University Quantum Mechanics II · Quantum Scattering Theory · 15.8

The Lippmann-Schwinger equation and first Born approximation

Partial waves win when a few angular momenta matter. This lesson builds the method for everything else: rewrite the Schrödinger equation as an integral equation with the outgoing wave built in, then guess the wave inside the potential. The simplest guess makes f a Fourier transform of V, the working formula of high-energy scattering.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Scattering is a boundary-value problem: a plane wave comes in, and far away nothing else may appear but outgoing waves. The outgoing Green's function G₊ = −e^{(ik|r − r′|)}/(4π|r − r′|) answers a point source with a wave that only radiates outward, so treating (2m/ħ²)Vψ as a source turns the Schrödinger equation and that boundary condition into one integral equation, the Lippmann-Schwinger equation. Read at a distant detector it gives the amplitude exactly, f = −(m/2πħ²) ∫ e^{(−ik′⋅r)} V(r) ψ(r) d³r, an integral over the range of V alone.

The formula is exact and still implicit, because ψ inside the potential is the unknown; every approximation is a guess at it. The first Born approximation guesses the undisturbed incident wave. Then f is the Fourier transform of V at the momentum transfer ħq, with q = 2k sin(θ/2): linear in V, for a central V real and a function of q alone, and equal to what Fermi's golden rule gives at the same order.

Iterating the equation adds one more scattering per order, the Born series. The guess fails where the potential reshapes the wave inside its range: in a well strong enough to nearly bind, the exact scattering length runs off to infinity while the Born value stays finite, and a real first-order amplitude can never carry the imaginary forward part the optical theorem requires.

Simple definition: The first Born approximation computes the scattering amplitude by replacing the exact wave inside the potential with the undisturbed incident plane wave, which makes f equal to −m/(2πħ²) times the Fourier transform of V at the momentum transfer.
Example: An attractive square well with g = 2mV₀R²/ħ² = 0.30 has f_B(0) = gR/3 = 0.100R. At kR = 3.0 and θ = 90°, qR = 2kR sin 45° = 4.24, just short of the first zero at 4.49, and f_B = 0.0040R: 4% of the forward value.

Outgoing Green's functionG₊(r) = −e^(ikr)/(4πr), solving (∇² + k²) G₊ = δ³(r)

Derived here. With the time factor e^(−iEt/ħ), the +ikr makes every wavelet travel outward: the scattering boundary condition is chosen once, inside G₊.

G₊ in m⁻¹; k in m⁻¹ with E = ħ²k²/2m; δ³ in m⁻³; checked with ∇²(1/r) = −4πδ³(r)

Lippmann-Schwinger equationψ(r) = e^(ik⋅r) − (m/2πħ²) ∫ e^{(ik|r − r′|)} V(r′) ψ(r′) d³r′/|r − r′|

Derived: the Schrödinger equation and its outgoing boundary condition in one line. Exact but implicit, with ψ on both sides; iterating it gives the Born series.

ψ dimensionless, unit incident amplitude; m the reduced mass in kg; V in J; the integral runs only where V ≠ 0

Exact amplitude, read at the detectorf(θ, φ) = −(m/2πħ²) ∫ e^{(−ik′⋅r′)} V(r′) ψ(r′) d³r′

Derived, and still exact: it needs ψ only inside the range of V. Every approximation to f is an approximation to ψ there.

f in m; k′ = kr̂ the outgoing wavevector toward the detector; needs r ≫ R and r ≫ kR², R the range of V

First Born approximationf_B(q) = −(m/2πħ²) ∫ e^(−iq⋅r) V(r) d³r

Derived by putting ψ ≈ e^(ik⋅r) inside the range. The amplitude is the Fourier transform of V: first order in V, a function of q alone, and real when V is real and central.

ħq = ħk′ − ħk the momentum transfer; q = 2k sin(θ/2) in m⁻¹; f_B in m

Born amplitude for a central potentialf_B(θ) = −(2m/ħ²q) ∫₀^∞ r V(r) sin(qr) dr

Derived from ∫ e^{(−iqr cos α)} dΩ = 4π sin(qr)/(qr). For a square well of depth V₀ it gives (2mV₀/ħ²q³)(sin qR − qR cos qR).

r in m, V(r) in J; as q → 0 it tends to −(m/2πħ²) ∫ V d³r, minus the Born scattering length

Golden-rule cross-sectiondσ/dΩ = (m²/4π²ħ⁴) |∫ e^(−iq⋅r) V(r) d³r|²

Derived here: it equals |f_B|², an independent check on the Born amplitude. Being a rate, it returns only |f|², blind to the sign of V and to any phase.

dσ/dΩ in m² sr⁻¹; from Unit 12's rate between box-normalised plane waves, divided by the incident current; the box size cancels

A point source that radiates only outward

With E = ħ²k²/2m the Schrödinger equation reads (∇² + k²)ψ = (2m/ħ²)Vψ. Read the right side as a source. If G solves (∇² + k²)G = δ³(r), the response to any source is a superposition of G's, so one point source settles everything. Try G₊ = −e^(ikr)/(4πr). For r > 0 it is a multiple of u/r with u = e^(ikr), and u″ + k²u = 0, so only the origin needs care. Write G₊ = e^(ikr) × (−1/4πr). The electrostatics identity ∇²(1/r) = −4πδ³(r) gives ∇²(−1/4πr) = δ³(r), and since e^(ikr) = 1 at the origin that term is the whole δ function. The cross term 2∇e^(ikr)⋅∇(−1/4πr) = 2ik e^(ikr)/(4πr²) cancels the 2ik/r piece of −(1/4πr)∇²e^(ikr), where ∇²e^(ikr) = (2ik/r − k²)e^(ikr), and what is left is −k²G₊. So (∇² + k²)G₊ = δ³(r), derived. Dimensions: δ³ is m⁻³ and ∇² is m⁻², so G is m⁻¹, as 1/4πr is. The sign in the exponent is physics: −e^(−ikr)/(4πr) solves the same equation but, with e^(−iEt/ħ), converges on the target.

The Lippmann-Schwinger equation, read at the detector

Since ∇² + k² annihilates e^(ik⋅r), the incident wave can be added freely: ψ(r) = e^(ik⋅r) + ∫ G₊(r − r′)(2m/ħ²)V(r′)ψ(r′) d³r′. That is the Lippmann-Schwinger equation: exact, with the outgoing condition inside, and implicit, ψ on both sides. Put the detector far outside the range R, where |r − r′| = r − r̂⋅r′ + O(r′²/r). The two places it appears need different care. In the denominator, 1/|r − r′| → 1/r costs a fraction R/r. In the phase, k r̂⋅r′ = k′⋅r′, with k′ = kr̂ the outgoing wavevector, is of order kR and must stay; only the next term, of order kR²/r, may go, which needs r ≫ kR². For 100 eV electrons, k = 5.12 × 10¹⁰ m⁻¹, and an atom with R = 0.10 nm, kR² = 0.51 nm, so a detector 0.2 m away clears it by eight orders of magnitude. What survives is ψ → e^(ik⋅r) + f e^(ikr)/r with f = −(m/2πħ²) ∫ e^{(−ik′⋅r′)}V(r′)ψ(r′) d³r′, derived and still exact. It needs ψ only where V is non-zero. Dimensions: m/ħ² is J⁻¹ m⁻², so (m/ħ²)V d³r′ is a length, as f must be.

One scattering: the amplitude becomes a Fourier transform

If V barely disturbs the wave inside its range, put ψ ≈ e^(ik⋅r) in the exact formula. Then f_B = −(m/2πħ²) ∫ e^(−iq⋅r)V(r) d³r, with ħq = ħk′ − ħk the momentum the potential hands over. Elastic scattering keeps |k′| = k, so q² = 2k² − 2k² cos θ = 4k² sin²(θ/2): q = 2k sin(θ/2), zero forward and 2k backward. Three properties follow, all derived. The amplitude depends on energy and angle only through q, so for a central V data at many energies fall on one curve. It is real for a real central V, and at θ = 0, where it is the volume integral of V, real for any real V: it carries no Im f(0), the part the optical theorem ties to σ; that arrives at second order. And it is linear in V: reversing the sign of V reverses f_B and leaves dσ/dΩ = |f_B|² unchanged. Forward, f_B(0) = −(m/2πħ²) ∫ V d³r, positive for an attractive well, and at low energy, where qR ≪ 1 at every angle, that constant is the whole amplitude: minus the Born scattering length. For a charged probe V carries e²/4πε₀ = αħc, so f_B is first order in α and dσ/dΩ second order.

Central potentials reduce to one radial integral

For V(r), point q along the polar axis. The angular integral ∫ e^{(−iqr cos α)} dΩ = 4π sin(qr)/(qr) gives f_B = −(2m/ħ²q) ∫₀^∞ rV(r) sin(qr) dr, derived. The attractive square well, V = −V₀ for r < R, needs ∫₀^R r sin(qr) dr = (sin qR − qR cos qR)/q², so f_B = (2mV₀/ħ²q³)(sin qR − qR cos qR) = gR(sin x − x cos x)/x³, with x = qR and g = 2mV₀R²/ħ². Checks: f_B/R is a pure number, and as x → 0, sin x − x cos x → x³/3, so f_B(0) = gR/3, which is (m/2πħ²)V₀(4πR³/3), the volume integral. The amplitude vanishes where tan x = x, at qR = 4.49 and 7.73, whatever V₀ is. Since q never exceeds 2k, the first zero needs kR ≥ 2.25; slower particles scatter smoothly, and for kR ≪ 1 isotropically, the s-wave limit of partial waves. Expanding sin(qr) gives f_B(q) = f_B(0)[1 − q²⟨r²⟩/6 + …], with ⟨r²⟩ weighted by V: the fall-off at small q measures size. That slope is how electron scattering measures charge radii, the proton's included, whose measured value, 0.841 fm, is quoted in the Hard example. The figure plots f_B² for this well.

Unit 12's golden rule gives the same cross-section

An independent route checks the result. In a box of side L take plane waves ⟨r|k⟩ = e^(ik⋅r)/L³⁄², so ⟨k′|V|k⟩ = Ṽ(q)/L³, with Ṽ(q) = ∫ e^(−iq⋅r)V d³r. States with wavevector in d³k′ number L³d³k′/(2π)³, and with E = ħ²k′²/2m that is ρ = L³mk/(8π³ħ²) per unit energy per steradian. The golden rule, quoted from Unit 12, gives the rate into dΩ as w = (2π/ħ)(|Ṽ|²/L⁶)ρ dΩ. The beam holds one particle per L³ moving at ħk/m, a current ħk/(mL³). So dσ/dΩ = (2π/ħ)(|Ṽ|²/L⁶)(L³mk/8π³ħ²)(mL³/ħk) = m²|Ṽ|²/(4π²ħ⁴). Every L cancels, as it must for a property of the target, and the result is exactly f_B². Two first-order calculations agreeing is the point: the golden rule is the first Born approximation written as a rate. It returns only |f|², so it cannot see the sign of V or any phase, and it says nothing about higher orders. For those, iterate the integral equation.

Iterating: the Born series against an exact result

Write the Lippmann-Schwinger equation as ψ = φ + G₊Uψ, with φ = e^(ik⋅r) and U = 2mV/ħ². Substituting it into itself gives ψ = φ + G₊Uφ + G₊UG₊Uφ + …, so f = f⁽¹⁾ + f⁽²⁾ + …, with f⁽ⁿ⁾ proportional to Vⁿ: n scatterings joined by free outgoing propagation. The square well tests this at k → 0, where the scattering-length lesson derived aₛ = R[1 − tan(KR)/(KR)], with (KR)² = g. Expanding tan x/x = 1 + x²/3 + 2x⁴/15 + 17x⁶/315 + … gives −aₛ = R(g/3 + 2g²/15 + 17g³/315 + …). The g term is the first Born result, and since a power series in V₀ is unique, 2g²R/15 is the second Born term at zero energy. At g = 0.1 the exact amplitude is 1.042 times Born, close to 1 + 2g/5 = 1.04; at g = 1 it is 1.67 times. The series converges only for g < (π/2)² = 2.47, exactly where the well first binds a state: no finite number of Born terms can build a bound state. The next lesson turns this into a working test.

Change one variable at a time

Make the relationship visible.

Interactive model

Energy, as kR3.00

Well strength g = 2mV₀R²/ħ²0.30

Detector angle θ90 °

At kR = 0.5 every angle has qR below 1 and the curve is flat: s-wave scattering. Raise kR past 2.25 and the first zero enters at 180°, then sweeps forward as the window widens. Drag g: the curve slides up or down and no zero moves, since f_B is linear in V.

qR at the detector4.24

f_B(θ) ÷ f_B(0)0.040

f_B(0) ÷ R = g/30.100

Exact ÷ Born, f(0) at k → 01.14

Live interpretationqR at the detector: 4.24. f_B(θ) ÷ f_B(0): 0.040. f_B(0) ÷ R = g/3: 0.100. Exact ÷ Born, f(0) at k → 0: 1.14

Catch the common trap

Explain before calculating.

Misconception check

The tempting shortcut in the far-field step is to set |r − r′| ≈ r everywhere, since the detector is far away. In the denominator that is right: it changes 1/|r − r′| by a fraction R/r. In the phase it is wrong, because what matters there is not r′/r but kr′, the number of wavelengths across the target, which is of order kR and not small. Drop k′⋅r′ and the amplitude becomes −(m/2πħ²) ∫ Vψ d³r′, the same in every direction: isotropic scattering whatever the energy. That is correct only when kR ≪ 1, which is exactly why slow particles scatter in the s wave. The angular structure of every pattern, the square well's zeros at qR = 4.49 and the forward peak narrowing as the energy rises, lives entirely in that phase. Keep r̂⋅r′ to first order in the exponent; the far-field limit discards only the next term, at most kr′²/2r.

Particles with kR = 3.0 scatter from a weak attractive square well of radius R whose first Born zero is at qR = 4.49. In the first Born approximation, where is the first minimum of dσ/dΩ, and what happens to it if the well is made twice as deep but still weak?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyElectrons of 100 eV scatter from a model atom treated as an attractive square well of depth V₀ = 5.0 eV and radius R = 0.050 nm; the atom is heavy, so m is the electron mass. In the first Born approximation, find f_B and dσ/dΩ at θ = 0° and at θ = 90°, and the angle of the first zero.

Wavenumber: k = √(2mE)/ħ = √(2 × 9.109 × 10⁻³¹ kg × 1.602 × 10⁻¹⁷ J)/(1.0546 × 10⁻³⁴ J s) = 5.12 × 10¹⁰ m⁻¹, so kR = 2.56.
Strength: g = 2mV₀R²/ħ² = 2 × 9.109 × 10⁻³¹ kg × 8.01 × 10⁻¹⁹ J × (5.0 × 10⁻¹¹ m)²/(1.0546 × 10⁻³⁴ J s)² = 0.328, a pure number as it must be. Born is a fair guess here: V₀R/ħv = g/(2kR) = 0.064 ≪ 1, so the well barely shifts the phase of the wave crossing it.
Forward: f_B(0) = gR/3 = 0.328 × 0.050 nm/3 = 5.47 × 10⁻¹² m, positive for an attractive well, and dσ/dΩ = f_B² = 2.99 × 10⁻²³ m² sr⁻¹.
At 90°: x = qR = 2kR sin 45° = 3.62, and (sin x − x cos x)/x³ = (−0.463 + 3.21)/47.5 = 0.0578, so f_B = gR × 0.0578 = 9.48 × 10⁻¹³ m and dσ/dΩ = 9.0 × 10⁻²⁵ m² sr⁻¹, 3.0% of the forward value.
First zero: qR = 4.493 needs sin(θ/2) = 4.493/(2kR) = 4.493/5.123 = 0.877, so θ/2 = 61.3° and θ = 123°. Its angle is fixed by kR alone; a deeper well would not move it in this approximation.

Answerf_B(0) = 5.47 pm and dσ/dΩ = 2.99 × 10⁻²³ m² sr⁻¹ forward; f_B = 0.948 pm and dσ/dΩ = 9.0 × 10⁻²⁵ m² sr⁻¹ at 90°; first zero at θ ≈ 123°.

MediumTwo particles with reduced mass mc² = 469.5 MeV interact through a model attractive square well of radius R = 2.0 fm and depth V₀ = 5.18 MeV. At zero energy, compare the first Born scattering length and cross-section with the exact ones, and find how much of the gap the second Born term closes.

Strength: g = 2mc²V₀R²/(ħc)² = 2 × 469.5 MeV × 5.18 MeV × (2.0 fm)²/(197.33 MeV fm)² = 0.500. Born needs g well below 1, so expect errors of tens of per cent.
First Born: f_B(0) = gR/3 = 0.333 fm, so a_B = −0.333 fm and σ_B = 4πa_B² = 1.39 fm².
Exact, from the scattering-length lesson: KR = √g = 0.7069 and tan(0.7069)/0.7069 = 1.2083, so aₛ = R[1 − tan(KR)/(KR)] = 2.0 fm × (−0.2083) = −0.417 fm, and σ = 4πaₛ² = 2.18 fm².
Series: −aₛ = R(g/3 + 2g²/15 + 17g³/315 + …) = 0.333 + 0.067 + 0.013 + … fm. The second Born term, 0.0666 fm, closes 80% of the 0.0835 fm gap, and the third brings the sum to 0.413 fm.
Cross-sections go as a², so a 20% shortfall in the amplitude becomes σ_B/σ = (0.333/0.417)² = 0.64: the first Born cross-section is 36% low.

Answera_B = −0.333 fm and σ_B = 1.39 fm², against the exact aₛ = −0.417 fm and σ = 2.18 fm²; the second Born term, 0.067 fm, closes 80% of the gap.

HardModel the proton's charge as ρ(r) = e^(−r/b)/(8πb³), normalised to one. In the first Born approximation an extended charge multiplies the point-charge amplitude by F(q) = ∫ e^(−iq⋅r)ρ d³r. Find F(q), fix b from the measured dipole fit F = (1 + ħ²q²/Λ²)⁻² with Λ² = 0.71 GeV²/c², compare the rms radius with the measured 0.841 fm, and find the q at which |F|² has fallen to 0.90.

Why a factor: the electron's potential energy is the point-charge Coulomb term convolved with ρ, and the Fourier transform of a convolution is the product of transforms, so f_B picks up F(q) and dσ/dΩ picks up |F|². For fast electrons the point-charge cross-section takes a relativistic form, quoted, and the factor |F|² survives.
The central formula with ρ in place of V gives F = (4π/q) ∫₀^∞ rρ sin(qr) dr = (1/2qb³) ∫₀^∞ r e^(−r/b) sin(qr) dr. With ∫₀^∞ r e^(−λr) sin(qr) dr = 2λq/(λ² + q²)² and λ = 1/b, F = 1/(1 + q²b²)². Check: F(0) = 1, the whole charge.
Matching the dipole: b = ħ/Λ = ħc/(Λc) = 0.19733 GeV fm/0.8426 GeV = 0.2342 fm.
Small q: F ≈ 1 − 2q²b², and the general expansion F ≈ 1 − q²⟨r²⟩/6 then gives ⟨r²⟩ = 12b², so rᵣₘₛ = √12 × 0.2342 fm = 0.811 fm. That is 3.5% below the measured 0.841 fm: the dipole is a fit over a wide range of q, while the radius is the slope at q = 0.
|F|² = (1 + q²b²)⁻⁴ = 0.90 gives q²b² = 0.90^(−1/4) − 1 = 0.0267, so qb = 0.163 and q = 0.698 fm⁻¹, a momentum transfer ħqc = 138 MeV. Since q ≤ 2k, even backscattered electrons need about 69 MeV, neglecting recoil, to see a 10% dip.

AnswerF(q) = (1 + q²b²)⁻² with b = 0.234 fm; rᵣₘₛ = √12 b = 0.811 fm, 3.5% below the measured 0.841 fm; |F|² = 0.90 at q = 0.698 fm⁻¹, ħqc ≈ 138 MeV.