University Quantum Mechanics II · Quantum Scattering Theory · 15.9

Yukawa and Rutherford scattering, and when Born holds

Here the Born approximation earns its keep and meets its limits. One Fourier transform of the Yukawa potential gives a cross-section for nuclear forces and screened atoms; remove the screening and Rutherford's formula appears exactly. Then comes the question every Born calculation owes an answer to: was the potential weak enough?

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

The first Born approximation turns scattering into a Fourier transform, and the Yukawa potential V = β e^(−μr)/r is the case where the transform takes one line: f(θ) = −2mβ/[ħ²(μ² + q²)], with ħq = 2ħk sin(θ/2) the momentum transfer. The range 1/μ sets the angular scale, the amplitude flat while q ≪ μ and falling as 1/q² beyond. Let the range grow without limit with β = Z₁Z₂e²/4πε₀ and ħ cancels from the cross-section, leaving Rutherford's classical formula exactly.

The exact Coulomb amplitude, quoted, differs from the Born one only by a phase, so the agreement is a property of 1/r and no evidence that the approximation was entitled to it. Entitlement is judged by the scattered wave the Born amplitude itself predicts at the centre of the potential, which must be small beside the incident wave. At low energy that demands |V₀| ≪ ħ²/(mR²), a well too weak to bind; at high energy |V₀|R ≪ ħv, which for Coulomb reads η = Z₁Z₂αc/v ≪ 1.

Classical orbits need the opposite, η ≫ 1, so 5.0 MeV α particles on gold, at η ≈ 22, sat far outside the Born regime. Where the tests fail, the later terms of the Born series are not small, and the second is needed in any case to supply the Im f(0) that the optical theorem demands.

Simple definition: The first Born approximation can be trusted when the scattered wave it predicts at the centre of the potential is small beside the incident wave: for depth V₀ and range R, |V₀| ≪ ħ²/(mR²) when kR ≪ 1 and |V₀|R ≪ ħv when kR ≫ 1.
Example: For Coulomb scattering V₀R becomes Z₁Z₂e²/4πε₀ and the test is Z₁Z₂αc/v ≪ 1. Electrons at v = 0.20c give 0.22 on a bare carbon nucleus, a fair case for Born, but 2.9 on bare gold, where it fails.

Born amplitude for the Yukawa potentialf(θ) = −2mβ / [ħ²(μ² + q²)]

Derived here from ∫₀^∞ e^(−μr) sin(qr) dr = q/(μ² + q²). Flat while q ≪ μ, halved at q = μ, falling as 1/q² beyond: the range alone shapes the angular distribution.

V = β e^(−μr)/r with β in J m; μ and q = 2k sin(θ/2) in m⁻¹; m the reduced mass in kg; f in m

Rutherford cross-section, the limit μ → 0dσ/dΩ = (Z₁Z₂e²/4πε₀)² / [16E² sin⁴(θ/2)]

Derived: ħ cancels and the classical orbit result returns exactly. Second order in α, since the amplitude carries Z₁Z₂αħc. The total diverges, because 1/r has no range.

E the centre-of-mass kinetic energy in J; dσ/dΩ in m² sr⁻¹; Z₁Z₂e²/4πε₀ = Z₁Z₂ × 1.440 MeV fm

Exact Coulomb amplitudef_C(θ) = −[η/(2k sin²(θ/2))] exp[−iη ln sin²(θ/2)] Γ(1 + iη)/Γ(1 − iη)

Quoted from the exact solution in parabolic coordinates, not derived. Both phase factors have modulus 1, so |f_C|² equals the Born cross-section for every η.

η = Z₁Z₂e²/(4πε₀ħv) = Z₁Z₂αc/v, dimensionless; k in m⁻¹; Γ the gamma function; f_C in m

Scattered wave at the centre of the potentialψ_sc(0) = −[2m/(ħ²k)] ∫₀^∞ e^(ikr) sin(kr) V(r) dr

Derived from the Lippmann–Schwinger equation at r = 0. Born puts the incident wave where ψ belongs, so it is sound only while |ψ_sc(0)| ≪ 1.

dimensionless, relative to an incident wave of unit amplitude; V in J, r in m, k in m⁻¹, m in kg

Low-energy test, kR ≪ 1|V₀| ≪ ħ²/(mR²)

Derived. Binding needs |V₀| ≥ π²ħ²/(8mR²) = 1.23ħ²/(mR²) for a square well and g ≥ 1.68 for a Yukawa (quoted), so a well Born can handle at low energy cannot bind.

V₀ the depth in J, R the range in m; from (2m/ħ²)|∫₀^∞ rV dr| ≪ 1, which is g = 2m|β|/(ħ²μ) ≪ 1 for a Yukawa

High-energy test, kR ≫ 1|V₀|R ≪ ħv

Derived: ψ_sc(0) → −(i/ħv)∫₀^∞ V dr, the phase gathered on the way in to the centre. In energies it reads |V₀|/E ≪ 2/(kR), stricter than E ≫ |V₀|.

v = ħk/m the relative speed in m s⁻¹; for Coulomb V₀R becomes Z₁Z₂e²/4πε₀ and the test reads η = Z₁Z₂αc/v ≪ 1

Fourier-transform the Yukawa potential

Quote the previous lesson's first Born amplitude for a central potential, f(θ) = −[2m/(ħ²q)] ∫₀^∞ rV(r) sin(qr) dr, with q = 2k sin(θ/2). For V = β e^(−μr)/r the factor r cancels the 1/r, and ∫₀^∞ e^(−μr) sin(qr) dr = q/(μ² + q²), so f(θ) = −2mβ/[ħ²(μ² + q²)]. Dimensions: β is an energy times a length, so 2mβ/ħ² is an inverse length, and dividing by μ² + q² leaves a length, as f must be. The amplitude is real, negative for repulsion, and its shape is set by the range alone: flat while q ≪ μ, halved at q = μ, falling as 1/q² beyond. Integrating |f|² with dΩ = 2πq dq/k² gives σ = 16πm²β²/[ħ⁴μ²(μ² + 4k²)]. Limit: as k → 0 every angle has q = 0, the scattering is isotropic, and σ → 4πa² with a = 2mβ/(ħ²μ²), the Born scattering length, positive for repulsion as a hard sphere's is. Yukawa proposed this form in 1935 for a force carried by a particle of mass ħμ/c; a range of 1.41 fm corresponds to 140 MeV/c², the pion's measured mass.

Remove the screening and Rutherford's law appears

Let the range 1/μ grow without limit and set β = Z₁Z₂e²/4πε₀. The amplitude becomes f = −2mβ/(ħ²q²) = −β/[4E sin²(θ/2)], using ħ²k² = 2mE, and so dσ/dΩ = (d/4)²/sin⁴(θ/2) with d = β/E, the head-on distance of closest approach. Planck's constant has cancelled: this is the classical orbit result of the unit's first lesson, reached this time through a Fourier transform. Order in α: writing β = Z₁Z₂αħc shows the amplitude is first order in α and the cross-section second, with no ħ left over, since αħc = e²/4πε₀ contains none. Geiger and Marsden tested the 1/sin⁴(θ/2) law with α particles on gold between 5° and 150°, over which it changes by a factor of 2.4 × 10⁵. The price of infinite range is the total. The Yukawa σ tends to 4πη²/μ² once k ≫ μ, with η = β/ħv, and so grows without bound as μ → 0; in a real target the atomic electrons screen the nucleus and keep it finite.

The exact Coulomb amplitude differs only by a phase

Coulomb's potential falls off too slowly for the asymptotic form e^(ikz) + f e^(ikr)/r: even far away the waves carry a phase that grows as η ln(kr), with η = β/ħv = Z₁Z₂αc/v, which is why the amplitude lesson set it aside. The exact solution in parabolic coordinates, quoted here and not derived, gives f_C(θ) = −[η/(2k sin²(θ/2))] exp[−iη ln sin²(θ/2)] Γ(1 + iη)/Γ(1 − iη). Since η/2k = β/4E, the prefactor is exactly the Born amplitude, and the other two factors have modulus 1, because Γ(1 − iη) is the complex conjugate of Γ(1 + iη). So |f_C|² equals the first Born |f|² for every η, while the phases differ at first order in η. It is a property of 1/r, not of the approximation: 1/r is the potential whose extra conserved Runge–Lenz vector makes the Schrödinger equation separable in parabolic coordinates. The phase is not idle: wherever two amplitudes interfere, as f(θ) and f(π − θ) do for identical nuclei, or Coulomb and nuclear amplitudes do at large angles, it is measured, and Born gets it wrong.

Test the approximation at the centre of the potential

Born puts the incident wave inside the integral where ψ belongs, so it is sound only if the scattered wave it predicts is small where V acts. The Lippmann–Schwinger equation at r = 0, with the angular integral of e^(ik⋅r) done, gives ψ_sc(0) = −[2m/(ħ²k)] ∫₀^∞ e^(ikr) sin(kr) V(r) dr. At low energy, kR ≪ 1, this becomes −(2m/ħ²)∫₀^∞ rV dr: mV₀R²/ħ² for a square well of depth V₀, so the test is |V₀| ≪ ħ²/(mR²), and g = 2m|β|/(ħ²μ) for a Yukawa. Compare binding. As the scattering-length lesson found, the square well's scattering length diverges, and a bound state appears, at KR = π/2, that is |V₀| = π²ħ²/(8mR²) = 1.23ħ²/(mR²); the Yukawa binds at g = 1.68, quoted from numerical integration. So a potential that binds can never be treated by Born at low energy. The exact R[1 − tan(KR)/(KR)] = −(R/3)(KR)²[1 + (2/5)(KR)² + …] shows it: Born gives the first term, and the relative correction, (4/5)mV₀R²/ħ², is the test itself. The deuteron confirms it: the measured triplet neutron–proton scattering length is +5.42 fm, while Born gives a negative one for every attractive well.

At high energy the test becomes |V₀|R ≪ ħv

For kR ≫ 1 write e^(ikr) sin(kr) = (e^(2ikr) − 1)/2i. The e^(2ikr) part oscillates many times across the potential and averages away, leaving ψ_sc(0) ≈ −(i/ħv)∫₀^∞ V dr with v = ħk/m: the phase −(1/ħ)∫V dt that the wave gathers on its straight path in to the centre. For a square well the test is |V₀|R ≪ ħv, or |V₀|/E ≪ 2/(kR), stricter than E ≫ |V₀|. For the Yukawa the integral, done exactly with ∫₀^∞ (e^(−ar) − e^(−br)) dr/r = ln(b/a), gives ψ_sc(0) = −iη ln(1 − 2ik/μ), the curve in the figure, which for k ≫ μ is −η[π/2 + i ln(2k/μ)]. The screening enters only through a logarithm in the imaginary part, the Coulomb phase; the real part, the change in the wave's size, is −πη/2 whatever μ is. The exact Coulomb density at the origin, 2πη/(e^(2πη) − 1), quoted, expands to 1 − πη and confirms it. So for Coulomb the test is η = Z₁Z₂αc/v ≪ 1. Classical orbits need the reverse: d/ƛ = 2η with ƛ = 1/k, so a sharp hyperbola needs η ≫ 1. Protons on carbon cross η = 1 at v = 0.044c, a centre-of-mass energy of 0.83 MeV.

Higher Born terms, and what the optical theorem forces

Iterating the Lippmann–Schwinger equation gives the Born series f = f⁽¹⁾ + f⁽²⁾ + …, with f⁽ⁿ⁾ proportional to Vⁿ, so the tests above estimate the ratio of successive terms. One piece of f⁽²⁾ is compulsory. For a real potential f⁽¹⁾ is real, so Im f⁽¹⁾(0) = 0 although σ is positive, and the optical theorem, taken order by order as in its own lesson, demands Im f⁽²⁾(0) = kσ_B/4π. With the Yukawa σ above, that is a fraction 2km|β|/[ħ²(μ² + 4k²)] of |f⁽¹⁾(0)|, derived: g kR at low energy, |η|/2 at high energy, and never more than g/4, reached at kR = ½. At high energy that is the size, of order η, the test predicts; at low energy it is smaller by the factor kR, and the correction of order g sits in the real part of f⁽²⁾, the (2/5)(KR)² term of the square-well expansion above. For bare Coulomb, f⁽²⁾ diverges logarithmically as μ → 0, a result quoted from Dalitz (1951): the Coulomb phase once more, invisible in |f|². Named and left out: absorption, spin, the exchange symmetry of identical particles, and relativistic corrections of order v²/c².

Change one variable at a time

Make the relationship visible.

Interactive model

Strength g = 2m|β|/(ħ²μ)0.50

Marker energy, log₁₀(kR)1.0

At g = 0.50 the curve drops below 0.1 only past kR ≈ 7.9. Lower g to 0.20 and the crossing moves down to kR ≈ 1.9; raise g past 1.68, where an attractive Yukawa binds, and no low energy is safe, yet at kR = 100 even g = 2.00 leaves only 0.055.

kR AT THE MARKER10.00

|ψ_sc(0)| AT THE MARKER0.084

|β|/ħv = g/(2kR)0.025

Im f⁽²⁾(0) / |f⁽¹⁾(0)|0.012

Live interpretationkR AT THE MARKER: 10.00. |ψ_sc(0)| AT THE MARKER: 0.084. |β|/ħv = g/(2kR): 0.025. Im f⁽²⁾(0) / |f⁽¹⁾(0)|: 0.012

Catch the common trap

Explain before calculating.

Misconception check

The first Born approximation reproduces Rutherford's formula exactly, so it is tempting to conclude that Born is accurate for Coulomb scattering, and in particular that it described Geiger and Marsden's experiment. It did not. For 5.0 MeV α particles on gold, η = Z₁Z₂αc/v ≈ 22, so the wave near the nucleus is nothing like the incident wave, and the Born series is useless term by term. The agreement is a property of 1/r, whose exact amplitude differs from the Born one only by a phase, and the same property lets the classical orbit, valid for η ≫ 1, give the same answer. An approximation is certified by its own error estimate, not by one agreement with experiment: screen the potential, add the nuclear force, or make the particles identical, and anything that depends on the phase comes out wrong.

Neutrons of 20 MeV meet a nucleus modelled as an attractive square well of depth 4.0 MeV and radius 7.0 fm, so the energy is five times the depth and kR = 6.9. Can the first Born approximation be trusted here?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA nucleon–nucleon force is modelled by a Yukawa potential of range 1/μ = 1.41 fm, the pion's reduced Compton wavelength. At a centre-of-mass energy of 25.0 MeV, with reduced mass 469.5 MeV/c², use the first Born amplitude to find the angle at which the amplitude has fallen to half its forward value, and the ratio of forward to backward dσ/dΩ.

Wavenumber: k = √(2mc²E)/ħc = √(2 × 469.5 MeV × 25.0 MeV)/(197.33 MeV fm) = 153.2 MeV/(197.33 MeV fm) = 0.7765 fm⁻¹. Range: μ = 1/(1.41 fm) = 0.7092 fm⁻¹.
The amplitude is f = −2mβ/[ħ²(μ² + q²)], so f/f(0) = μ²/(μ² + q²). It halves when μ² + q² = 2μ², that is at q = μ.
With q = 2k sin(θ/2): sin(θ/2) = μ/2k = 0.7092/1.5529 = 0.4567, so θ/2 = 27.17° and θ = 54.3°.
Backward, q = 2k, so the ratio of dσ/dΩ at 0° to that at 180° is [(μ² + 4k²)/μ²]² = [(0.5030 + 2.4115)/0.5030]² = 5.794² = 33.6.
β never entered: the strength scales the distribution and the range shapes it. Limits: as E → 0 the ratio tends to 1, isotropic s-wave scattering, and as μ → 0 it grows without bound, Rutherford's forward divergence. These are Born's predictions, and trusting them needs the centre-of-potential test: at kR = 1.09 a Yukawa of this range strong enough to bind the deuteron, g ≥ 1.68, gives |ψ_sc(0)| ≥ 1.1, so for the real nucleon–nucleon force neither the shape nor the size can be trusted.

AnswerThe amplitude halves at θ = 54.3°, where sin(θ/2) = μ/2k = 0.457, and dσ/dΩ forward is 33.6 times its backward value; both are Born predictions fixed by the range alone, untrustworthy for a force strong enough to bind the deuteron.

MediumA neutron and a proton, reduced mass 469.5 MeV/c², interact through an attractive square well of radius 2.00 fm. Compare the first-Born scattering length with the exact aₛ = R[1 − tan(KR)/(KR)], K = √(2mV₀)/ħ, for a depth of 5.00 MeV and for 36.6 MeV, the depth that binds the deuteron by 2.2 MeV. The measured triplet scattering length is +5.42 fm.

The energy scale of the test: ħ²/(mR²) = (197.33 MeV fm)²/(469.5 MeV × 4.00 fm²) = 20.73 MeV. The Born parameter mV₀R²/ħ² is 5.00/20.73 = 0.241 for the shallow well and 36.6/20.73 = 1.765 for the deep one.
First Born at k → 0: f(0) = −(2m/ħ²)∫₀^R r²V dr = (2mV₀/ħ²)(R³/3), and f → −aₛ, so aₛ = −(2/3)(mV₀R²/ħ²)R. That is −(2/3)(0.241)(2.00 fm) = −0.322 fm for the shallow well and −2.35 fm for the deep one.
Exact, shallow well: (KR)² = 2mV₀R²/ħ² = 0.482, KR = 0.6945 and tan(KR) = 0.8329, so aₛ = 2.00 fm × (1 − 1.199) = −0.399 fm. Born is 19% short, the size of the correction (2/5)(KR)² = 0.193 that the series predicts.
Exact, deep well: KR = √(2 × 1.765) = 1.879, beyond π/2 = 1.571, so a bound state exists; tan(KR) = −3.142 and aₛ = 2.00 fm × (1 + 3.142/1.879) = +5.34 fm, close to the measured +5.42 fm.
Born's −2.35 fm has the wrong sign. Its parameter, 1.77, exceeds the square-well binding value π²/8 = 1.23, and no low-energy Born treatment of a force that binds can succeed.

AnswerShallow well: Born −0.322 fm against exact −0.399 fm, 19% short. Deuteron well: Born −2.35 fm against exact +5.34 fm and measured +5.42 fm, wrong in sign, because the well binds.

HardModel a helium atom, as a fast electron sees it, by the screened potential V = −(2e²/4πε₀) e^(−r/a)/r with a = 0.30 Å. For 10.0 keV electrons find η, the strength g, the first-Born scattered wave at the nucleus, the Born total cross-section, and the Im f(0) that the optical theorem forces into the second Born term. Treat the electron nonrelativistically and the atom as infinitely heavy.

Speed: v/c = √(2E/mₑc²) = √(20.0 keV/511.0 keV) = 0.1978, 1.5% above the relativistic 0.1950. So |η| = Zα/(v/c) = 2 × 0.007297/0.1978 = 0.0738, and k = (v/c)mₑc²/ħc = 0.1978 × 511.0 keV/(1.9733 keV Å) = 51.2 Å⁻¹, giving ka = 15.4.
Strength: g = 2mₑ|β|a/ħ² = 2|η|ka = 2 × 0.0738 × 15.37 = 2.27. That exceeds 1.68, so this model potential binds and Born is hopeless at low energy; only high energy can rescue it.
At the nucleus ψ_sc(0) = −iη ln(1 − 2ika), with η negative for attraction. Real part |η| atan(2ka) = 0.0738 × 1.538 = 0.1135, near π|η|/2 = 0.116: the wave is 11% larger there. Imaginary part (|η|/2) ln(1 + 4k²a²) = 0.0369 × 6.852 = 0.253, mostly the screening logarithm, a phase. Check: exact Coulomb density 2π|η|/(1 − e^(−2π|η|)) = 1.25 against 1 + π|η| = 1.23.
Cross-section: σ = 16πm²β²/[ħ⁴μ²(μ² + 4k²)] = 4πa²g²/(1 + 4k²a²) = 4π × 0.0900 Å² × 5.142/946 = 6.15 × 10⁻³ Å², that is 6.15 × 10⁻²³ m². Its high-energy form 4πη²a² is larger only by the factor (1 + 4k²a²)/(4k²a²) = 1.001, and it is second order in α.
Optical theorem: f⁽¹⁾(0) = ga = 2.268 × 0.30 Å = 0.680 Å, positive for attraction, and Im f⁽²⁾(0) = kσ/4π = 51.2 Å⁻¹ × 6.15 × 10⁻³ Å²/12.57 = 0.0251 Å, a fraction 0.037 of the first term: |η|/2, as the high-energy limit predicts.

Answer|η| = 0.074, g = 2.27, |ψ_sc(0)| = 0.28 with size part 0.11, σ = 6.15 × 10⁻²³ m², Im f⁽²⁾(0) = 0.025 Å, 3.7% of f⁽¹⁾(0). Born should hold to roughly ten percent at 10 keV but fails at low energy; exchange and polarisation are omitted.