ORIGIN
- 0.7411
- DENSITY POWER AT THE ORIGIN
- -0.518
- THRESHOLD Z FOR SECOND ROOT
- 118.7
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. Trial energy E/mc²: 0.74; Nuclear charge Z: 92; Dirac label |κ| = j + ½: 1
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The wavefunction must be finite at the origin, so a Dirac s₁/₂ state, which grows there as r(γ−1), must be wrong or must be thrown away. Finiteness was never Quantum Mechanics II's criterion; it only looked like one because the accepted power, rl with l ≥ 0, never had a negative exponent. A state needs a finite norm and finite energy expectation values. Finiteness would reject every s₁/₂ and p₁/₂ state, since for |κ| = 1 both r(γ−1) and r(−γ−1) diverge. Normalisability is the right test only up to Zα = √3/2, beyond which r(−γ) is square-integrable. The test that works across the whole range is a finite ⟨V⟩ = −Zαħc ∫ (G² + F²) r⁻¹ dr: it keeps rγ for any positive γ and rejects r(−γ) at every Zα. QM-II's s wave already showed this: ψ ∝ 1/r is square-integrable and was rejected because ∇²(1/r) = −4πδ³(r), not because it is infinite.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
For hydrogen-like iron, Fe²⁵⁺ (Z = 26), find γ for the s₁/₂, p₁/₂ and p₃/₂ states, say which component dominates near the nucleus in each, and give the power of r with which that component's radial function, G/r or F/r, behaves there. Compare with Quantum Mechanics II's rl.
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.
- |κ| = 1 for s₁/₂ (κ = −1) and p₁/₂ (κ = +1): γ = √(1 − 0.035998) = 0.98184. |κ| = 2 for p₃/₂ (κ = −2): γ = √(4 − 0.035998) = 1.99098.
- s₁/₂: F/G → (γ + κ)/(Zα) = −0.01816/0.18973 = −0.0957, so G leads and G/r ∝ r(γ−1) = r(−0.0182). QM-II's s state goes as r⁰; the Dirac one diverges, slightly.
- p₁/₂: same γ, but F/G → (γ + 1)/(Zα) = 1.98184/0.18973 = 10.4, so F, with l′ = 0, leads: F/r ∝ r(−0.0182), where QM-II's p state went as r¹.
- p₃/₂: κ = −2, so G leads with G/r ∝ r(γ−1) = r0.991, within 1% of QM-II's r¹ for l = 1.
Answer. γ = 0.9818 for s₁/₂ and p₁/₂, 1.9910 for p₃/₂. Near the nucleus G/r (s₁/₂) and F/r (p₁/₂) both go as r(−0.0182); G/r for p₃/₂ goes as r0.991.
MediumExample 2
The measured 1s binding energy of hydrogen-like uranium, U⁹¹⁺, is 131.82 keV (quoted). With mc² = 511.00 keV and ħc = 197.327 keV pm, find the decay constant λ and the decay length 1/λ from the exact far-field formula, compare with √(2mB)/ħ at the same B, and find F/G at large r.
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.
- E = mc² − B = 511.00 − 131.82 = 379.18 keV, so mc² + E = 890.18 keV.
- (ħcλ)² = (mc² − E)(mc² + E) = 131.82 × 890.18 = 1.1734 × 10⁵ keV², so ħcλ = 342.55 keV and λ = 342.55/197.327 = 1.7360 pm⁻¹: 1/λ = 0.5760 pm.
- Non-relativistic: ħcλ = √(2mc²B) = √(2 × 511.00 × 131.82) = 367.04 keV, so 1/λ = 0.5376 pm, 6.7% too short. The ratio is √(1 − B/2mc²) = √(1 − 0.12898) = 0.9333: at this binding the correction is not small.
- F/G → −√((mc² − E)/(mc² + E)) = −√(131.82/890.18) = −0.385: far out the small component is 38% of the large one.
- Check: the point-nucleus Dirac binding, 132.28 keV (quoted from topic 7.07), gives ħcλ = 343.06 keV and 1/λ = 0.5752 pm, which is a₀/92. The measured level, less bound by 0.46 keV, has a tail 0.15% longer.
Answer. λ = 1.736 pm⁻¹, so 1/λ = 0.576 pm; √(2mB)/ħ would give 0.538 pm, 6.7% short. F/G → −0.385 at large r.
HardExample 3
Take a hypothetical bare point nucleus with Z = 122 and an s₁/₂ electron. Find γ, decide whether normalisability can reject the r(−γ) root, show which condition does reject it, and say how fast the rejected root's ⟨V⟩ diverges as a cut-off radius δ shrinks. Then check the p₃/₂ states at the same Z.
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 = 122/137.036 = 0.89028, (Zα)² = 0.79259, so for |κ| = 1, γ = √(1 − 0.79259) = 0.45542, below ½.
- Norm integrands near r = 0: G² + F² ∝ r(2γ) = r0.911 for rγ and r(−2γ) = r(−0.911) for r(−γ). Both exponents exceed −1, so both integrals converge: normalisability cannot choose. At Z = 118 the second exponent is −1.017 and it could.
- ⟨V⟩ integrands, (G² + F²)/r: r(2γ−1) = r(−0.089) for the rγ root, which converges, and r(−2γ−1) = r(−1.911) for r(−γ), which diverges. Finite mean potential energy keeps rγ alone.
- With the integral cut off at r = δ, the r(−γ) root's ⟨V⟩ grows as δ(−2γ) = δ(−0.911): each tenfold cut in δ multiplies it by 10(2γ) = 8.14, so no finite energy can be assigned to it.
- The accepted root is no tamer to the eye: |ψ|² ∝ r(2γ−2) = r(−1.089), steeper than 1/r, yet ∫|ψ|² r² dr converges.
- p₃/₂ (|κ| = 2): γ = √(4 − 0.79259) = 1.7909, so r(−γ) has norm integrand r(−3.58) and is already excluded by normalisability.
Answer. γ = 0.455 for s₁/₂. Both roots are normalisable, so only finite ⟨V⟩ (r(−0.089) against r(−1.911)) rejects r(−γ). For p₃/₂, γ = 1.79 and normalisability suffices.
Quick check
Select an answer and explain it before checking.
Review & continue
Near a point Coulomb charge the Dirac radial functions go as rγ with γ = √(κ² − (Zα)²), far away they fall as e(−λr) with ħcλ = √(m²c⁴ − E²), and the acceptable solution is the one with a finite norm and a finite mean potential energy.
- 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.