University Quantum Mechanics II · Quantum Scattering Theory · 15.5

The optical theorem: σ = (4π/k) Im f(0)

This is the audit every amplitude in the unit has to pass. One number, the imaginary part of the forward amplitude, must account for every particle scattered in every direction. You will prove why, find where in space the accounting happens, and use its failure to judge the Born approximation.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

A scattered wave is not new flux: it is flux the beam no longer carries, and the optical theorem makes that bookkeeping exact. Far from the potential, ψ = e^(ikz) + f(θ)e^(ikr)/r, and the flux out through a large sphere has three parts. The plane wave's own part nets to zero. The scattered part is (ħk/m)σ, quadratic in f.

The cross term between them oscillates away at every angle except inside a forward cone of half-angle about √(2π/kr), where it is linear in f(0) and removes flux from the beam. Conservation of probability sets that loss equal to what went out everywhere else, so σ = (4π/k) Im f(0). For a real central potential the partial-wave sum proves it in two lines, because Pₗ(1) = 1 turns f(0) into a sum whose imaginary part is the same sum of sin²δₗ that makes up σ. The theorem fixes Im f(0) and nothing else: Re f(0), which shifts the phase of the transmitted beam, is free.

It survives absorption only if σ is read as the total, elastic plus absorbed, a form this lesson quotes. It is empty for the Coulomb potential, where f(0) and σ are both infinite, and no purely real amplitude can satisfy it, which is why the first Born approximation appears to break it and the second Born term must repair it.

Simple definition: The optical theorem states that the total cross-section is 4π/k times the imaginary part of the forward scattering amplitude, σ = (4π/k) Im f(0), because the flux scattered in all directions is removed from the beam by interference in the forward direction.
Example: A lone s wave at δ₀ = 90° with k = 0.50 fm⁻¹ has f(0) = i/k = 2.0i fm, so (4π/k) Im f(0) = 25.13 fm × 2.0 fm = 50.3 fm² = 0.503 b, exactly the s-wave unitarity limit 4π/k².

The optical theorem, elastic scatteringσ = (4π/k) Im f(0)

Derived here for a real central potential, and outlined for any finite-range potential by flux balance. One forward number audits the whole angular distribution.

σ in m²; k in m⁻¹ with E = ħ²k²/2m, m the reduced mass; f(0) the forward amplitude, in m

The forward partial-wave amplitudeIm f(0) = (1/k) Σ (2l + 1) sin²δₗ

Derived. No term is negative, so Im f(0) is positive for any potential that scatters at all, attractive or repulsive, and term by term it is k/4π times σ.

k in m⁻¹, Im f(0) in m, δₗ the real phase shift of wave l; θ = 0 sets every Pₗ(cos θ) to Pₗ(1) = 1

One partial wave on its unitarity circleIm aₗ = |aₗ|²aₗ = e^(iδₗ) sin δₗ

Derived: both sides equal sin²δₗ. The point aₗ runs round a circle of radius ½; its top, δₗ = 90°, is the unitarity limit 4π(2l + 1)/k².

aₗ dimensionless, with f(θ) = (1/k) Σ (2l + 1) aₗ Pₗ(cos θ); the same statement as |aₗ − i/2| = ½

Interference current on a large spherejᵢₙₜ = (ħk/m)(1 + cos θ) Re[f(θ) e^{(ikr − ikr cos θ)}]/r

Derived. Over the sphere it integrates to −(ħk/m)(4π/k) Im f(0), all from θ ≲ √(2π/kr): the forward shadow that takes the scattered flux out of the beam.

radial current in m s⁻¹ for ψ scaled to unit incident density; r far outside the potential, kr ≫ 1

With absorption: the general formσₑₗ + σ_abs = (4π/k) Im f(0)

Quoted, not derived here. Since σ_abs ≥ 0, (4π/k) Im f(0) ≥ σₑₗ for every physical amplitude; a computed result below that bound creates flux and is wrong.

f the elastic amplitude in m; σ_abs in m² counts every exit from the elastic channel, capture, excitation or reaction

The theorem at second order in VIm f⁽²⁾(0) = (k/4π) ∫ |f⁽¹⁾(θ)|² dΩ

Derived here from the theorem taken order by order. The first Born amplitude cannot satisfy it alone; the second Born term must carry this imaginary part.

f⁽ⁿ⁾ the n-th Born term, in m, proportional to Vⁿ; Im f⁽¹⁾(0) = 0 for any real potential, and f⁽¹⁾ is wholly real for a real central one

Set θ = 0 in the partial-wave sum

Quote the phase-shift lesson: for a real central potential f(θ) = (1/k) Σ (2l + 1) e^(iδₗ) sin δₗ Pₗ(cos θ) and σ = (4π/k²) Σ (2l + 1) sin²δₗ. The forward direction is where every Legendre polynomial agrees, since Unit 1 normalised them to Pₗ(1) = 1. So f(0) = (1/k) Σ (2l + 1) e^(iδₗ) sin δₗ, with real part (1/k) Σ (2l + 1) sin δₗ cos δₗ and imaginary part (1/k) Σ (2l + 1) sin²δₗ. Term by term that imaginary part is k/4π times σ, so σ = (4π/k) Im f(0). Dimensions: 1/k and f are both lengths, so the right side is an area. Limit: a lone s wave at δ₀ = 90° gives f(0) = i/k, and both sides give 4π/k², the unitarity limit. Nothing about the potential entered beyond its being real and central, so the identity holds for any set of real phase shifts, however many partial waves are kept and however they were obtained.

Each partial wave lives on a circle

Write aₗ = e^(iδₗ) sin δₗ, so f(θ) = (1/k) Σ (2l + 1) aₗ Pₗ(cos θ). Its real part is ½ sin 2δₗ and its imaginary part ½(1 − cos 2δₗ), so as δₗ runs through 180° the point aₗ goes once round the circle of radius ½ centred at i/2. In algebra, Im aₗ = |aₗ|²: the height of the point equals the square of its distance from the origin, the optical theorem for one wave. The top of the circle, δₗ = 90° and aₗ = i, is the unitarity limit, and no real phase shift can leave the circle. The forward amplitude k f(0) = a₀ + 3a₁ + 5a₂ + … is a chain of such links, link l on a circle of radius (2l + 1)/2. The heights of the links add, and so do the terms (2l + 1)|aₗ|² that build σ, so the height reached by the end of the chain is k²σ/4π: a whole angular distribution read off one point. The figure draws this chain for an s and a p wave.

Where the flux goes: the forward shadow

The partial-wave proof hides the mechanism, which needs neither l nor a central potential. On a sphere of radius r far outside the range, with ψ = e^(ikz) + f(θ)e^(ikr)/r and incident current ħk/m, the radial current has three parts. The plane wave's own part integrates to zero, since what enters one side leaves the other. The scattered part integrates to (ħk/m)σ. The cross term is jᵢₙₜ = (ħk/m)(1 + cos θ) Re[f(θ) e^{(ikr − ikr cos θ)}]/r. For kr ≫ 1 its phase swings so fast that the angular integral cancels everywhere except near θ = 0; integrating by parts leaves −(ħk/m)(4π/k) Im f(0), the backward end killed by the factor 1 + cos θ. This step is outlined, not proved: a careful version uses wave packets. Zero net outflow then gives the theorem again, with f(0) the amplitude along the beam. The cone is narrow: for 1.8 Å neutrons and r = 1.0 m, √(2π/kr) = 1.3 × 10⁻⁵ rad, 13 μm off axis at the detector, inside the beam.

What one forward number fixes, and what it leaves free

The theorem pins down Im f(0) and nothing else. Re f(0) shifts the phase of the beam behind the target rather than its strength; in a dilute medium of N scatterers per unit volume the complex f(0) sets the refractive index n = 1 + 2πN f(0)/k², quoted here, so Re f(0) fixes the real part of n while, with the theorem, the imaginary part of n attenuates the beam as e^(−Nσz), the measured transmission law. That is why the theorem is called optical. Two consequences hold for every elastic amplitude. Im f(0) is never negative, even for a repulsive potential whose phase shifts are all negative, because it is a sum of sin²δₗ. And the forward intensity is bounded below: dσ/dΩ at θ = 0 is |f(0)|² ≥ (kσ/4π)², so a large cross-section forces a strong forward peak. As V → 0 both sides of the theorem vanish together, as V², which the last section uses.

Absorption: the theorem with other channels open

When a collision can end elsewhere, a neutron captured, an atom excited, the elastic outgoing wave carries less flux than came in. In partial waves e^(2iδₗ) becomes ηₗe^(2iδₗ) with 0 ≤ ηₗ ≤ 1, so aₗ = (ηₗe^(2iδₗ) − 1)/2i and the point falls inside its circle. The general theorem, quoted here, keeps the forward elastic amplitude on the right and puts the total on the left: σₑₗ + σ_abs = (4π/k) Im f(0). With the quoted partial-wave forms σₑₗ = (4π/k²) Σ (2l + 1)|aₗ|² and σ_abs = (π/k²) Σ (2l + 1)(1 − ηₗ²), one line, Im aₗ − |aₗ|² = (1 − ηₗ²)/4, checks that they obey it. A black disc of radius R, ηₗ = 0 for l up to kR ≫ 1, absorbs πR² and diffracts another πR² into its shadow, a ratio σₑₗ/σₜₒₜ of ½. Measured: TOTEM, using the relativistic form of the theorem, found σₜₒₜ = 110.6 ± 3.4 mb and σₑₗ = 31.0 ± 1.7 mb for proton–proton collisions at 13 TeV, a ratio of 0.28: a grey disc, not a black one.

A real amplitude cannot pass: the Born series

If f were real, Im f(0) would vanish while σ = ∫ |f|² dΩ stayed positive, so no nonzero real amplitude satisfies the theorem. The first Born amplitude, derived later in the unit and quoted here as f⁽¹⁾(θ) = −(2m/ħ²q) ∫ r V(r) sin(qr) dr from r = 0 to ∞, is exactly such an amplitude for a real central potential. The resolution is order counting. In the Born series f = f⁽¹⁾ + f⁽²⁾ + … each f⁽ⁿ⁾ is proportional to Vⁿ, so σ starts at order V², and the theorem at that order demands Im f⁽²⁾(0) = (k/4π) ∫ |f⁽¹⁾|² dΩ. Partial waves show the same without Born machinery: for a weak potential δₗ is small and proportional to V, and aₗ = δₗ + iδₗ² + O(δₗ³), real at first order, imaginary only at second. For a screened Coulomb potential, V ∝ α, so f⁽¹⁾ is of order α while Im f(0) and σ start at α². The ratio Im f(0)/Re f(0) is then a cheap estimate of how far the first Born term can be trusted.

Change one variable at a time

Make the relationship visible.

Interactive model

s-wave phase shift δ₀60 °

p-wave phase shift δ₁20 °

s-wave inelasticity η₀1.00

Keep η₀ = 1 and drag δ₀ to 90°: a₀ reaches the top of its circle, the unitarity limit, and the bar follows the dot. Make δ₁ negative: the dot still rises. Then lower η₀: the dot sinks inside the circle and the bar gains an absorbed block, but its top stays level with the dot.

Im k f(0), dot height1.101

k²σₑₗ/4π1.101

k²σ_abs/4π0.000

Re k f(0), left free1.397

Live interpretationIm k f(0), dot height: 1.101. k²σₑₗ/4π: 1.101. k²σ_abs/4π: 0.000. Re k f(0), left free: 1.397

Catch the common trap

Explain before calculating.

Neutrons with k = 2.0 fm⁻¹ scatter elastically from a real, purely repulsive central potential, so every phase shift is negative. The measured total cross-section is 0.40 b, and 1 b = 100 fm². What is the imaginary part of the forward amplitude f(0)?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyNeutrons with k = 0.40 fm⁻¹ scatter from a light nucleus small enough that only the s and p waves matter, with phase shifts δ₀ = 50° and δ₁ = 12°. Find Im f(0), use the optical theorem to get σ, and confirm the result against the partial-wave sum.

Since Pₗ(1) = 1, f(0) = (1/k)[e^(iδ₀) sin δ₀ + 3e^(iδ₁) sin δ₁], whose imaginary part is (1/k)[sin²δ₀ + 3 sin²δ₁].
sin²50° = 0.5868 and 3 sin²12° = 3 × 0.04323 = 0.1297, so the bracket is 0.7165 and Im f(0) = 0.7165/(0.40 fm⁻¹) = 1.791 fm.
Optical theorem: σ = (4π/k) Im f(0) = (31.42 fm)(1.791 fm) = 56.3 fm² = 0.563 b.
Partial-wave sum: σ = (4π/k²)(0.7165) = (78.54 fm²)(0.7165) = 56.3 fm², the same, as it must be term by term. The p wave supplies 0.1297/0.7165 = 18% of it.
What σ leaves free: Re f(0) = (1/k)[sin 50° cos 50° + 3 sin 12° cos 12°] = (0.4924 + 0.6101)/0.40 fm⁻¹ = 2.756 fm, so |f(0)|² = 10.8 fm² per steradian, above the bound (kσ/4π)² = 3.21 fm².

AnswerIm f(0) = 1.79 fm and σ = 56.3 fm² = 0.563 b, the same from the theorem and from the partial-wave sum.

MediumNeutrons at 2200 m s⁻¹ are absorbed by boron-10 with σ_abs = 3835 b, while its scattering cross-section is only about 3 b. Treating the nucleus as fixed, so that k is the neutron's own wave number, and ignoring spin, use the optical theorem with absorption to find Im f(0), compare it with tables, and find how far the s-wave η₀ falls below 1.

λ = h/(mₙv) = (6.626 × 10⁻³⁴ J s)/(1.675 × 10⁻²⁷ kg × 2200 m s⁻¹) = 1.798 × 10⁻¹⁰ m, so k = 2π/λ = 3.494 × 10¹⁰ m⁻¹ = 3.494 × 10⁻⁵ fm⁻¹. With a nuclear radius near 3 fm, kR ≈ 10⁻⁴, so only the s wave matters.
Quoted theorem: σₑₗ + σ_abs = 3 b + 3835 b = 3838 b = 3.838 × 10⁵ fm², so Im f(0) = kσₜₒₜ/4π = (3.494 × 10⁻⁵ fm⁻¹)(3.838 × 10⁵ fm²)/4π = 1.067 fm.
Tables give boron-10 a coherent scattering length, the negative of f(0), whose imaginary part is −1.066 fm, obtained exactly this way from σ_abs alone; including the 3 b of scattering moves it by 0.08%.
Inelasticity: σ_abs = (π/k²)(1 − η₀²), and π/k² = π/(3.494 × 10⁻⁵ fm⁻¹)² = 2.573 × 10⁹ fm² = 2.573 × 10⁷ b. So 1 − η₀² = 3835/(2.573 × 10⁷) = 1.49 × 10⁻⁴ and η₀ = 0.99993.
Reading: 3835 b dwarfs the nucleus's geometric area of about 0.3 b, yet the outgoing s wave loses only 0.015% of its flux. As k → 0, f(0) tends to a constant, so σ_abs = (4π/k) Im f(0) − σₑₗ grows as 1/k: the 1/v absorption law boron-10 is measured to follow over several decades of energy.

AnswerIm f(0) = 1.07 fm, matching the tabulated −1.066 fm imaginary part of the scattering length; η₀ = 0.99993, a loss of 1.5 × 10⁻⁴ of the s-wave flux.

HardA neutron of energy E = 40 MeV, treated nonrelativistically, scatters from a fixed Gaussian well V(r) = V₀e^(−r²/b²), V₀ = −15 MeV and b = 1.5 fm, a model of a short-range target held in place, so that the reduced mass is the neutron's own, mc² = 939.6 MeV. Take the first Born amplitude as given: f⁽¹⁾(θ) = −(√π mV₀b³/(2ħ²)) e^{(−q²b²/4)}, with q = 2k sin(θ/2). Find the Im f(0) that the second Born term must supply, and compare it with f⁽¹⁾(0).

f⁽¹⁾ is real, so the theorem at order V² requires Im f⁽²⁾(0) = kσ⁽¹⁾/4π, with σ⁽¹⁾ = ∫ |f⁽¹⁾|² dΩ.
Scales: ħ²/m = (197.33 MeV fm)²/(939.6 MeV) = 41.44 MeV fm², k = √(2E/(ħ²/m)) = √(80/41.44) fm⁻¹ = 1.389 fm⁻¹, and g = m|V₀|b²/ħ² = (15 × 2.25)/41.44 = 0.8144.
Forward value: f⁽¹⁾(0) = (√π/2) g b = 0.8862 × 0.8144 × 1.5 fm = 1.083 fm, positive for an attractive well.
With dΩ = 2π d(cos θ), cos θ running from −1 to 1, and q² = 2k²(1 − cos θ), the integral gives σ⁽¹⁾ = (π²/2) g²(1 − e^(−2k²b²))/k² = 4.935 × 0.6633 × 0.9998/(1.930 fm⁻²) = 1.695 fm² = 17.0 mb.
Im f⁽²⁾(0) = kσ⁽¹⁾/4π = (1.389 fm⁻¹)(1.695 fm²)/4π = 0.1874 fm, so |f⁽²⁾(0)|/f⁽¹⁾(0) ≥ 0.1874/1.083 = 0.17.
Check: the ratio is (√π/4)(g/kb)(1 − e^(−2k²b²)), and g/kb = |V₀|b/ħv with ħv = ħ²k/m = 57.6 MeV fm, so it is 0.443 × 22.5/57.6 = 0.173 again: the high-energy Born condition |V₀|b ≪ ħv, recovered from probability conservation alone.

AnswerThe second Born term must supply Im f(0) = 0.187 fm, 17% of f⁽¹⁾(0) = 1.08 fm, so near θ = 0 the first Born amplitude alone is in error by at least about 17% at 40 MeV.