Skip to main content

Advanced Quantum Mechanics course sequence · Interactive lesson

Exact Klein-Gordon Coulomb levels and pionic atoms

Topic 3.06 ended on a radial equation that is hydrogen's with two parameters altered. Here the old series is run once more and the exact spectrum falls out in a line. Then it is held against hydrogen, where it misses by precisely the spin terms, and against pionic atoms, where it works well enough to help fix the pion's mass.

UAQM-03.07 · Lesson 7 of 9 in this unit. Position is not a completion score.
Starting model: Exact Klein-Gordon Coulomb levels and pionic atomsShell n = 2 of hydrogen, levels drawn below the Bohr energy to order α⁴ in units of mc²α⁴/n³ = 43.8 GHz. Left: a spinless electron's Klein-Gordon ns and np. Right: the same levels with the Darwin and spin-orbit terms switched on at strength λ = 0.00, which at 1 gives Dirac's levels. Now ns − np₁/₂ = -29.20 GHz; the weighted np centroid never moves.below Bohr in units mc²α⁴/n³ = 43.8 GHzBohrKlein-Gordonspin terms × 0.002s2p2p₁/₂2p₃/₂2s₁/₂
Start with a picture. Change the model in the lab below.
1

Build the model

Exact for a spinless particle, so wrong for the electron and right for the pion.

The key idea

The Klein-Gordon Coulomb spectrum is the set of exact bound energies of a spinless charged particle around a point nucleus, E = mc²[1 + (Zα)²/N²](−1/2), where N = n − (l + ½) + √((l + ½)² − (Zα)²) depends on l as well as on n.

Example. With the electron's mass and Z = 1 the formula puts 2s 120.8 μeV (29.2 GHz) below 2p. In real hydrogen 2s₁/₂ lies 4.37 μeV (1.06 GHz) above 2p₁/₂, so the spinless prediction has the wrong sign and is 28 times too large.

The complete physical model and assumptions

A spinless particle of charge −e around a point nucleus of charge Ze obeys the minimally coupled Klein-Gordon equation, and topic 3.06 reduced it to Quantum Mechanics II's radial Coulomb equation with two edits: E/c² stands where m stood in the Coulomb term, and l becomes the non-integer l′ = −½ + √((l + ½)² − (Zα)²). The series solution never needed integer l, so its termination condition carries over and gives the spectrum in closed form, E = mc²[1 + (Zα)²/N²](−1/2) with N = n − (l + ½) + √((l + ½)² − (Zα)²). That is Dirac's formula with j + ½ replaced by l + ½, and the one swap is the whole story. Expanded, the formula returns the Bohr levels and then a (Zα)⁴ term equal, state by state, to the expectation value of −p⁴/(8m³c²): relativistic kinetic energy and nothing else. Hydrogen needs two more terms, spin-orbit coupling for l ≥ 1 and the Darwin term for s states, so the spinless spectrum puts 2s 29 GHz below 2p, while the measured 2s₁/₂ sits 1.06 GHz above 2p₁/₂. A negative pion has spin 0, so the same formula, with the reduced mass and vacuum polarisation added, fits pionic atoms to parts per million on circular orbits, and what is left over in the lowest levels measures the strong interaction. The formula stops where N turns complex, at Zα = l + ½, and wherever the orbit reaches a nucleus that is not a point.

1Run Quantum Mechanics II's series with l′

Units are SI with ħ and c kept, and the metric is diag(1, −1, −1, −1) as fixed in topic 3.01, though only the radial equation is needed now. Topic 3.06 brought the bound problem to u″ + [−κ² + 2ZαE/(ħc r) − l′(l′ + 1)/r²]u = 0, with κ = √(m²c⁴ − E²)/(ħc). Quantum Mechanics II's equation was the same with 2Zαmc/(ħr) in the middle, so with ρ = κr its analysis applies line for line: ρ₀ = 2ZαE/(ħcκ), u = ρ(l′+1) e(−ρ) v(ρ), and the recursion for v ends, as it must for a normalisable u, only if ρ₀ = 2(nᵣ + l′ + 1) with nᵣ = 0, 1, 2, … Integer l was never used, only the choice of the ρ(l′+1) root at the origin, which topic 3.08 examines. So ZαE = N√(m²c⁴ − E²), where N = nᵣ + l′ + 1 = n − (l + ½) + √((l + ½)² − (Zα)²) and n = nᵣ + l + 1. The right side is positive, so E must be too, and squaring gives E²[N² + (Zα)²] = N²m²c⁴, that is E = mc²[1 + (Zα)²/N²](−1/2). Derived.

2Checks: units, the Bohr limit and a broken degeneracy

Dimensions: Zα and N are pure numbers, so E is mc² times a number. Limit: as Zα → 0, N → n and E → mc² − mc²(Zα)²/(2n²), the Bohr level plus the rest energy, with its l-degeneracy intact. What is new is that N depends on l. The (Zα)²/r² piece of the squared potential, an extra attraction that topic 3.06 folded into l′, matters most where the centrifugal barrier is weakest: for hydrogen's 1s, l′ = −½ + √(¼ − α²) = −5.3 × 10⁻⁵, while for 2p it falls only 1.8 × 10⁻⁵ below 1. Low-l states are drawn further in and bind more, and the accidental l-degeneracy of the Coulomb problem is gone. Now set Dirac's formula, quoted in Quantum Mechanics II and derived in Unit 7, beside it: its N is n − (j + ½) + √((j + ½)² − (Zα)²). The two are identical except that the spinless particle carries l + ½, a half-integer, where the electron carries j + ½, an integer. Sommerfeld's 1916 formula also had the integer, quoted, which is why it fitted the fine structure the spinless formula misses.

3The (Zα)⁴ term is relativistic kinetic energy alone

Write δ = l + ½ − √((l + ½)² − (Zα)²) ≈ (Zα)²/(2l + 1), so N = n − δ and 1/N² ≈ 1/n² + 2(Zα)²/((2l + 1)n³). Expanding [1 + y](−1/2) = 1 − y/2 + 3y²/8 with y = (Zα)²/N² gives E ≈ mc² − mc²(Zα)²/(2n²) − (mc²(Zα)⁴/(2n⁴))[n/(l + ½) − ¾], derived. Quantum Mechanics II evaluated ⟨−p⁴/(8m³c²)⟩ in the hydrogen state nl by writing p² = 2m(Eₙ − V), and found −(Eₙ²/(2mc²))[4n/(l + ½) − 3]; with Eₙ = −mc²(Zα)²/(2n²) that is the same expression, state by state. Topic 3.03 showed that the stationary equation shifts levels by exactly this operator at first order, and the exact solution now confirms that nothing else appears: no contact term and no l⋅s, because a scalar wave function has no spin to couple. The cost sits in the coefficient, n/(l + ½) where hydrogen's fine structure has n/(j + ½). For hydrogen's 1s the spinless term is −(1.25/2)mc²α⁴ = −906 μeV; Dirac's, with 1 − ¾ in the bracket, is −181 μeV.

4Hydrogen by state: Darwin for s, spin-orbit for the rest

Set the two n = 2 spectra side by side in units of mc²α⁴ = 1.449 meV, which is h × 350.4 GHz. Spinless: 2s at −13/128 (35.6 GHz below Bohr) and 2p at −7/384 (6.39 GHz below), so 2s lies 1/12 of the unit, 120.8 μeV or 29.2 GHz, below 2p. Dirac: 2s₁/₂ and 2p₁/₂ together at −5/128 (13.7 GHz below), 2p₃/₂ at −1/128 (2.74 GHz below). The gaps are Quantum Mechanics II's quoted terms, computed: 2s₁/₂ − 2s = +1/16, the Darwin term mc²α⁴/(2n³); 2p₃/₂ − 2p = +1/96 and 2p₁/₂ − 2p = −1/48, the spin-orbit term with its Thomas factor. Weighted 4 : 2 the spin-orbit shifts cancel, so the spinless 2p is the centroid of hydrogen's doublet and only the s level is wrong on average. Measurement decides, quoted: 2p₃/₂ − 2p₁/₂ = 10.97 GHz, and 2s₁/₂ sits 1.06 GHz above 2p₁/₂. Schrödinger met this clash with Sommerfeld's fine structure in the winter of 1925 to 1926, shelved the relativistic equation and published the non-relativistic one, quoted.

5Pionic atoms: the system the formula does describe

A negative pion has spin 0, quoted, so the Klein-Gordon equation is its equation, with the reduced mass μ = mπM/(mπ + M) in place of m. Its orbits are small: in pionic hydrogen μc² = 121.5 MeV and the Bohr radius ħ/(μcα) is 223 fm, inside the electron's reduced Compton wavelength of 386 fm. There the pion sees the proton's charge less screened by virtual electron-positron pairs, and vacuum polarisation, which the course quotes rather than derives, adds 3.24 eV to the 1s binding against 0.22 eV from relativity. With both included, circular orbits, l = n − 1, which keep clear of the nucleus, give X-ray lines such as pionic nitrogen's 5g → 4f near 4.055 keV, measured with crystal spectrometers to a few ppm; lines like it fix mπc² = 139.570 39(18) MeV, quoted. The lowest levels do reach the nucleus, and whatever the electromagnetic calculation leaves unexplained measures the strong interaction: pionic hydrogen's 1s is 7.086 eV more bound than electromagnetism predicts and 0.856 eV wide, quoted. Topic 3.09 explains why these energies survive although the one-particle reading does not.

2

Key equations

Read the units, assumptions and limits before substituting.

Exact Klein-Gordon Coulomb levels

Eₙₗ = mc² [1 + (Zα)²/N²](−1/2)

Derived: the series ends when ZαE/(ħcκ) = nᵣ + l′ + 1 = N, with κ = √(m²c⁴ − E²)/(ħc); squaring gives this, and only positive E solves the unsquared condition.

E and mc² in J or eV; α = e²/(4πε₀ħc) and Zα are pure numbers; n = nᵣ + l + 1 with nᵣ = 0, 1, 2, …

Effective principal quantum number

N = n − (l + ½) + √((l + ½)² − (Zα)²)

Put j + ½ in place of l + ½ and this is Dirac's formula, quoted in Quantum Mechanics II: spin 0 and spin ½ differ by that one label.

Dimensionless; N → n as Zα → 0. Real only while Zα ≤ l + ½, the limit topic 3.08 examines.

Expansion to order (Zα)⁴

E ≈ mc² − mc²(Zα)²/(2n²) − (mc²(Zα)⁴/(2n⁴))[n/(l + ½) − ¾]

Derived. The last term equals ⟨−p⁴/(8m³c²)⟩ in the Schrödinger state nl: relativistic kinetic energy, with no contact term and no spin.

Each term in J. For the electron mc²α⁴ = 1.449 meV, which is h × 350.4 GHz; the second term is the Bohr level.

Gap to hydrogen at order (Zα)⁴

EDirac − EKG = (mc²(Zα)⁴/(2n³))[1/(l + ½) − 1/(j + ½)]

Computed from the two expansions. Weighted by 2j + 1 it sums to zero for l ≥ 1, so only the s levels are wrong on average.

j = l ± ½, energies in J. For l = 0 it is +mc²(Zα)⁴/(2n³), the Darwin term; for l ≥ 1, the spin-orbit term.

Spinless 2s − 2p splitting

E(2s) − E(2p) = −mc²(Zα)⁴/12

The number that sank the equation for hydrogen: 2s below 2p, on the wrong side and 28 times the size of the measured Lamb shift.

−120.8 μeV, or −29.2 GHz, for the electron's mass and Z = 1. For 2s₁/₂ − 2p₁/₂ Dirac gives 0 and experiment +4.37 μeV.

Pion mass from a circular-orbit line

δmπ/mπ = (1 + mπ/M) δE/E

A line known to 1 ppm fixes mπ to about 1 ppm, but only after the 111 ppm relativistic and 308 ppm vacuum-polarisation terms are computed.

E the X-ray energy, M the nuclear mass; Bohr and (Zα)⁴ terms both scale with μ = mπM/(mπ + M). 1.0107 for ¹⁴N.
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

0.00
0 to 1; steps of 0.05.
2
2 to 4; steps of 1.

Try this. Start at 0, where 2s lies 29.2 GHz below 2p, and drag the spin terms to 1: s rises by the Darwin term, p splits with its 2 : 4 weighted centroid fixed, and 2s₁/₂ lands on 2p₁/₂. A pionic atom, spin 0, stays at 0 for good.

Live results

GAP ns − np₁/₂
-29.20 GHz
SPLITTING np₃/₂ − np₁/₂
0.00 GHz
ns SHIFT FROM BOHR
-35.59 GHz
np CENTROID SHIFT
-6.39 GHz

Results follow this lesson’s model and assumptions. Displayed values are rounded.

Interactive Advanced Quantum Mechanics modelShell n = 2 of hydrogen, levels drawn below the Bohr energy to order α⁴ in units of mc²α⁴/n³ = 43.8 GHz. Left: a spinless electron's Klein-Gordon ns and np. Right: the same levels with the Darwin and spin-orbit terms switched on at strength λ = 0.00, which at 1 gives Dirac's levels. Now ns − np₁/₂ = -29.20 GHz; the weighted np centroid never moves.below Bohr in units mc²α⁴/n³ = 43.8 GHzBohrKlein-Gordonspin terms × 0.002s2p2p₁/₂2p₃/₂2s₁/₂
Shell n = 2 of hydrogen, levels drawn below the Bohr energy to order α⁴ in units of mc²α⁴/n³ = 43.8 GHz. Left: a spinless electron's Klein-Gordon ns and np. Right: the same levels with the Darwin and spin-orbit terms switched on at strength λ = 0.00, which at 1 gives Dirac's levels. Now ns − np₁/₂ = −29.20 GHz; the weighted np centroid never moves.

Predict → change one variable → check

Plan a change before running it. Every other input stays at the saved baseline.

Saved baseline. Spin terms switched on, λ: 0.00; Principal quantum number n: 2

Baseline 0.00; choose 0 to 1, in steps of 0.05.
4. Compared with the saved baseline, this result will…

This experiment is kept only while this page is open. It does not change saved course progress.

4

Catch the common trap

The tempting diagnosis is that a spinless equation can only go wrong through spin-orbit coupling, and since that vanishes for l = 0, its s levels should be the ones to trust. The opposite is true. For l ≥ 1 the spin-orbit shifts of the j = l ± ½ pair, weighted by their 2j + 1 states, sum to zero, so the Klein-Gordon level sits exactly at the centroid of hydrogen's doublet: wrong in detail, right on average. For l = 0 there is no spin-orbit term to average away, but there is the Darwin term, +mc²(Zα)⁴/(2n³), which acts only at the nucleus and only on s states, and which a scalar equation lacks entirely. So the spinless 2s lies 21.9 GHz below hydrogen's 2s₁/₂, the largest single error in the n = 2 shell. Nor does the failure show that the equation is bad relativity: the same formula is right for the pion.

5

Practice & worked examples

3 authored examples. Make an attempt, then reveal the reasoning one step at a time.

EasyExample 1

A hypothetical spinless particle with the electron's mass is bound to a proton. From the (Zα)⁴ term of the exact Klein-Gordon formula, find how far its 3s, 3p and 3d levels lie below the Bohr level, in GHz, and the spread of the n = 3 shell. Compare it with the spread of hydrogen's Dirac levels from 3s₁/₂ to 3d₅/₂.

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.

MediumExample 2

A negative pion (mπc² = 139.570 MeV, spin 0) is bound to a ¹⁴N nucleus (Mc² = 13 040.20 MeV, Z = 7). Use the exact Klein-Gordon formula with the reduced mass to find the 5g → 4f X-ray energy for a point nucleus, the part of it that is relativistic, and the bias in mπ if a measured line were fitted with the Bohr formula instead.

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.

HardExample 3

In pionic hydrogen (mπc² = 139.570 MeV, mpc² = 938.272 MeV), build the 3p → 1s X-ray energy from the Bohr term with the reduced mass, the Klein-Gordon (Zα)⁴ correction and the vacuum-polarisation shifts −3.241 eV for 1s and −0.011 eV for 3p (Uehling potential, quoted). Add the measured strong-interaction shift, 1s 7.086 ± 0.009 eV more bound, quoted. Then find the error in that shift if the pion were wrongly given hydrogen's spin-½ fine structure.

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.

6

Quick check

Select an answer and explain it before checking.

A spinless particle with the electron's mass is bound to a proton. To order α⁴, how does its 2s level compare with hydrogen's 2s₁/₂, and why?
7

Review & continue

The Klein-Gordon Coulomb spectrum is the set of exact bound energies of a spinless charged particle around a point nucleus, E = mc²[1 + (Zα)²/N²](−1/2), where N = n − (l + ½) + √((l + ½)² − (Zα)²) depends on l as well as on n.

  • 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.

Extra challenging questions

Connect several ideas, explore supplied data, and compare your reasoning with a worked solution.