University Quantum Mechanics II · Quantum Scattering Theory · 15.2

The scattering amplitude and dσ/dΩ = |f(θ)|²

Every number a scattering experiment reports is a count in a detector ten orders of magnitude farther away than the target is wide. This topic fixes what the wavefunction must look like out there, reads the particle current off it, and turns the count into the squared modulus of a single function of angle.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

A scattering experiment never looks at the target. It counts particles in a detector a metre away, ten orders of magnitude or more beyond the range R of the force, so the theory needs the stationary state only there, and four requirements fix it. Outside R the particle is free with E = ħ²k²/(2m), m the reduced mass, and ψ is the incident plane wave e^(ikz) plus whatever the target sends out. That wave must travel outward, keep the same k because the collision is elastic, fall as 1/r so that every large sphere passes the same flux, and depend on θ alone because a central V singles out no azimuth.

What is left is ψ → A[e^(ikz) + f(θ)e^(ikr)/r], with one unknown function f whose dimension is length. Feed the two waves into the probability current and, as this lesson derives, the incident flux |A|²ħk/m and the radial flux |A|²(ħk/m)|f|²/r² give dσ/dΩ = |f(θ)|². The model has three edges. The potential must fall faster than 1/r, or its tail keeps bending the phase: for Coulomb the drift grows as ln r and no fixed f exists.

The cross term between the two waves is dropped, which is right only for a detector outside the beam; in the forward direction it survives and becomes the optical theorem. And a steady beam replaces wave packets, a substitution stated here rather than proved.

Simple definition: The scattering amplitude f(θ) is the coefficient, with the dimension of a length, of the outgoing wave e^(ikr)/r that a stationary scattering state carries far from the target, and its squared modulus is the differential cross-section.
Example: If |f(30°)| = 12 fm, then dσ/dΩ = 144 fm² sr⁻¹ = 1.44 b sr⁻¹, and a 1.0 cm² counter 0.50 m away, subtending 4.0 × 10⁻⁴ sr, catches from each nucleus the particles that cross 5.8 × 10⁻³² m² of the beam.

The far-field boundary conditionψ → A [e^(ikz) + f(θ) e^(ikr)/r]

Derived here: e^(+ikr) because with e^(−iωt) its crests move out, 1/r so every large sphere passes the same flux, no φ because a central V and a beam along z pick out no azimuth.

k = √(2mE)/ħ in m⁻¹, m the reduced mass in kg; f in m; A in m^(−3/2); θ the CM angle; for r ≫ R and r ≫ kR²

Radial current of the outgoing wavejᵣ = |A|² (ħk/m) |f(θ)|² / r²

Derived from j = (ħ/m) Im(ψ*∇ψ), quoted from Quantum Mechanics I. Multiplied by the detector area r² dΩ it gives a rate with no r left in it.

jᵣ in m⁻² s⁻¹ with |A|² a number density in m⁻³; the θ component falls as 1/r³ and carries no flux to a far detector

Differential cross-sectiondσ/dΩ = |f(θ)|²

Derived: the rate jᵣ r² dΩ divided by the incident current |A|²ħk/m. The speed cancels because it is the same in and out; a channel leaving at speed v′ would keep a factor v′/v.

m² sr⁻¹, or b sr⁻¹ with 1 b = 10⁻²⁸ m²; elastic scattering, detector outside the beam

Total cross-sectionσ = 2π ∫₀^π |f(θ)|² sin θ dθ

Integrates dσ/dΩ over the sphere. An isotropic f gives 4π|f|²: the 20.4 b measured for slow neutrons on protons means a spin-averaged |f| ≈ 12.7 fm, six times the force's range.

σ in m²; the 2π is the φ integral, allowed because f has no φ dependence for a central V

Phase gathered in the tail of VΔφ(r) ≈ −(1/ħv) ∫^r V(r′) dr′

Derived from k(r) ≈ k − mV/(ħ²k). It settles to a constant only if V falls faster than 1/r; Coulomb gives −η ln r with η = Z₁Z₂αc/v, so no fixed f exists.

v = ħk/m in m s⁻¹, V in J, Δφ in rad relative to a free outgoing wave; valid where |V| ≪ E

The cross term between the two wavesjᵢₙₜ = |A|² (ħk/m)(1 + cos θ) Re[f(θ) e^{(ikr − ikr cos θ)}] / r

Derived. Absent outside a real beam; within θ_F = √(2π/kr) of the axis its phase stays below π, and there it survives to remove the incident flux through an area (4π/k) Im f(0): the optical theorem, quoted.

radial, m⁻² s⁻¹; first order in f, falling only as 1/r: up to 2r/|f| times the scattered current, 2 × 10¹⁴ for |f| = 10 fm at 1 m

Two waves far from the target, and why each has its form

Quantum Mechanics I read one-dimensional scattering off a steady beam, e^(ikx) plus a reflected wave on one side and a transmitted wave on the other, with the probabilities read off currents. Three dimensions keep that logic. Outside the range R of a central V the equation is free, (∇² + k²)ψ = 0 with E = ħ²k²/(2m), m the reduced mass and θ the centre-of-mass angle. The incident part is e^(ikz); the part the target adds must meet four demands. It is outgoing: with the time factor e^(−iωt), e^{(ikr − iωt)} has crests moving to larger r, while e^(−ikr) would converge on the target. It keeps k, because the collision is elastic and V has died away. It falls as 1/r, because in a steady state the flux through a sphere cannot depend on its radius. And it depends on θ alone, because a central V and a beam along z are unchanged by rotations about z. Check that f(θ)e^(ikr)/r solves the free equation: the radial part of ∇² returns −k² exactly, and the angular part leaves a remainder smaller by about 1/(kr)², 8 × 10⁻²² for 25 meV neutrons on a heavy nucleus (k ≈ 3.5 × 10¹⁰ m⁻¹) at 1 m.

The tail of V must fall faster than 1/r

Far out, where |V| ≪ E, the local wave number is k(r) = √(k² − 2mV/ħ²) ≈ k − mV/(ħ²k), so relative to a free outgoing wave the particle gathers the extra phase Δφ(r) ≈ −(1/ħv)∫^r V dr′, with v = ħk/m. For V ∝ 1/rⁿ with n > 1 the integral converges, the phase settles to a constant, and the constant is absorbed into f. For the Coulomb potential Z₁Z₂e²/(4πε₀r) it does not: Δφ = −η ln r + constant, with η = Z₁Z₂e²/(4πε₀ħv) = Z₁Z₂αc/v, first order in α. The estimate reproduces the logarithm in the exact Coulomb wave, e^{(i[kr − η ln 2kr])}/r, quoted from the Rutherford topic. For a 5.0 MeV alpha particle on gold, v = 0.0518c and η = 22.3, so moving the detector from 0.10 m to 0.20 m shifts the phase by η ln 2 = 15.4 rad: there is no fixed coefficient of e^(ikr)/r to call f. A neutral atom screens its nucleus and its potential falls exponentially, so electron–atom scattering has a proper f. A finite total cross-section asks more, a fall faster than 1/r², quoted from partial-wave analysis.

From two currents to dσ/dΩ = |f|²

The probability current j = (ħ/m) Im(ψ*∇ψ) is quoted from Quantum Mechanics I. For the incident wave ∇e^(ikz) = ik ẑ e^(ikz), so j_inc = |A|²ħk/m along z: a number density |A|² in m⁻³ times a speed. For the outgoing wave ∂/∂r [f e^(ikr)/r] = f e^(ikr)(ik/r − 1/r²), so ψ_sc*∂ψ_sc/∂r = |A|²|f|²(ik/r² − 1/r³), whose imaginary part gives jᵣ = |A|²(ħk/m)|f|²/r². The θ component comes from (1/r)∂/∂θ and falls as 1/r³, so through an area growing as r² it carries nothing to a distant detector. A counter of area dA at distance r subtends dΩ = dA/r² and receives jᵣ r² dΩ = |A|²(ħk/m)|f|² dΩ per second, independent of r, as the 1/r promised. Counting, from the previous topic, defines dσ through rate = j_inc dσ. Divide, and |A|² and ħk/m cancel: dσ/dΩ = |f(θ)|². The speed cancels only because it is the same before and after; for a channel leaving at speed v′ the same two lines give (v′/v)|f|², a case this unit does not need.

The cross term: dropped at the detector, kept in the beam

The current of a sum is not the sum of the currents. The mixed products ψ_inc*∇ψ_sc + ψ_sc*∇ψ_inc give a radial cross term jᵢₙₜ = |A|²(ħk/m)(1 + cos θ) Re[f e^{(ikr − ikz)}]/r, with z = r cos θ. It is first order in f and falls only as 1/r, so it is not small: for |f| = 10 fm at r = 1 m it outweighs the scattered current by 2r/|f| = 2 × 10¹⁴. It is dropped because a real beam is a plane wave only inside its cross-section, a few millimetres across, and a counter at 30° and 1 m sits 500 mm off the axis, where there is no incident wave to interfere with. Along the axis the story changes. The phase kr(1 − cos θ) ≈ krθ²/2 stays below π inside θ_F = √(2π/kr), which for 25 meV neutrons at 1 m is about 1.3 × 10⁻⁵ rad, a disc about 13 μm in radius deep inside the beam. There the cross term does not oscillate away, and its integral over the forward directions removes exactly (4π/k) Im f(0) times j_inc from the beam, which must equal what was scattered. That is the optical theorem, σ = (4π/k) Im f(0), derived in its own topic and quoted here.

A steady beam standing in for wave packets

A real projectile is a wave packet, and the stationary state an idealisation, as in Quantum Mechanics I's one-dimensional scattering. Its justification is stated here, not proved. Transversely, the packet's width w, its own coherence width and never more than the beam's, must be much larger than R, so the target sits in uniform plane wave, and much smaller than r sin θ, so the detector sits outside the beam. Longitudinally, its spread Δk must be small enough that f barely changes across it: a packet much longer than R and, near a resonance, longer than v times the resonance's lifetime. The counter must sit at r ≫ kR², where the phase difference kR²/r between paths from different parts of the target is negligible. When these hold, averaging over the random impact parameters of many packets gives each the probability |f|² dΩ/S of reaching the counter, S the beam area, as the steady beam does. For 25 meV neutrons in a 1 mm beam on nuclei with R ≈ 5 fm, any w from 1 nm up to the beam's 1 mm gives w/R from 2 × 10⁵ to 2 × 10¹¹, and a counter at 10° and 1 m is 174 mm off the axis: all hold easily. The steady picture gives up timing, such as a resonance's delay, which returns in the resonance topic.

Checks: dimensions, limits, α and a measurement

Dimensions: e^(ikz) and f e^(ikr)/r are added, so f is a length, |f|² is an area per steradian, and |A|²ħk/m is m⁻³ × m s⁻¹ = m⁻² s⁻¹, a flux. Limits: switch V off and nothing is scattered, so f → 0 and σ → 0. At low energy, when kR ≪ 1, the outgoing wave cannot resolve the target's shape and f tends to a constant, −a, with a the scattering length of a later topic, quoted here: the scattering is isotropic and σ = 4πa². Order in α: for an electromagnetic interaction the tail's phase is set by η = Z₁Z₂αc/v, first order in α, and for a screened charge the amplitude itself is first order in α, quoted from the Born-approximation topic, so dσ/dΩ = |f|² is second order. A measurement: slow neutrons on protons scatter isotropically in the centre-of-mass frame, which is why proton-recoil counters see a flat recoil spectrum, and σ = 20.4 b is measured. Averaged over the spin states this unit leaves aside, 4π|f|² = 20.4 b gives |f| = 12.7 fm, six times the 2 fm range of the force: σ is about 160 times π(2 fm)², a strength and not a size.

Change one variable at a time

Make the relationship visible.

Interactive model

Detector distance kr60

Beam half-width a/r0.20

Phase of f90 °

Step the phase of f from 15° to 165°: the fringes inside the beam slide and flip, yet the flat level outside it never moves, because a counter reads |f|² and not the phase. Then narrow the beam to 0.10: past 30° the solid curve lies on |f|²/r², all that a detector there counts.

FORWARD ZONE θ_F18.6 °

BEAM EDGE, 1% INTENSITY25.4 °

CROSS TERM ON AXIS0.00 × |f|²/r²

CROSS-TERM SIZE AT 30°0.176 × |f|²/r²

Live interpretationFORWARD ZONE θ_F: 18.6 °. BEAM EDGE, 1% INTENSITY: 25.4 °. CROSS TERM ON AXIS: 0.00 × |f|²/r². CROSS-TERM SIZE AT 30°: 0.176 × |f|²/r²

Catch the common trap

Explain before calculating.

Misconception check

Its name invites reading f(θ) as a probability amplitude, so that |f|² would be a probability per steradian and ∫|f|²dΩ would equal 1. It is neither. f carries the dimension of length, because it multiplies e^(ikr)/r to make something that adds to the pure number e^(ikz), and ∫|f|²dΩ is the total cross-section σ, an area, capped neither at 1 nor at the target's shadow. The probability that one particle in a beam of area S scatters is σ/S, tiny because S is macroscopic. Nor is σ the target's size: slow neutrons on protons give 20.4 b, about 160 times π(2 fm)², and a hard sphere at low energy scatters from 4πR², four times its shadow, a result of the hard-sphere topic quoted here. Normalise a count by the incident flux, never by the number of incident particles, and read f as a length.

Set at θ = 30° and 0.50 m from a thin target, a counter records slow neutrons scattered out of a beam 2 mm wide. The current built from e^(ikz) + f(θ)e^(ikr)/r contains a cross term between the two waves. Why is it right to leave that term out when the count is turned into dσ/dΩ?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyAt r = 0.80 m and θ = 40°, the outgoing part of a scattering state has modulus 1.5 × 10⁻¹⁴ times that of the incident plane wave. Find |f(40°)| and dσ/dΩ there. A counter with a 1.5 cm² face sits at that point, facing a thin target of 4.0 × 10²⁰ independent scatterers in a beam of flux 2.5 × 10¹⁴ m⁻² s⁻¹. Find its count rate, and say what changes if it is moved out to 1.60 m.

Far out |ψ_sc|/|ψ_inc| = (|A||f|/r)/|A| = |f|/r, so |f(40°)| = 0.80 m × 1.5 × 10⁻¹⁴ = 1.2 × 10⁻¹⁴ m = 12 fm. The normalisation A has cancelled, as it must.
dσ/dΩ = |f|² = (1.2 × 10⁻¹⁴ m)² = 1.44 × 10⁻²⁸ m² sr⁻¹ = 1.44 b sr⁻¹.
The face subtends ΔΩ = 1.5 × 10⁻⁴ m² ÷ (0.80 m)² = 2.34 × 10⁻⁴ sr. Each scatterer sends it Φ(dσ/dΩ)ΔΩ = 2.5 × 10¹⁴ × 1.44 × 10⁻²⁸ × 2.34 × 10⁻⁴ = 8.4 × 10⁻¹⁸ particles per second, which is jᵣ r² ΔΩ with |A|²ħk/m = Φ.
The scatterers are independent, so their counts add: 4.0 × 10²⁰ × 8.4 × 10⁻¹⁸ s⁻¹ = 3.4 × 10³ s⁻¹.
At 1.60 m the ratio halves to 7.5 × 10⁻¹⁵, but |f| = 1.60 m × 7.5 × 10⁻¹⁵ = 12 fm again: f belongs to the collision, not to the counter. The solid angle falls fourfold, so the rate falls to 8.4 × 10² s⁻¹.

Answer|f(40°)| = 12 fm and dσ/dΩ = 1.44 b sr⁻¹; the counter records about 3.4 × 10³ s⁻¹ at 0.80 m and 8.4 × 10² s⁻¹ at 1.60 m, with f unchanged.

MediumAt one energy, the elastic amplitude for a spinless projectile on a spinless target is fitted as f(θ) = f₀ + f₁ cos θ, with f₀ = −2.0 fm and |f₁| = 1.2 fm. Find dσ/dΩ at 0°, 90° and 180°, and σ, first for f₁ = +1.2 fm and then for f₁ = +1.2i fm. Which part of f can no counter measure?

Square the modulus: |f|² = |f₀|² + |f₁|² cos²θ + 2 Re(f₀f₁*) cos θ. The last term is interference between the two pieces of f, and it depends on their relative phase.
For f₁ = +1.2 fm, 2 Re(f₀f₁*) = 2 × (−2.0)(1.2) = −4.80 fm², so dσ/dΩ = 4.00 − 4.80 cos θ + 1.44 cos²θ in fm² sr⁻¹: 0.64 at 0°, 4.00 at 90° and 10.24 at 180°. The two pieces cancel forward and reinforce backward.
For f₁ = +1.2i fm, f₀f₁* = (−2.0)(−1.2i) = 2.4i fm² is purely imaginary, the interference term vanishes, and dσ/dΩ = 4.00 + 1.44 cos²θ: 5.44, 4.00 and 5.44 fm² sr⁻¹, symmetric about 90°.
σ = 2π ∫₋₁¹ |f|² d(cos θ) = 4π(|f₀|² + |f₁|²/3) = 4π × 4.48 fm² = 56.3 fm² = 0.563 b in both cases, because the interference term is odd in cos θ and integrates to zero.
Multiplying f by any e^(iγ) leaves every |f|², and so every count, unchanged, and so does replacing f by its complex conjugate: counters fix |f₀|, |f₁| and only the cosine of their relative phase, so f₁ = −1.2i fm gives the same counts as +1.2i fm. The overall phase is not free, though. The optical theorem σ = (4π/k) Im f(0), quoted, needs Im f(0) = kσ/4π > 0, while the first fit read literally has f(0) = −2.0 + 1.2 = −0.8 fm, real, so Im f(0) = 0 against σ = 0.563 b. A fit to counts gives |f|; the phase shows only in the forward cross term.

AnswerWith f₁ = 1.2 fm: 0.64, 4.00 and 10.24 fm² sr⁻¹; with f₁ = 1.2i fm: 5.44, 4.00 and 5.44 fm² sr⁻¹. σ = 56.3 fm² = 0.563 b either way. Counters miss the overall phase of f and the sign of the relative phase, and the optical theorem shows that the first fit, with f(0) real, cannot be the literal amplitude.

HardA 150 eV electron scatters from a bare proton, and separately from a ground-state hydrogen atom treated as a static charge cloud, with potential energy V(r) = −(e²/4πε₀)(1/r + 1/a₀)e^(−2r/a₀); ignore exchange. Using Δφ ≈ −(1/ħv)∫V dr, find the phase change of the outgoing wave when a detector is moved from 0.20 m to 0.40 m from the proton, and bound the whole phase the atom adds beyond 5a₀. In which case does f exist?

Speed: v/c = √(2 × 150 eV ÷ 510 999 eV) = 0.02423, so the Coulomb parameter is η = Z₁Z₂αc/v = −1/(137.036 × 0.02423) = −0.3012, negative because the force attracts, and first order in α.
Bare proton: −(1/ħv)∫V dr = (e²/4πε₀ħv)∫dr/r = |η| ln(r₂/r₁). From 0.20 m to 0.40 m that is 0.3012 × ln 2 = 0.209 rad, about 12°, and every further doubling adds the same again.
So the coefficient of e^(ikr)/r depends on where it is measured, and no fixed f exists; this is the ln r in the exact Coulomb wave, quoted from the Rutherford topic. From 5a₀ = 2.65 × 10⁻¹⁰ m out to 0.40 m the drift is 0.3012 × ln(1.51 × 10⁹) = 0.3012 × 21.14 = 6.37 rad. The expansion is safe there: at 5a₀ |V| = 5.4 eV against 150 eV.
Atom: −(1/ħv)∫V dr = |η| ∫(1/r + 1/a₀)e^(−2r/a₀) dr. Beyond 5a₀, 1/r ≤ 1/(5a₀), so the integral is at most (6/5a₀)(a₀/2)e⁽⁻¹⁰⁾ = 0.6e⁽⁻¹⁰⁾ = 2.72 × 10⁻⁵, and the phase is at most 0.3012 × 2.72 × 10⁻⁵ = 8.2 × 10⁻⁶ rad, whatever the detector distance.
The screened phase converges, so it is absorbed into a fixed f; at 5a₀ the atom's |V| is only 1.5 × 10⁻³ eV. The exact integral, using the exponential integral E₁(10) = 4.16 × 10⁻⁶, gives 8.1 × 10⁻⁶ rad, inside the bound.

AnswerBare proton: 0.209 rad (12°) per doubling of the distance and 6.37 rad from 5a₀ to 0.40 m, so no fixed f. Atom: at most 8.2 × 10⁻⁶ rad beyond 5a₀ in all, so f exists. η = −0.301, first order in α.