
Advanced Quantum Mechanics course sequence · Interactive lesson
Mott's formulas for identical charged particles
The exact Coulomb amplitude and the symmetrised cross-section, both already yours, predict together what neither can alone: at 90° spinless identical nuclei scatter at exactly twice the classical rate at every energy, unpolarised spin-half ones at half, with fringes either side. We derive it, time the fringes, and see why counters wash them out.
UAQM-02.08 · Lesson 8 of 9 in this unit. Position is not a completion score.Build the model
Exchange cancels every Coulomb phase but one logarithm, and that logarithm draws fringes.
The key idea
Mott's formula is the centre-of-mass cross-section for pure Coulomb scattering of two identical particles: Rutherford's term for the scattered particle, Rutherford's term for the recoil, and an exchange term whose size is set by the spin and whose phase is η ln tan²(θ/2).
Example. For ¹²C on ¹²C at 4.00 MeV in the centre of mass, η = 6.94 and (η/2k)² = 10.5 fm². At 90° the spin-0 formula gives 16 × 10.5 fm² sr⁻¹ = 1.68 b sr⁻¹, twice the classical 0.840 b sr⁻¹, and the fringes there repeat every 180°/η = 25.9°.
The complete physical model and assumptions
Two identical nuclei that repel each other scatter by an amplitude no Born approximation supplies: topic 2.07's exact fC(θ) carries a phase −η ln sin²(θ/2) that is not small, plus a constant 2σ₀. Symmetrisation compares the amplitude at θ with the one at π − θ. In that comparison the constant cancels, the two logarithms combine into η ln tan²(θ/2), and the moduli return Rutherford's 1/sin⁴(θ/2) for the particle that turned and 1/cos⁴(θ/2) for the partner that recoiled into the same counter. So Mott's cross-section is the classical sum, what a counter blind to which nucleus is which would record classically, plus one interference term. Statistics fixes its coefficient, +2 for spinless bosons and −1 for unpolarised spin-half fermions; η = Z²e²/(4πε₀ħv), half the head-on closest approach measured in reduced wavelengths, fixes its phase, and it is the only place ħ enters. The term does not shrink as η grows. At 90° its cosine is 1 at every energy, and close to 90° it oscillates either side with angular period π/η, so the classical sum returns only when a detector's finite resolution averages the fringes away. The formula fails where its inputs fail: when the nuclear force reaches the pair, as the proton-proton data of 1936 showed, when atomic electrons screen the field at small angles, and when the particles are relativistic.
1Exchange keeps one logarithm of the Coulomb phase
Units are SI and angles are in the centre-of-mass frame unless marked; nothing here is relativistic, so the course metric diag(1, −1, −1, −1) never enters. Topic 2.07 derived the exact amplitude for two charges Ze, fC(θ) = −(η/2k) exp[−iη ln sin²(θ/2) + 2iσ₀]/sin²(θ/2), with σ₀ = arg Γ(1 + iη); it is used here, not re-derived. Exchange needs it at π − θ, which swaps sin(θ/2) for cos(θ/2). Multiply fC(θ) by the complex conjugate of fC(π − θ): the moduli give (η/2k)²/[sin²(θ/2) cos²(θ/2)], the two minus signs cancel, the constant 2σ₀ cancels, and the logarithms combine into −η ln tan²(θ/2). So Re[fC(θ) fC*(π − θ)] = (η/2k)² cos(η ln tan²(θ/2))/[sin²(θ/2) cos²(θ/2)], derived. The first Born amplitude cannot stand in: it is real, so its version of this cosine is 1 at every angle, whereas at η = 3 and θ = 60° the exact one is cos(3 ln(1/3)) = cos(−3.30) = −0.99, the opposite sign.
2Mott's two formulas, and the one term that holds ħ
Topic 2.05 gave spinless bosons |f + f′|² = |f|² + |f′|² + 2 Re(f f′*), and topic 2.06 gave unpolarised spin-half fermions |f|² + |f′|² − Re(f f′*), with f′ = f(π − θ). Insert the Coulomb pieces and both read dσ/dΩ = (η/2k)²[1/sin⁴(θ/2) + 1/cos⁴(θ/2) + c cos(η ln tan²(θ/2))/(sin²(θ/2) cos²(θ/2))], with c = +2 for bosons and −1 for fermions: Mott's 1930 result, derived. Dimensions: ħk = μv makes η/2k = Z²e²/(16πε₀E), with E the centre-of-mass energy, and e²/(4πε₀) = 1.440 MeV fm is an energy times a length, so η/2k is a length and dσ/dΩ comes out in m² sr⁻¹. The first two terms are Rutherford for the particle that turned through θ and for its partner recoiling into the same counter; set c = 0, distinguishable particles, and that classical sum is all that remains. Only the third term carries ħ, inside η. Topic 2.06's spin counting extends c to (−1)(2s) × 2/(2s + 1), so deuterons, spin 1, get +2/3.
3At 90° the interference is complete at every energy
At θ = 90°, tan(θ/2) = 1 and its logarithm vanishes, so the cosine is 1 whatever η is. The reason is that π − θ equals θ there: the direct and exchange amplitudes are one number, logarithmic phase included. Each Rutherford term is 4(η/2k)² and the interference term c × 4(η/2k)², so dσ/dΩ(90°) = 16(η/2k)² for spinless bosons, twice the classical 8(η/2k)², and 4(η/2k)² for unpolarised spin-half fermions, half of it, both derived. Dividing Mott by the classical sum at any angle gives the cleaner ratio 1 + (c/2)[sin²θ/(1 + cos²θ)] cos(η ln tan²(θ/2)), derived, whose envelope is 1 at 90°, 0.6 at 60° and 120° and 1/7 at 30° and 150°: exchange matters most where the two Rutherford terms are equal. For equal masses a centre-of-mass angle θ is θ/2 in the laboratory and dΩcm/dΩlab = 4 cos θlab, both quoted from two-body kinematics, so 90° is the counter at 45°, where the two nuclei leave symmetrically, 45° either side of the beam; at every angle equal masses leave at right angles to each other.
4The fringe period π/η, and the Born limit
Away from 90° the phase φ(θ) = η ln tan²(θ/2) grows, and dφ/dθ = 2η/sin θ. One full oscillation therefore spans Δθ = π sin θ/η, derived: π/η at 90°, shrinking toward 0° and 180°, where the envelope has already died. For η = 10 the period near 90° is 18.0° in the centre of mass and 9.0° in the laboratory, and the cosine first reaches −1 at exactly 2 arctan(e(π/20)) = 99.0°, as the linear estimate 90° + 9° says. Counting is quicker still: between 60° and 120° there are η ln 9/(2π) = 0.35η fringes. The opposite limit checks the derivation. As η → 0 the cosine tends to 1 at every angle and Mott's formula becomes the symmetrised first Born result, built from the real amplitude −(η/2k)/sin²(θ/2). Protons at 1 MeV in the laboratory have η = 0.158, a period near 90° of 1140° and a phase of only −0.42 rad at 30°, so their Mott interference is one smooth dip to 1/2 at 90°, with no fringes at all.
5Why a classical experiment sees the classical sum
Large η is the classical regime: η = kd/2, where d = Z²e²/(4πε₀E) is the head-on closest approach, so η ≫ 1 means an orbit many reduced wavelengths across. Yet the interference term does not shrink as η grows; at 90° it is always c/2 times the classical sum. What changes is its period, π/η, and a real counter accepts a spread of angles. Model the spread as a Gaussian of rms width δθ and take the phase as linear across it, φ(θ + u) ≈ φ(θ) + (2η/sin θ)u. The Gaussian average of cos φ is then cos φ(θ) × exp(−2η²δθ²/sin²θ), so the interference term is multiplied by a visibility V, derived, equal to exp(−2η²δθ²) at 90°. With δθ = 1° in the centre of mass, V = 0.94 at η = 10, 0.58 at η = 30 and 0.002 at η = 100, and V = 1/2 at ηδθ = 0.59. The classical sum returns because no counter resolves angles of order 1/η once η is large, not because the quantum term goes away: here the correspondence limit is an average, not a vanishing.
6Measured: helium, and a force between protons
The measurements that test Mott's formulas are quoted here, not derived. In 1930 Chadwick scattered slow α-particles in helium and found the rate at 45° in the laboratory, 90° in the centre of mass, close to twice the classical value: the spin-0 factor derived above. Measured α–α angular distributions at a few hundred keV follow the spin-0 form, its swings above and below the classical sum included, and depart from it as the energy rises; the departures are read as nuclear phase shifts. Protons are the sharper test, because c = −1 and η = 0.158 at 1 MeV predict one smooth dip. In 1936 Tuve, Heydenburg and Hafstad measured proton-proton scattering below 1 MeV and found departures from Mott's formula far larger than their errors. Breit, Condon and Present read them as an s-wave phase shift: an attractive nuclear force between two protons, close in strength to the neutron-proton singlet force, and by topic 2.06 acting in the singlet alone. The formula also fails where atomic electrons screen the field at small angles, and where relativity adds the Mott factor of Unit 5.
Key equations
Read the units, assumptions and limits before substituting.
Exchange interference of two Coulomb amplitudes
Re[fC(θ) fC*(π − θ)] = (η/2k)² cos(η ln tan²(θ/2)) / [sin²(θ/2) cos²(θ/2)]
The constant phase 2σ₀ = 2 arg Γ(1 + iη) cancels between the two amplitudes, so only the difference of logarithms, η ln tan²(θ/2), survives.
fC in m, k in m⁻¹, η dimensionless, θ the centre-of-mass angle; derived here from topic 2.07's exact fCMott's formula, unpolarised
dσ/dΩ = (η/2k)² [1/sin⁴(θ/2) + 1/cos⁴(θ/2) + c cos(η ln tan²(θ/2)) / (sin²(θ/2) cos²(θ/2))]
Rutherford for the scattered particle, Rutherford for its recoil, and one exchange term. ħ enters only through η, inside the cosine.
m² sr⁻¹, counting either particle; c = +2 for spin 0, −1 for spin ½, +2/3 for spin 1; η/2k = Z²e²/(16πε₀E) in m, E the centre-of-mass energy; derivedSommerfeld parameter for two charges Ze
η = Z²e²/(4πε₀ħv) = Z²αc/v
η = kd/2, with d the head-on closest approach: η ≫ 1 is a classical orbit, η ≪ 1 the Born regime. 0.600 MeV α on helium gives 1.63.
v the relative speed in m s⁻¹, which is the beam speed on a target at rest; dimensionless; quoted from topic 2.07Ratio to the classical sum
dσ/dΩ ÷ classical sum = 1 + (c/2) [sin²θ/(1 + cos²θ)] cos(η ln tan²(θ/2))
At 90° the cosine is 1 for every η, so the ratio is 2 for spinless bosons and 1/2 for unpolarised spin-half fermions at any energy.
dimensionless; the envelope sin²θ/(1 + cos²θ) is 1 at 90°, 0.6 at 60° and 120°, 1/7 at 30° and 150°; derivedAngular period of the fringes
Δθ = π sin θ / η
π/η at 90°: η = 10 gives 18° there in the centre of mass and 9° in the laboratory. Fringes crowd toward 0° and 180°, where the envelope dies.
Δθ in rad, centre-of-mass frame; from dφ/dθ = 2η/sin θ for φ = η ln tan²(θ/2); halve it in the laboratory for equal massesVisibility through a finite resolution
V(θ) = exp(−2η²δθ²/sin²θ)
Multiplies the interference term. At 90°, V = 1/2 when ηδθ = 0.59: the classical sum returns once δθ ≫ 1/η, not because the term shrinks.
δθ the rms Gaussian resolution in rad, centre-of-mass frame; derived with the phase taken as linear across the windowInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Hold δθ at 0 and raise η from 1 to 20: the 90° point never moves while the open circles, half a period either side, close in as π/η. Then set δθ to 4°: by η = 15 the solid curve stays within 0.12 of the classical line. Set the spin to 0.5 and every swing flips sign and halves.
Live results
- INTERFERENCE COEFFICIENT c
- 2.00
- FRINGE PERIOD AT 90°, π/η
- 22.5 °
- VISIBILITY AT 90°
- 0.86
- RATIO RECORDED AT 90°
- 1.86
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. Sommerfeld parameter η: 8.0; Spin s of each particle: 0.0; Resolution δθ, rms, centre of mass: 2.00 °
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The tempting argument is that large η is the classical limit, so the quantum interference term must fade as η grows and leave the classical sum behind. It does not fade. At 90° its cosine is exactly 1 for every η, so the exact cross-section there is always twice the classical sum for spinless bosons and half of it for unpolarised spin-half fermions, however heavy and slow the nuclei. What η controls is the period, π/η: at η = 100 the boson ratio falls from 2 to almost 0 within 0.9° of 90°. The classical answer appears only after averaging over the angles a real counter accepts, which at 90° multiplies the term by exp(−2η²δθ²). A related slip is to call Rutherford's 1/sin⁴(θ/2) alone the classical answer: a counter that cannot tell the nuclei apart also records the recoils, so the classical sum carries 1/cos⁴(θ/2) as well.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
α-particles of laboratory kinetic energy 0.600 MeV strike helium gas. Treating the force as pure Coulomb, find η and Mott's cross-section at 90° in the centre of mass, and compare it with the classical sum and with Rutherford for the projectile alone. Take mα c² = 3727 MeV and e²/(4πε₀) = 1.440 MeV fm.
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.
- Equal masses share the energy equally in the centre of mass: E = 0.600/2 = 0.300 MeV. Then η/2k = Z²e²/(16πε₀E) = 4 × 1.440 MeV fm/(4 × 0.300 MeV) = 4.80 fm, so (η/2k)² = 23.04 fm².
- The relative speed is the beam speed: v/c = √(2 × 0.600/3727) = 0.01794, so η = Z²α c/v = 4 × 0.007297/0.01794 = 1.63. Check through η = kd/2: k = μv/ħ = 1863.5 × 0.01794/197.3 fm⁻¹ = 0.1694 fm⁻¹ and d = 4 × 1.440/0.300 fm = 19.2 fm give 1.63 ✓.
- At 90°, 1/sin⁴(45°) = 1/cos⁴(45°) = 4, 1/(sin²45° cos²45°) = 4 and the cosine is 1, so the spin-0 bracket is 4 + 4 + 2 × 4 = 16 and dσ/dΩ = 16 × 23.04 fm² sr⁻¹ = 369 fm² sr⁻¹ = 3.69 b sr⁻¹.
- The classical sum is 8 × 23.04 = 184 fm² sr⁻¹ = 1.84 b sr⁻¹, and the projectile's Rutherford term alone is 4 × 23.04 = 92.2 fm² sr⁻¹. Exchange doubles the classical count and quadruples the naive one.
- Dimensions: fm² sr⁻¹ is an area per solid angle ✓. The period π/η = 1.93 rad, about 110°, is wider than the whole region around 90°, so at this energy the interference is one broad hump, not a row of fringes.
Answer. η = 1.63; dσ/dΩ(90°) = 3.69 b sr⁻¹, twice the classical sum of 1.84 b sr⁻¹ and four times the projectile-only Rutherford value of 0.922 b sr⁻¹.
MediumExample 2
Two ¹³C nuclei, spin ½ and rest energy 1.211 × 10⁴ MeV each, collide with unpolarised spins at a centre-of-mass energy of 5.00 MeV. A counter sits at 50.0° in the laboratory. Treating the force as pure Coulomb, find η, Mott's cross-section at the counter in the centre of mass and in the laboratory, and its ratio to the classical sum, and compare that ratio with the one at 90° in the centre of mass.
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.
- Equal masses: θ = 2 × 50.0° = 100.0°. With μc² = 6055 MeV, v/c = √(2 × 5.00/6055) = 0.04064 and η = 36 × 0.007297/0.04064 = 6.464. The length scale is η/2k = 36 × 1.440 MeV fm/(4 × 5.00 MeV) = 2.592 fm, so (η/2k)² = 6.718 fm².
- Phase: tan(50.0°) = 1.1918, so η ln tan²(θ/2) = 6.464 × 0.3509 = 2.268 rad, and its cosine is −0.642.
- Bracket with c = −1: 1/sin⁴(50°) = 2.904, 1/cos⁴(50°) = 5.858 and 1/(sin²50° cos²50°) = 4.124, so the interference term is −(−0.642) × 4.124 = +2.648 and the bracket is 11.41. Then dσ/dΩ = 11.41 × 6.718 fm² sr⁻¹ = 76.7 fm² sr⁻¹ = 0.767 b sr⁻¹.
- Laboratory: dΩcm/dΩlab = 4 cos θlab = 2.571 for equal masses, quoted from kinematics, so the counter sees 197 fm² sr⁻¹ = 1.97 b sr⁻¹, recoils included.
- Ratio to the classical sum 2.904 + 5.858 = 8.762: 11.41/8.762 = 1.30, as 1 + (−1/2)(0.941)(−0.642) also gives. At 90° the ratio is exactly 1/2, so 10° away these fermions scatter 30% above classical: c = −1 flips the sign of the swings, it does not remove them.
- Pure Coulomb is safe: with d = 36 × 1.440/5.00 fm = 10.37 fm, the classical closest approach at 100° is (d/2)(1 + 1/sin 50°) = 12.0 fm, twice the 5.64 fm sum of radii 1.2A¹⁄³ fm.
Answer. η = 6.46; at 100° in the centre of mass dσ/dΩ = 0.767 b sr⁻¹, or 1.97 b sr⁻¹ at the 50.0° laboratory counter, 1.30 times the classical sum against exactly 0.50 times at 90°.
HardExample 3
Two ¹⁶O nuclei, spin 0 and rest energy 1.490 × 10⁴ MeV each, collide at a centre-of-mass energy of 8.00 MeV, below their Coulomb barrier. Find η and the fringe period near 90° in the centre of mass and in the laboratory. A detector has a Gaussian laboratory resolution of 1.0° rms: what ratio to the classical sum does it record at 90°, and what laboratory resolution would cut the visibility to one half?
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.
- μc² = 7450 MeV, v/c = √(2 × 8.00/7450) = 0.04634 and η = Z²α c/v = 64 × 0.007297/0.04634 = 10.08. Check: d = 64 × 1.440/8.00 fm = 11.52 fm and k = μv/ħ = 7450 × 0.04634/197.3 fm⁻¹ = 1.750 fm⁻¹, so kd/2 = 10.08 ✓.
- Period at 90°: π/η = 0.3117 rad = 17.9° in the centre of mass. For equal masses θlab = θ/2, so the fringes sit 8.93° apart in the laboratory.
- Resolution: 1.0° rms in the laboratory is 2.0° = 0.03491 rad in the centre of mass. Then ηδθ = 0.352 and V = exp(−2 × 0.352²) = exp(−0.248) = 0.781.
- With c/2 = +1 for spin 0, the 90° ratio is 1 + V = 1.78, against 2.00 for a perfect detector and 1.00 classically.
- Visibility one half needs 2η²δθ² = ln 2, so δθ = √(ln 2/2)/η = 0.5887/10.08 = 0.0584 rad = 3.35° in the centre of mass, or 1.67° rms in the laboratory.
- Pure Coulomb holds: at 90° the classical closest approach is (d/2)(1 + 1/sin 45°) = 13.9 fm, more than twice the 6.05 fm sum of nuclear radii 1.2A¹⁄³ fm.
Answer. η = 10.1; the period is 17.9° in the centre of mass and 8.93° in the laboratory; with 1.0° laboratory resolution the 90° ratio is 1.78, not 2.00; a 1.67° rms laboratory resolution halves the visibility.
Quick check
Select an answer and explain it before checking.
Review & continue
Mott's formula is the centre-of-mass cross-section for pure Coulomb scattering of two identical particles: Rutherford's term for the scattered particle, Rutherford's term for the recoil, and an exchange term whose size is set by the spin and whose phase is η ln tan²(θ/2).
- 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.